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Chapter 4 Inaccessible and Mahlo Cardinals
CHAPTER 4
INACCESSIBLE AND MAHLO CARDINALS $1. Properties of V, The work of the previous chapter has prepared the way for us to give more answers to the question: what is true in a given set? In particular we want to see which axioms hold in the sets V,. We have already noted that (since V, is transitive for each u) the axioms of extensionality and foundation will hold. We can now say more: 1.1. Theorem (ZFC). For every ordinal u > 0, the axioms A1 (extensionality), A2 (foundation), A3 (subsets), A4 (empty set), A7 (union) and A10 (choice), are true in V,. Further, if u is a limit ordinal, axioms A5 (pairs) and A6 (power set) are true in V,; ifcc > w , then A8 (injinity) is true in V. (So that, e.g., V , , , is a model of the theory ZC.) (Note that we are writing “true in X” in place of “true in ( X , E(X))”.) Proof. For any a > 0, V, is transitive, and Al, A2 are both logically equivalent to d oformulas. But these are absolute between any transitive structures, and so they will hold in V, since they hold in V. (Note that this repeats the proof given in ch. 3 $1.2.) All of the remaining axioms (except A10) will require certain terms to be legitimate in V,. In the case of A4, the set required (the empty set) is A,, i.e., defined by a bounded formula, and so the definition is absolute for any transitive set containing 0;but 8 E V, for any a > 0. Similarly the axioms A5, A7 and A8 all require sets which have dodefinitions. For A8, the set required (w) will be in V, iff a > w; so V, satisfies A8 iff a > w. For A7, the set required is U a , and since p ( u a ) < p(a), we have a E V, + U a E V,; so A7 will hold in all V,, a > 0. For A5, however, the set required is { a , b}; and p({a, b}) = = max(p(a), p(b)) 1. So if a = @ 1 and p(a) or p(b) = @, then {a, b } $ V,, and A5 will fail; but if u is a limit ordinal, then a$ E V, + {a, b) E V,, and A5 will hold. For the other axioms we no longer have dodefinitions, and we must look more closely. For A6, we do have a ITl definition of P ( a ) , and ITl formulas are preserved under restriction; so if x = P( a ) holds in V
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+
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$1
(i.e., if x is a ( a ) ) ,then (provided a,x E V,), x = P(a) will hold in V,. Since p ( g ( a ) ) = p(a) + 1 , this means that A6 will hold if u is a limit. For A3, we must consider all formulas 4, and we have no hope of reducing the definition { x E a I +(x)} to simple form. But we can prove in ZF that b = {x E a I V, I= +(x)} will be a set, using the definability of satisfaction; and for a E V,, we shall have b E V, since b c a So we shall have V, k Vx ( x E b t-) x
E
a
A
+(x)).
Hence V, satisfies A3. For A10 we have no definition of the choice function to use. But if a is a disjointed set, and b is a choice set for a, we can assume that b c U a ; hence if a E V,, then b E V,. But the statement that b is a choice set for a can be written as a A,-formula VY E U [ 3 2 ( 2 E Y ) -+ 32 E Y
(2 E
b
A
VW E Y
(WE
b -+ W = Z))]
and so this will be absolute between Vand V,, i.e., b will be a choice set for a in V,. So A10 will hold in V,. (We are, of course, assuming that every set has a choice function, i.e., that the axiom of choice holds in V.) 0 1.2. Note. We have set out this proof only semi-formally, yet it is a proof which can be carried out within ZFC,just as the proofs of ch. 2 could be carried out in the various systems claimed. Thus using the notions defined in ch. 3 $5, we have shown that, e.g.,
ZFC 1 Va ( a > 0
--f
Vx E "V, Sat['AlO',
V,, x ] ) ,
where n is the rank of the Godel-set 'A10'. It is worth noting that this proof relies heavily on the fact that satisfaction can be defined in ZF (which we showed in ch. 3 $5.4) and it also relies on using the properties of absoluteness of ch. 3 $4, particularly ch. 3 $4.14. These last were not presented as proofs within ZFC (they could not be until after the formalization of the language in ch. 3 $9,and we do not intend to go over them again to show that they can be presented within ZFC.It should be clear that they can, and we leave this as an exercise to check. (It may be noted that we used this fact before in proving ch. 3 $7.4; the proof that a subset closed under a set of Skolem functions is an elementary substructure, so that everything is absolute, has to be regarded there as being proved in ZFC.)
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One point to be observed in this is that we do not have a general notion of satisfaction in V ; but we do have the notion for restricted classes of formulas (from Exercises ch. 3 $5.11(5) and (6)), and this is all we need in the proof of 1.1. In fact the work of ch. 3 $4 on absoluteness can be carried out in ZFC in two distinct senses: one is for absoluteness between transitive structures which are sets, and the other between V and transitive sets. If we carry out both these formalizations, we can then use ch. 3 $4.8 with Godel’s second incompleteness theorem (as in ch. 3 $5.10) to deduce that we cannot define a collection of Skolem functions for V within ZFC (in contrast with ch. 3 $5.7, which shows that such a collection must exist for sets). We want now to move to higher realms, and prove:
1.3. Theorem (ZFC). If. is strongly inaccessible, then V, != ZFC. Proof. We only have to prove the axiom of replacement, A9, since K is a limit ordinal > w . Suppose that y(x, z) is a formula such that
and let b be the set
We must show for each such y that for each a E V,, b E V,; then the case of the replacement axiom using y will be true in V,. It is clear that p(b) < K ; we must show that p(b) < K . Put g(c) = p(d) 1, where V, k y(c, d ) , if such a d exists in V,, otherwise put g(c) = 0. Then we shall have p(b) = Sup{g(c) I c ~ a } . But a € V,, and so since K is a limit ordinal, a E V, for some IX < K . Then Fa < 1,,and so d < 1,; and since K is strongly inaccessible and a < K , we have 2, < K by Exercise ch. 2 $7.8(2). Hence since g : a + K and d < K and K is regular, we have Sup{g(c) I c E a} < K , i.e., p(b) < K and b E V,, as required. 0
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1.4. Corollary. Unless ZFC is inconsistent, we cannot prove the existence of strongly inaccessible cardinals in ZFC.
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Proof. By Godel’s second incompleteness theorem, ch. 3 $5.10, since 1.3 gives ZFC I- 3~ (Inac(K)) + 3 X ( X k ZFC).
0
Models of the form V, for K strongly inaccessible are called natural models for ZF. 1.5. The assumption that there are sets A such that ( A , E(A)) is a
model of ZF (or ZFC) is much weaker than the assumption that there are inaccessible cardinals. We could justify the assumption that ZFC has transitive models which are sets by the following heuristic argument: Suppose we had a countable collection F of Skolem functions for V. Then the closure of (0)under F would be a countable elementary submodel of V ; we could use Mostowski’s isomorphism theorem to collapse this to a transitive set A , and we should have A k ZFC since V b ZFC. (Of course this is the argument mentioned in $1.2, which shows that we cannot define a collection of Skolem functions for V, or prove that they exist, in ZFC.) But this argument will hold whether or not there are inaccessible cardinals. Now suppose that a is the first inaccessible cardinal, then we can find a model A of ZFC by taking the closure of (8) under a set of Skolem functions for V, and collapsing to a transitive set. Since A will be hereditarily countable, we shall have A E V,, c V,, so V, k 3 a countable model of ZFC. But if a is the first inaccessible, then also Inac(B)) (see Exercise 1.6(3)) since Inac(p) is absolute for V,, and if /? < a, then /?is not strongly inaccessible; so V, is a model of Va
k VB
ZFC
(7
+ Vp (7
InacG))
+ 3 a model of ZFC.
1.6. Exercises (1) Models of Z. V , + o) is the most natural model to think of for ZC, and we could use it to show that many of the uses we have made of the replacement axiom are essential. But we can find smaller models: ,W,. Show that (i) Define W,, = o, W,+, = @( W J , W , = W , is transitive, (ii) W , k ZC, (iii) V , # W , (although V , c W,).
u,,
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For another example, let qn = { . . . {n} . . . } with n braces, i.e., a rank 2n set (if t ( x ) = { x } , then q , = t"(n)).Then put Q = {q, I n < w} and Ql = UQ, Q"" = UQ". Now define U,, = w v Q, Untl = = g ( U , ) V U , V Q", U = U,. Show that (i) U is transitive (although U , is not, for each n), (ii) U k ZC, (iii) Q E U but TC(Q) $ U (although TC(Q) c U ) . Hence some use of replacement is needed before transitive closure can be defined. [These examples show that ZC is not a particularly nice theory, and better theories, such as were noted in ch. 3 $6.5(5), can be given for ZFC minus replacement.]
u,,,
(2) Show that if K > w is a cardinal, then H ( K )is a model of the axioms A l , A2, A3, A4, A5, A7, A8 and A10; and if K is regular, then it is a model of A9 also (and hence all of ZFC except the power set axiom). [If K is strongly inaccessible, then H(K)= V, and so is a model of all of ZFC.]
<
(3) Show that if a is a limit ordinal, then cf(/?) = 6, /? 26, and Inac(/?) are all absolute for V,. [Use completeness as in ch. 3 54.9. If / ? , 6 ~V,, then the functions required will be in V,.] (4) Show that the first a for which V, k ZFC has cf(a) = w, and hence the converse to 1.3 does not hold. [Suppose that V6 k ZFC, and let F be a set of Skolem functions for V,. Given /3 < 6, let g(/?) be the least ordinal such that if xl,. . ., x, E V, and f E F, then f(xl, . . ., x,) E V,,,,. Let a = Sup{gn(/?)I n < w}. Show that a < 6 and V, I= ZFC (in fact V, < V6),but cf(a) = w.] Show that if a is the first ordinal for which V, I= ZFC, then V, is a model of ZFC 3 X ( X k ZFC) V/?(- Vb k ZFC).
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(5) Hyperinaccessible cardinals. Suppose that the strongly inaccessible cardinals have been enumerated in order as 8, for ordinal a. Then a cardinal K is hyperinaccessible if 8, = K , i.e., if K is strongly inaccessible and there are K strongly inaccessible cardinals below K . Show that if K is the first hyperinaccessible cardinal, then V, is a model of ZFC the axiom of inaccessibles (of ch. 2 $7.7) "there are no hyperinaccessiblecardinals".
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Similarly K is hyper-hyperinaccessible if K is hyperinaccessible and there are K hyperinaccessible cardinals below K . Show that if K is the first hyper-hyperinaccessible cardinal, then V, is a model of
ZFC +Vg 3 y > ( y is hyperinaccessible) + “there are no hyper-hyper inaccessible cardinals”. (6) Natural models f o r von-Neumann-Bernays-Godel set theory and for Bernays-Morse theory. These systems have variables for classes as well as for sets, and the axioms A3 (subsets) and A9 (replacement) can be regarded as being single axioms with a free class variable taking the place of the arbitrary formula. Then a comprehension axiom is added for classes, as noted in ch. 1 $5.3. Suppose that K is strongly inaccessible, and interpret class variables as ranging over V,,, (i.e. 9 ( V , ) ) and set variables as ranging over V,. Show that this makes (V,+l, V,, E(V,+,)) into a model of the full Bernays-Morse theory. Show that the converse is now true: Show that if ( Va+,,V,, E(V,+J) is a model of even the von-Neumann-BernaysGodel set theory, then M is strongly inaccessible. [Indeed what is needed here has nothing to do with the comprehension axiom assumed; it is the result of taking A3 and A9 in second-order form, since the class variables are being interpreted as ranging over P(V,). Then A9 says that the image of a set under any (set-valued) function is a set, and A3 says that any subcollection of a set is a set (and they now really do mean any). If these second-order axioms are true in a given transitive set which is a model of Z, show that that set must be of the form V, for strongly inaccessible M.]
(7) Show that if u < /?are ordinals such that V, < Vs, then u must be a strong limit ordinal and both V, and V, must be models of ZFC. [Show first that u must be a limit ordinal. To see that u must be a strong limit, suppose not: let f : P(8)+ u be onto, for 6 < u. Then there is a set X c Y(d) x 8 ( 6 ) which well-orders Y ( 6 ) in order-type u ; X will be in V,,, (it could be coded in V,,,) and so in V,. Now in V,, there will be an ordinal similar toP(6) ordered by X ; since V, < V,, this must also be true in V,, contradicting the assumption that the ordertype was CI.To see that the axiom of replacement must hold in V,, suppose that $(x, y , a,, . . ., a,) defines a function f in V,, using parameters a,, . . ., a, from V,. Then (b must also define a function, say g, extendingf, in V o ;and for any set a E V,, the range of g 1 a c V, and
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so is a set in V,. Hence the range off 1 a must be a set (i.e., a member) in V,, and so V, satisfies replacement.] (8) Show that if a is strongly inaccessible, then there is a p < a such that V, .< V,. [Take a set F of Skolem functions for V,. Let g(6) be the Sup of the ranks of images of members of V6 under members of F, for 6 < a. Let g"(6) be Sup{gn(6) I n < o}, and show that if 6 < a then gw(6) < a, and if p = g"(6) then V, < V,.]
$2. Normal functions
We want to present a very powerful way of postulating the existence of large numbers of inaccessible, hyperinaccessible, hyper-hyperinaccessible,etc., cardinals of all sorts, which is due to Mahlo. This relies on the idea of a normal function, or a closed unbounded class (or set). We shall present this first for functions which are defined for all ordinals, and these must be proper classes; so we shall work in this section with variables for proper classes. We shall be able to restrict all that we do to ZFC if we wish to by regarding classes as always being definable by a formula (possibly with parameters), so that statements about classes are to be read as statements about all defining formulas. We shall not use different letters for proper classes, because we have already used most of the kinds available, and also because we shall later want to take what we do here and restrict it to sets, and it is easier to use the same notation in both cases. 2.1. Definition. A function f :On + On is a normal function iff (i) f is increasing, i.e., if a < /? then f ( a ) < f ( p ) ;and (ii) f i s continuous, i.e., if Lim(a) then f ( a ) = f(E). Here On is the class of all ordinals, {x I Ord(x)}. Note that (i) implies that for all a, a < f ( a ) (by induction on a ; there cannot be a first u such that f ( a ) < a if (i) holds). (ii) says that f is continuous with respect to the order topology on the ordinals; this is given by saying that a Y EX. subset X c On is closed if, whenever Y c X , Y # 0, then Given this definition of closed, we have:
uZ<,
u
2.2 Lemma. The range of a normal function is closed and unbounded in On; and every closed unbounded class of ordinals is the range of a unique normal function (its enumerating function).
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Proof. The range is unbounded since a 6 f ( a ) , it is closed because
f is continuous and On is closed. Given a closed unbounded class
X,
its enumeration (which exists by transfinite induction) will be a normal function, and it will be defined for all ordinals since X must be a proper class, by the axiom of replacement. 0 One of the important properties of normal functions concerns fixed points. 2.3. Definition.
is afixedpoint for f iff f ( a ) = u.
M.
Theorem. I f f is a normal function, then the fixed points o f f form a closed unbounded class. Proof. First we show that it is unbounded. Given a, define
f “ 4 = a, fn+Ya) = f(f”(4), f Y 4
=
un
If u is a fixed point, then all of these will be u ; if not, then a < f ( a ) , and so by (i) the sequencef”(u) will be strictly increasing for n < m. Hence f W(u)is a limit ordinal, so by (ii), f(fw(E>)
=
Ut
f(Q*
But if E < f “(a),then 5 < f”(a) for some n < o,so f ( t ) < f””(a) and
f ( f “(a))
u n
f”+YQ)= f “(4-
Hence f ” ( a ) must be a fixed point; a 6 f w(a)and so the fixed points are unbounded. (We have actually shown that f W ( u 1) is the next fixed point above a, for each u.) To see that the fixed points are closed, suppose that Y is a set of fixed points and Y 4 Y. Then Y must be a limit, and a < Y iff 3~ E Y ( M . < 6). S O f ( U Y ) = U F EfY( 6 ) = UtEYE (since Y is a set of fixed points); i.e., f ( u Y ) = Y and Y is a fixed point. Hence the class of fixed points is closed.
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u
u
u
u
u
2.4. Corollary. For any normal function f we can form the derived function f’, which enumerates the fixed points off; and f’ is also a normal unction. 0 Two important normal functions which we have already met are the functions K and 1.Fixed points of these are ordinals a such that K, = a or 1, = a ; it is clear from the construction in the proof of 2.3 that these will usually be singular (indeed, the first fixed point above
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any ordinal will always have cofinality 0).Regular fixed points of these two functions are the inaccessibles (strong or weak). So we can see the following axiom as a natural generalization of the axiom of inaccessibles: 2.5. Axiom F. Every normal function has a regular fixed point.
(Since the normal functions are proper classes, this axiom can be stated only as a scheme in the language of ZFC, with one instance for each formula, saying that if that formula defines a normal function, then it has a fixed point.) To see that this is much stronger than the axiom of inaccessibles, first note that we can enumerate the beth numbers greater than a given u by a normal function, and this must have a regular fixed point which must therefore be a strong inaccessible >u. So the axiom of inaccessibles follows. But then we can enumerate the class of strong inaccessibles by a function which is continuous at limits (i.e., we enumerate the closure of the class of strong inaccessibles), and this will be a normal function whose regular fixed points will be hyperinaccessibles. So (since we can start the enumeration of strong inaccessibles above any point we choose) we can deduce the existence of a proper class of hyperinaccessibles, and repeat the process to get hyperhyperinaccessibles, etc. This axiom does not seem to have received a name in the literature, probably because stronger axioms were proposed at the same time. We shall derive another form of axiom F in $4; meanwhile to give the stronger axioms, we need to relativize the notion of a normal function to a set, which in practice will be a cardinal. 2.6. Definition. For ordinal u, a functionf :6 + u is a normal function on u iff is increasing, continuous, and its range is unbounded in u, i.e., URange(f) = u. We shall really only be interested in this notion when u is a regular cardinal, thus we have: 2.7. Theorem (ZFC). The following are equivalent: (i) Every normal function on u has afixedpoint. (ii) u is regular and u > w. Proof. If u is regular and f :6 + u is unbounded in u, then we must have 6 3 u ; iff is normal on u, then in fact we must have 6 = u.
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Then if u > m, we can construct a fixed point off exactly as in the proof of 2.3. To prove the converse, note that w certainly has normal functions without fixed points (any increasingf: m --f m will be counted as a normal function on m). If u is not a limit, then no function can be unbounded in u, and if cf(u) < u, say cf(u) = 6, then we can find f : 6 + a which is unbounded, increasing and continuous, and such that f(q) > 6 for all 11 < 6; this f cannot have a fixed point. 0 We set out some of the properties of normal functions on regular cardinals in the exercises; in the next section we use them t o define Mahlo cardinals. 2.8. Exercise
(1) (a) Given a normal function f (on a regular cardinal K > cc) or on On), let f' be the derived function off. Then define f" by f o = f, f"" = (f*)',and if 1is a limit ordinal, f is the function enumerating all ordinals which are fixed points of each f' for 6 < 1. Show that iff is a normal function on On, thenf" is a normal function for each a ; iff is a normal function on K , then so isf" for ci < K. (b) We can also define the function g by
d.1
=f"(O) Show that g is also a normal function (in the same sense as f, i.e. on On or on K ) . (c) We can also define the diagonal function f A as the function which enumerates all ordinals u such that cc = f J ( a ) for all 6 < a. Show thatfA is also a normal function (in the same sense as f ) .
$3. Mahlo cardinals The definitions which Mahlo gave are effectively equivalent to the following: 3.1. Definition. K is weakly Muhlo iff K is an ordinal such that every normal function on K has a regular fixed point. K is strongly Muhlo iff every normal function on K has a strongly inaccessible fixed point. In view of 2.7, it is clear that weakly or strongly Mahlo ordinals must in fact be cardinals, since they must be regular, and they must be > m . We shall use Muhlo for strongly Mahlo in accordance with the
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general practice (though originally Mahlo himself considered only weakly Mahlo cardinals). First we show that these cardinals must themselves be inaccessible: 3.2. Lemma. If K is weakly Mahlo, then K is weakly inaccessible; i f K is strongly Mahlo, then K is strongly inaccessible. Proof. We have seen that K is regular. Suppose that K is not a limit cardinal, then K is 6 + (i.e. N(6)) for some 6, and f : K + K given by f(u) = 6 u is a normal function on K whose fixed points cannot be regular. So K cannot be weakly Mahlo. If K is not a strong limit cardinal, then 26 2 K for some 6 < K , and again f : K + K given by f(u) = 6 u is a normal function; in this case its fixed points cannot be strong limit cardinals since they are between 6 and 2d, so they are not strongly inaccessible and K is not strongly Mahlo. 0
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We can now go further using the following: 3.3. Definition. K is 0-weakly hyperinaccessible iff K is weakly inacces1-weakly hyperinaccessible iff K is regular and there are sible. K is u K u-weakly hyperinaccessible cardinals below K . For limit A, K is Aweakly hyperinaccessible iff K is u-weakly hyperinaccessible for all u < A. Similarly for u-strongly hyperinaccessible. Now we can show:
+
3.4. Theorem (ZFC). If
K is weakly Mahlo, then K is u-weakly hyperinaccessible for all u < K ; if K is strongly Mahlo, then K is u-strongly hyperinaccessible for all u < K . Proof. The proof is identical for the strong or weak case, and we shall write u-hyp for either u-weakly or a-strongly hyperinaccessible according to which case is being considered. First we note that if K is weakly Mahlo, then any normal function on K must have K regular fixed points. For supposef:K -,K is normal on K . Then for any fixed p < K , if g(u) = f(p a), then g : --f~K is also normal on K , and so g has a regular fixed point 6. But then 6 = = f(p + 6), and so since f(p 6) 2 p 6, we have p 6 = 6 = = f(p + a), i.e., 6 is a regular fixed point off greater than p. Since K is regular, this means that there must be K such regular fixed points. This gives the induction step from u 1 to u 2 immediately. Suppose that K is a + 1-hyp, then K is a regular fixed point of fa, wherefx is the normal function enumerating the closure of the class of u-hyp cardinals. Hence
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K is a normal function on K , and so has K regular fixed points. Each of these is then a 1-hyp and less than K , i.e., K is a 2-hyp. The induction step for A-hyp is trivial for limit A, since we have defined A-hyp as: a-hyp for all a < I . It remains to show that if K is A-hyp, 1-hyp. where A is a limit < K , then K is I So suppose that K is A-hyp, and I < K . Then for 6 < I , let X , be the closure of {a < K [ a is 8-hyp). Then since K is 8 1-hyp, card(Xd) = K for each 6 < I , and all regular members of X , are d-hyp. We need to show that X, = X , has cardinal K also. Then, since X , is also closed, the function enumerating X , will be normal on K and so will 1-hyp. have K regular fixed points; these will all be I-hyp, and so K is A So suppose that p < K , and set
fa
+
+
+
+
nd<,
+
a, = Inf(X, - p) for 6 < A,
u6<,
and let y = ad. Since K is regular and I < K, we have y < K . But if 6 < 6' < A, then ad,E X,,and so y is a limit of members of A', and since X,is closed, we have y E X , for each 8 < A, i.e., y E X,. Hence X,is cofinal in K , and since K is regular, it must have cardinal K , as required. 0 This theorem is not the strongest we can prove in this connection; indeed, much of the interest in Mahlo cardinals arises because there seems to be no strongest statement which can be made, and the Mahlo cardinals are larger than any which can be given, using methods like those of $3.3. Of course this is an imprecise statement; a little more justification for it is given in Exercise 3.7(2). Another way of looking at Mahlo cardinals is in terms of stationary sets:
3.5. Definition. A subset x of K is called stationary in K (sometimes Mahlo in K ) if, for every closed, unbounded subset y of K , x n y # 8. For cardinals cofinal with w , this notion is not of much interest; a subset will be stationary only if it contains a final segment. But if cf(K) > o,then the closed unbounded subsets of K generate a filter on K ; the intersection of any two closed unbounded subsets of K is again closed and unbounded in K . (To see this, if x and y are both closed, unbounded, in K , and a < K , put a. = Inf(x - a ) ; a1 = Inf(y - ao), as = Inf(x - al), etc., and a , = a,. Then amE X n y since it is a limit point of both, and a 6 a, < K since cf(K) > w . ) So for
u,,,
~13.4,$31
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119
cf(K) > o,any set of this filter (i.e., any set which contains a closed unbounded subset of K ) will be stationary, and the stationary sets are those which meet every member of this fdter. Now the definition of Mahlo cardinals can be rephrased as: K is weakly Mahlo if the regular cardinals less than K form a stationary subset of K ; K is strongly Mahlo if the strongly inaccessible cardinals less than K form a stationary subset of K.
3.6. Mahlo's operation In terms of stationary sets, we can set out the process which Mahlo used in passing from the inaccessible cardinals to the Mahlo cardinals: Definition. Suppose X is any class of cardinals. Then H ( X ) = {a E X [ X n a is stationary in a}. Then H is Mahlo's operation: If X is the class of regular cardinals, then H ( X ) is the class of weakly Mahlo cardinals; if X is the class of strongly inaccessible cardinals, then H ( X ) is the class of strongly Mahlo cardinals. We can now iterate the operation, and define H a for ordinals a by:
HO(X) = X , Ha"(X) = H(H"(x)), H A ( X )= na,,Ha(X) if Lim(A). If X is the class of strongly inaccessible cardinals, then H"(X) is the class of strongly Mahlo cardinals of kind a. Thus the strongly Mahlo cardinals of the second kind will be just those cardinals K such that every normal function on K has a fixed point which is strongly Mahlo of the first kind. We can then diagonalize this operation, and define a E H*(X) iff Vg < a ( a E P ( X ) ) .
Then if X is the class of strongly inaccessible cardinals, H A ( X )is the class of (strongly) hyper-Mahlo cardinals, (HA)A(X)is the class of hyper-hyper-Mahlo cardinals, etc. (Then K is hyper-Mahlo iff it is Mahlo of kind K . ) It is clear that there is no end to the diagonalizations which can be performed on the operation H, to give higher and higher classes of Mahlo cardinals.
I20
[CH.4,
INACCESSIBLE AND MAHLO CARDINALS
$3
3.7. Exercises (1) Show that the filter generated by the closed unbounded subsets of K is cf(K)-closed; that is, the intersection of a set of < c f ( ~ ) sets of the filter is again a set of the filter. [Use the argument given in 3.5 to show that a set of < c f ( ~ )closed unbounded subsets of K has an intersection which is unbounded in K . ] (2) Show that theorem 3.4 can be strengthened. Show that if K is Mahlo, then K is not the first cardinal u such that u is u-hyperinaccessible. Show that there must be K cardinals u below K such that a is u-hyperinaccessible. [For cc < K , let f(u) be the first ordinal /3 such that there are a cardinals below fi which are cr-hyperinaccessible; show that fis a normal function on K . ]
+
(3) Show that if K is Mahlo, then V, is a model of ZFC axiom F. Show that the converse of this is not true. [Since axiom F is first-order, arguments such as used in Exercise 1.6(4) will show this.] However, show that if a second-order form of axiom F holds in V,, then K is Mahlo. [Compare Exercise 1.6(6).J (4) Regressive functions and stationary sets. If X c K and f:X --f K , then f is called regressive on X if f(u) < u for all u in X , a # 0. Establish the following characterization of stationary sets : If Cf(K) > w , then X c K is stationary in K iff for every function f regressive on X , there is some ,!? < K such that { % € X I f(a) < p} is unbounded in K . [If X is not stationary in K, there is a B c K - X which is closed, unbounded in K . Set f(a) = U ( B n u) for u E X , and show that f is regressive on X and for B < K , ( a E X < p} is bounded by Inf(B - (fi 1)). To prove the converse, set D = f”X n ( K - X ) and consider two cases : First if D is bounded below K, take an upper bound 8, of D and note that if we put f ‘((a) = a,f”f+’(u)= fcf“(u)) iff”(.) E X,f””(a) = 0 otherwise, then for each u there is a first n(u) such that f”‘”’(a) = 0. Take the least n such that {f”(u) I u E X , n(u) = n l} = Z is unbounded in K and show that f ” (2)is bounded by b. If D is unbounded in K , find a contradiction from the assumption: (i) for each u E X , there is a 4(a) E K such that /I> +(a) impliesf(p) > a.
+
).(fI
+
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REFLECTION PRINCIPLES FOR MAHLO CARDINALS
121
Set X' = f - ' ( D ) and defme a cofinal sequence in X', (at 16 < T ) , such that: (ii) t < T,I < T implies $(ae) < a,,. Then Y = { f ( a t ) I 5' < T } is unbounded in K and a subset of K - X; since X is stationary, X n Cl(Y) # 0 (where Cl(Y) is the closure of Y). Then if a* E X n C1( Y ) , f ( a * ) < a*, and this contradicts (i) and (ii).] As a corollary of this, deduce the following theorem, due to Fodor: Suppose that K is regular, > o,Xis stationary on K , andfis regressive on X.Then for some a < ~ , f - ~ (isa stationary ) on K . [Show first: (iii) if { Y, 1 a < T } is a sequence of pairwise disjoint, non-empty, non-stationary subsets of K , with T < K , such that if U = = {y, I a < T } , where y, = Y,, then U is not stationary also, then Y= < Y, is not stationary. Prove this by using the characterization of stationary sets above; show that a sequence of regressive functions for each Y, - {y,}, and one for U, can be fitted together to give a regressive function on Y. Now set Y, = { E E X I f ( t ) = a> and use (iii) on this to show that some Y, must be stationary.]
n
u,
$4. Reflection principles for Mahlo cardinals
Ltvy and Bernays have both given strengthenings of the reflection principle R,, and the first of these turns out to be equivalent to axiom F. For any formula with at most x,, . . ., x , free, we get an instance of the reflection principle V a 38 > a (Inac(B) A Vx,, . . ., x, E Vs (4 tf 4'0)). R1 4.1. Theorem (ZF). R1 is equivalent to axiom F. Proof. First we derive F from R1.We shall use the form of R1which reflects two formulas at once; this will be equivalent to R1exactly as in ch. 3 $6.4 for R,. Suppose that f is a definable normal function. We must show that f has an inaccessible fixed point. We use R1on the formulas
f(Y) = 6, VY 36 CfW = 4-
Then we get an inaccessible B such that, for y,6 < /?,
f(r) = f3wf '(74 = 6
(so the definition off will be absolute for V8),and
VY < B 36 < B CfV4r)= d),
122
INACCESSIBLE AND MAHLO CARDINALS
[CH.4,
$4
(since V y 38 ( f ( y ) = 6) holds in Y by hypothesis). But together these say VY < B 36 < B (fb) = 61,
and so since ,8 is a limit ordinal,f(/?) = Uy<,f(y) < /?,i.e., B is an inaccessible fixed point off. The proof in the opposite direction, F implies R,, is an extension of the proof of R, given in ch. 3 $6.3. In that proof we defined a function f on all ordinals, for a given formula 4, such that f was increasing and had the property that if X = V,,,, for any <, then X reflects 4. The f as defined was not necessarily continuous, so let g be the result of making f continuous, i.e., g enumerates the closure of Range(f). (Then for IZ < co, g(n) = f ( n ) , for a > co, g(a 1) = f ( a ) , and for limits A, g(4 =
+
Umf(a).>
Then in fact g will have the same property as$ We need to check that for limit 1, X = V,,,, reflects 4, but this follows just as for f, since if xl,. . ., x,, y,, . . ., yr E X, then x,, . . ., x,, y,, . . ., y r E V&,) for some a < 1. So g,(xl,. . ., x,, yl,. . ., y,.) < f ( a 1) < g(A), and so the induction on r will hold as before and so X will reflect 4. But now g is a definable normal function (with parameter a only) and so by F it will have an inaccessible fixed point. This will be a /I> cc such that /? is inaccessible and V, reflects 4, hence R, holds. 0
+
We can now strengthen R1 to: R2
V a 38 > a (/I is Mahlo
A
Vx,,
. . ., x, E
V,
(4 ++ CvB)),
which will be equivalent to a scheme asserting that every normal function has a Mahlo fixed point, and then again using hyper Mahlo to get RB,etc. LCvy has given a way of writing these reflection principles using the notion of a standard complete model of a theory (which will be Z or ZF or a theory extending ZF). The idea is to write the schemes of the theory in second-order form, so that we get the natural models of the theories. We then cannot put this in the form x k ZF, whichwas used in ch. 3 55.9, but we can write them directly: 4.2. Definition. SCMZ(x)for
Trans(x) A Vy E x Vz ( z c y + z E x) A x k Z “x is a standard complete model of Z ’
CH.4, $41
REFLECTION PRINCIPLES FOR MAHLO CARDINALS
123
SCMZF(x)for SCMZ(x) A V f v y ~ ~ ( C f : x - - , x ) ~ f " y ~ x ) . For this definition we shall have: 4.3. Theorem (ZF). SCMZF(x)c-) 3cr (Inac(or) A x Proof. Exercise.
Now we can re-write Ro and R1as
=
V,).
-
. . ., x , E y (4 p)); Vx 3y ( x € y A SCMzF(y) A Vx,, . . ., x, E Y (4 p)). RF If we then write ZF1 for ZF + R1,and define an appropriate notion R:
Vx 3y ( x E y
A
SCMz(y)
A
Vx,,
f-)
of SCMZF1,we can then write
vx 3y ( x E Y A SCMZF1(y)A Vx,, . . ., x , E Y (4 p)). R: We can then define ZF2 and repeat this, paralleling the applications of Mahlo's operation. We leave the details as an exercise. f+
4.4. A note on the plausibility of these cardinals Since we cannot prove in ZFC that there are strongly inaccessible cardinals, it is clear that we cannot prove that Mahlo cardinals exist; axioms would have to be added to say that they do exist, such as the schemes R1,R2,etc. How plausible are these axioms? There is no question of ever proving them consistent with ZFC, of course, since we cannot prove that ZFC is consistent, using methods available in ZFC. It may happen that they will one day be proved inconsistent with ZFC, but if they are in fact consistent, then we shall never have a proofwe shall never know by looking within ZFC. We can only hope for heuristic justification from outside. We look for justification for these axioms from the point of view of the cumulative type structure, where we want to say that the collection of levels, which is indexed by the ordinals, is a very rich structure with no conceivable end. An axiom saying that Mahlo cardinals of the first kind exist can then be regarded as saying that the class of strongly inaccessible cardinals is so rich that there are members K of the class such that no normal function on K can avoid this class; however we climb through K , provided we are continuous at limits (so that we are
124
INACCESSIBLE AND MAHLO CARDINALS
[CH.4
enumerating a closed subset of K ) , we shall eventually have to hit a strongly inaccessible. In this sense, axiom F can be regarded as the ultimate in saying that inaccessibles, hyperinaccessibles, etc., of all sorts exist, and the existence of Mahlo cardinals can be regarded as the next step after this-it always seems a plausible step, in view of the reflection principle, to take a property of the whole universe (such as axiom F), and postulate that it already holds at some level V,. (See Exercise 3.7(3) in this connection.) We shall find axioms in later chapters which imply that Mahlo cardinals of all sorts exist (and far more than that), but their justification from the point of view of the cumulative type structure is much more problematic; it cannot be given from the viewpoint of starting “from below” with simpler ideas, and saying as much as possible in that direction, then relativizing to a set and continuing the process. 4.5. Exercises
(1) Show that if Z F is consistent, then Z F is an essentially infinite extension of Z, that is, for any sentence (b such that ZF !- (by there is an axiom y of ZF such that Z #J Y y. [The method is to use R, to find an a such that V, I. Z 4, then if all axioms of ZF follow from Z -t (b, V, would be a model of ZF.] Show similarly that ZF, is an essentially infinite extension of ZF, and that ZF2 is an essentially infinite extension of ZF1, etc. (provided these theories are consistent.)
+
+
(2) Prove Theorem 4.3. However, show that SCMZ(y)as defined in 4.2 does not imply that y is V , for a limit ordinal il > o.(For example, the models given in Exercise 1.6(1) for Z will show this.) : and RT given in 4.3 are Show also that the forms of reflection R indeed equivalent to R, and R1,respectively. [These will be equivalent in the same sense as the other equivalent forms for Royas noted in ch. 3 §6.5(5).]
Notes to Chapter 4 Shepherdson [1951] gives most of the properties of V, as in $1. He uses the term complete for transitive, and supercomplete for transitive and closed under taking subsets (in connection with models of set
CH.~]
NOTES
125
theory). Other results are in Montague and Vaught [1959]. Mahlo’s work appeared in Mahlo [1911], [I9121 and [1913], though as noted only for weakly inaccessible cardinals. The reflection principles of $4 are in Bernays [1961] and LCvy [1960a]. The result of $1.6(6) is often taken as a better way of looking at classes than an impredicative class theory such as that of Morse [1965] (see ch. 1 55.3). If we consider that we have all classes to quantify over, then we have simply stopped forming sets at some strongly inaccessible cardinal (and we shouId have gone further). Note that the assumption of a collection of Skolem functions for V (as in $1.5) is not particularly easy to justify from the point of view of the cumulative type structure. Another argument for the existence of ordinals a such that V, < V is as follows. Take all the formulas of the language as a simple sequence 40, d2, . . .. Then using the reflection principle Ro, for each n find a normal function f, such that for each cc, VfnCa) reflects 4,,. (Note that ch. 3 51.6 does in fact prove that we can define such a normal function, for each formula 4.)Then if we can take the collection of allf, for n < w , we shall have common points of their ranges, and if a is in Range&) for all n < w , then V, < V. [We cannot of course define the collection {f,, 1 n < o} in ZFC, since the definitions of thef, will vary in length with n ( n is not a variable offn), but it seems a reasonable collection to assume, when each separatef, is definable. The proof can be carried out in the impredicative Bernays-Morse theory, but there it really proves the result asked for in 51.6(8). See ch. 5 $7.18(3) for more detail.]
INACCESSIBLE AND MAHLO CARDINALS $1. Properties of V, The work of the previous chapter has prepared the way for us to give more answers to the question: what is true in a given set? In particular we want to see which axioms hold in the sets V,. We have already noted that (since V, is transitive for each u) the axioms of extensionality and foundation will hold. We can now say more: 1.1. Theorem (ZFC). For every ordinal u > 0, the axioms A1 (extensionality), A2 (foundation), A3 (subsets), A4 (empty set), A7 (union) and A10 (choice), are true in V,. Further, if u is a limit ordinal, axioms A5 (pairs) and A6 (power set) are true in V,; ifcc > w , then A8 (injinity) is true in V. (So that, e.g., V , , , is a model of the theory ZC.) (Note that we are writing “true in X” in place of “true in ( X , E(X))”.) Proof. For any a > 0, V, is transitive, and Al, A2 are both logically equivalent to d oformulas. But these are absolute between any transitive structures, and so they will hold in V, since they hold in V. (Note that this repeats the proof given in ch. 3 $1.2.) All of the remaining axioms (except A10) will require certain terms to be legitimate in V,. In the case of A4, the set required (the empty set) is A,, i.e., defined by a bounded formula, and so the definition is absolute for any transitive set containing 0;but 8 E V, for any a > 0. Similarly the axioms A5, A7 and A8 all require sets which have dodefinitions. For A8, the set required (w) will be in V, iff a > w; so V, satisfies A8 iff a > w. For A7, the set required is U a , and since p ( u a ) < p(a), we have a E V, + U a E V,; so A7 will hold in all V,, a > 0. For A5, however, the set required is { a , b}; and p({a, b}) = = max(p(a), p(b)) 1. So if a = @ 1 and p(a) or p(b) = @, then {a, b } $ V,, and A5 will fail; but if u is a limit ordinal, then a$ E V, + {a, b) E V,, and A5 will hold. For the other axioms we no longer have dodefinitions, and we must look more closely. For A6, we do have a ITl definition of P ( a ) , and ITl formulas are preserved under restriction; so if x = P( a ) holds in V
+
+
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INACCESSIBLE AND MAHLO CARDINALS
ICH.4,
$1
(i.e., if x is a ( a ) ) ,then (provided a,x E V,), x = P(a) will hold in V,. Since p ( g ( a ) ) = p(a) + 1 , this means that A6 will hold if u is a limit. For A3, we must consider all formulas 4, and we have no hope of reducing the definition { x E a I +(x)} to simple form. But we can prove in ZF that b = {x E a I V, I= +(x)} will be a set, using the definability of satisfaction; and for a E V,, we shall have b E V, since b c a So we shall have V, k Vx ( x E b t-) x
E
a
A
+(x)).
Hence V, satisfies A3. For A10 we have no definition of the choice function to use. But if a is a disjointed set, and b is a choice set for a, we can assume that b c U a ; hence if a E V,, then b E V,. But the statement that b is a choice set for a can be written as a A,-formula VY E U [ 3 2 ( 2 E Y ) -+ 32 E Y
(2 E
b
A
VW E Y
(WE
b -+ W = Z))]
and so this will be absolute between Vand V,, i.e., b will be a choice set for a in V,. So A10 will hold in V,. (We are, of course, assuming that every set has a choice function, i.e., that the axiom of choice holds in V.) 0 1.2. Note. We have set out this proof only semi-formally, yet it is a proof which can be carried out within ZFC,just as the proofs of ch. 2 could be carried out in the various systems claimed. Thus using the notions defined in ch. 3 $5, we have shown that, e.g.,
ZFC 1 Va ( a > 0
--f
Vx E "V, Sat['AlO',
V,, x ] ) ,
where n is the rank of the Godel-set 'A10'. It is worth noting that this proof relies heavily on the fact that satisfaction can be defined in ZF (which we showed in ch. 3 $5.4) and it also relies on using the properties of absoluteness of ch. 3 $4, particularly ch. 3 $4.14. These last were not presented as proofs within ZFC (they could not be until after the formalization of the language in ch. 3 $9,and we do not intend to go over them again to show that they can be presented within ZFC.It should be clear that they can, and we leave this as an exercise to check. (It may be noted that we used this fact before in proving ch. 3 $7.4; the proof that a subset closed under a set of Skolem functions is an elementary substructure, so that everything is absolute, has to be regarded there as being proved in ZFC.)
CH.4,
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PROPERTIES OF
V,
109
One point to be observed in this is that we do not have a general notion of satisfaction in V ; but we do have the notion for restricted classes of formulas (from Exercises ch. 3 $5.11(5) and (6)), and this is all we need in the proof of 1.1. In fact the work of ch. 3 $4 on absoluteness can be carried out in ZFC in two distinct senses: one is for absoluteness between transitive structures which are sets, and the other between V and transitive sets. If we carry out both these formalizations, we can then use ch. 3 $4.8 with Godel’s second incompleteness theorem (as in ch. 3 $5.10) to deduce that we cannot define a collection of Skolem functions for V within ZFC (in contrast with ch. 3 $5.7, which shows that such a collection must exist for sets). We want now to move to higher realms, and prove:
1.3. Theorem (ZFC). If. is strongly inaccessible, then V, != ZFC. Proof. We only have to prove the axiom of replacement, A9, since K is a limit ordinal > w . Suppose that y(x, z) is a formula such that
and let b be the set
We must show for each such y that for each a E V,, b E V,; then the case of the replacement axiom using y will be true in V,. It is clear that p(b) < K ; we must show that p(b) < K . Put g(c) = p(d) 1, where V, k y(c, d ) , if such a d exists in V,, otherwise put g(c) = 0. Then we shall have p(b) = Sup{g(c) I c ~ a } . But a € V,, and so since K is a limit ordinal, a E V, for some IX < K . Then Fa < 1,,and so d < 1,; and since K is strongly inaccessible and a < K , we have 2, < K by Exercise ch. 2 $7.8(2). Hence since g : a + K and d < K and K is regular, we have Sup{g(c) I c E a} < K , i.e., p(b) < K and b E V,, as required. 0
+
1.4. Corollary. Unless ZFC is inconsistent, we cannot prove the existence of strongly inaccessible cardinals in ZFC.
110
INACCESSIBLE AND MAHLO CARDINALS
[CH.4, $1
Proof. By Godel’s second incompleteness theorem, ch. 3 $5.10, since 1.3 gives ZFC I- 3~ (Inac(K)) + 3 X ( X k ZFC).
0
Models of the form V, for K strongly inaccessible are called natural models for ZF. 1.5. The assumption that there are sets A such that ( A , E(A)) is a
model of ZF (or ZFC) is much weaker than the assumption that there are inaccessible cardinals. We could justify the assumption that ZFC has transitive models which are sets by the following heuristic argument: Suppose we had a countable collection F of Skolem functions for V. Then the closure of (0)under F would be a countable elementary submodel of V ; we could use Mostowski’s isomorphism theorem to collapse this to a transitive set A , and we should have A k ZFC since V b ZFC. (Of course this is the argument mentioned in $1.2, which shows that we cannot define a collection of Skolem functions for V, or prove that they exist, in ZFC.) But this argument will hold whether or not there are inaccessible cardinals. Now suppose that a is the first inaccessible cardinal, then we can find a model A of ZFC by taking the closure of (8) under a set of Skolem functions for V, and collapsing to a transitive set. Since A will be hereditarily countable, we shall have A E V,, c V,, so V, k 3 a countable model of ZFC. But if a is the first inaccessible, then also Inac(B)) (see Exercise 1.6(3)) since Inac(p) is absolute for V,, and if /? < a, then /?is not strongly inaccessible; so V, is a model of Va
k VB
ZFC
(7
+ Vp (7
InacG))
+ 3 a model of ZFC.
1.6. Exercises (1) Models of Z. V , + o) is the most natural model to think of for ZC, and we could use it to show that many of the uses we have made of the replacement axiom are essential. But we can find smaller models: ,W,. Show that (i) Define W,, = o, W,+, = @( W J , W , = W , is transitive, (ii) W , k ZC, (iii) V , # W , (although V , c W,).
u,,
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PROPERTIES OF V,
111
For another example, let qn = { . . . {n} . . . } with n braces, i.e., a rank 2n set (if t ( x ) = { x } , then q , = t"(n)).Then put Q = {q, I n < w} and Ql = UQ, Q"" = UQ". Now define U,, = w v Q, Untl = = g ( U , ) V U , V Q", U = U,. Show that (i) U is transitive (although U , is not, for each n), (ii) U k ZC, (iii) Q E U but TC(Q) $ U (although TC(Q) c U ) . Hence some use of replacement is needed before transitive closure can be defined. [These examples show that ZC is not a particularly nice theory, and better theories, such as were noted in ch. 3 $6.5(5), can be given for ZFC minus replacement.]
u,,,
(2) Show that if K > w is a cardinal, then H ( K )is a model of the axioms A l , A2, A3, A4, A5, A7, A8 and A10; and if K is regular, then it is a model of A9 also (and hence all of ZFC except the power set axiom). [If K is strongly inaccessible, then H(K)= V, and so is a model of all of ZFC.]
<
(3) Show that if a is a limit ordinal, then cf(/?) = 6, /? 26, and Inac(/?) are all absolute for V,. [Use completeness as in ch. 3 54.9. If / ? , 6 ~V,, then the functions required will be in V,.] (4) Show that the first a for which V, k ZFC has cf(a) = w, and hence the converse to 1.3 does not hold. [Suppose that V6 k ZFC, and let F be a set of Skolem functions for V,. Given /3 < 6, let g(/?) be the least ordinal such that if xl,. . ., x, E V, and f E F, then f(xl, . . ., x,) E V,,,,. Let a = Sup{gn(/?)I n < w}. Show that a < 6 and V, I= ZFC (in fact V, < V6),but cf(a) = w.] Show that if a is the first ordinal for which V, I= ZFC, then V, is a model of ZFC 3 X ( X k ZFC) V/?(- Vb k ZFC).
+
+
(5) Hyperinaccessible cardinals. Suppose that the strongly inaccessible cardinals have been enumerated in order as 8, for ordinal a. Then a cardinal K is hyperinaccessible if 8, = K , i.e., if K is strongly inaccessible and there are K strongly inaccessible cardinals below K . Show that if K is the first hyperinaccessible cardinal, then V, is a model of ZFC the axiom of inaccessibles (of ch. 2 $7.7) "there are no hyperinaccessiblecardinals".
+ +
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INACCESSIBLE AND MAHLO CARDINALS
[CH.4,
81
Similarly K is hyper-hyperinaccessible if K is hyperinaccessible and there are K hyperinaccessible cardinals below K . Show that if K is the first hyper-hyperinaccessible cardinal, then V, is a model of
ZFC +Vg 3 y > ( y is hyperinaccessible) + “there are no hyper-hyper inaccessible cardinals”. (6) Natural models f o r von-Neumann-Bernays-Godel set theory and for Bernays-Morse theory. These systems have variables for classes as well as for sets, and the axioms A3 (subsets) and A9 (replacement) can be regarded as being single axioms with a free class variable taking the place of the arbitrary formula. Then a comprehension axiom is added for classes, as noted in ch. 1 $5.3. Suppose that K is strongly inaccessible, and interpret class variables as ranging over V,,, (i.e. 9 ( V , ) ) and set variables as ranging over V,. Show that this makes (V,+l, V,, E(V,+,)) into a model of the full Bernays-Morse theory. Show that the converse is now true: Show that if ( Va+,,V,, E(V,+J) is a model of even the von-Neumann-BernaysGodel set theory, then M is strongly inaccessible. [Indeed what is needed here has nothing to do with the comprehension axiom assumed; it is the result of taking A3 and A9 in second-order form, since the class variables are being interpreted as ranging over P(V,). Then A9 says that the image of a set under any (set-valued) function is a set, and A3 says that any subcollection of a set is a set (and they now really do mean any). If these second-order axioms are true in a given transitive set which is a model of Z, show that that set must be of the form V, for strongly inaccessible M.]
(7) Show that if u < /?are ordinals such that V, < Vs, then u must be a strong limit ordinal and both V, and V, must be models of ZFC. [Show first that u must be a limit ordinal. To see that u must be a strong limit, suppose not: let f : P(8)+ u be onto, for 6 < u. Then there is a set X c Y(d) x 8 ( 6 ) which well-orders Y ( 6 ) in order-type u ; X will be in V,,, (it could be coded in V,,,) and so in V,. Now in V,, there will be an ordinal similar toP(6) ordered by X ; since V, < V,, this must also be true in V,, contradicting the assumption that the ordertype was CI.To see that the axiom of replacement must hold in V,, suppose that $(x, y , a,, . . ., a,) defines a function f in V,, using parameters a,, . . ., a, from V,. Then (b must also define a function, say g, extendingf, in V o ;and for any set a E V,, the range of g 1 a c V, and
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113
NORMAL FUNCTIONS
so is a set in V,. Hence the range off 1 a must be a set (i.e., a member) in V,, and so V, satisfies replacement.] (8) Show that if a is strongly inaccessible, then there is a p < a such that V, .< V,. [Take a set F of Skolem functions for V,. Let g(6) be the Sup of the ranks of images of members of V6 under members of F, for 6 < a. Let g"(6) be Sup{gn(6) I n < o}, and show that if 6 < a then gw(6) < a, and if p = g"(6) then V, < V,.]
$2. Normal functions
We want to present a very powerful way of postulating the existence of large numbers of inaccessible, hyperinaccessible, hyper-hyperinaccessible,etc., cardinals of all sorts, which is due to Mahlo. This relies on the idea of a normal function, or a closed unbounded class (or set). We shall present this first for functions which are defined for all ordinals, and these must be proper classes; so we shall work in this section with variables for proper classes. We shall be able to restrict all that we do to ZFC if we wish to by regarding classes as always being definable by a formula (possibly with parameters), so that statements about classes are to be read as statements about all defining formulas. We shall not use different letters for proper classes, because we have already used most of the kinds available, and also because we shall later want to take what we do here and restrict it to sets, and it is easier to use the same notation in both cases. 2.1. Definition. A function f :On + On is a normal function iff (i) f is increasing, i.e., if a < /? then f ( a ) < f ( p ) ;and (ii) f i s continuous, i.e., if Lim(a) then f ( a ) = f(E). Here On is the class of all ordinals, {x I Ord(x)}. Note that (i) implies that for all a, a < f ( a ) (by induction on a ; there cannot be a first u such that f ( a ) < a if (i) holds). (ii) says that f is continuous with respect to the order topology on the ordinals; this is given by saying that a Y EX. subset X c On is closed if, whenever Y c X , Y # 0, then Given this definition of closed, we have:
uZ<,
u
2.2 Lemma. The range of a normal function is closed and unbounded in On; and every closed unbounded class of ordinals is the range of a unique normal function (its enumerating function).
114
[CH.4,
INACCESSIBLE AND MAHLO CARDINALS
$2
Proof. The range is unbounded since a 6 f ( a ) , it is closed because
f is continuous and On is closed. Given a closed unbounded class
X,
its enumeration (which exists by transfinite induction) will be a normal function, and it will be defined for all ordinals since X must be a proper class, by the axiom of replacement. 0 One of the important properties of normal functions concerns fixed points. 2.3. Definition.
is afixedpoint for f iff f ( a ) = u.
M.
Theorem. I f f is a normal function, then the fixed points o f f form a closed unbounded class. Proof. First we show that it is unbounded. Given a, define
f “ 4 = a, fn+Ya) = f(f”(4), f Y 4
=
un
If u is a fixed point, then all of these will be u ; if not, then a < f ( a ) , and so by (i) the sequencef”(u) will be strictly increasing for n < m. Hence f W(u)is a limit ordinal, so by (ii), f(fw(E>)
=
Ut
f(Q*
But if E < f “(a),then 5 < f”(a) for some n < o,so f ( t ) < f””(a) and
f ( f “(a))
u n
f”+YQ)= f “(4-
Hence f ” ( a ) must be a fixed point; a 6 f w(a)and so the fixed points are unbounded. (We have actually shown that f W ( u 1) is the next fixed point above a, for each u.) To see that the fixed points are closed, suppose that Y is a set of fixed points and Y 4 Y. Then Y must be a limit, and a < Y iff 3~ E Y ( M . < 6). S O f ( U Y ) = U F EfY( 6 ) = UtEYE (since Y is a set of fixed points); i.e., f ( u Y ) = Y and Y is a fixed point. Hence the class of fixed points is closed.
+
u
u
u
u
u
2.4. Corollary. For any normal function f we can form the derived function f’, which enumerates the fixed points off; and f’ is also a normal unction. 0 Two important normal functions which we have already met are the functions K and 1.Fixed points of these are ordinals a such that K, = a or 1, = a ; it is clear from the construction in the proof of 2.3 that these will usually be singular (indeed, the first fixed point above
CH.4,
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NORMAL FUNCTIONS
115
any ordinal will always have cofinality 0).Regular fixed points of these two functions are the inaccessibles (strong or weak). So we can see the following axiom as a natural generalization of the axiom of inaccessibles: 2.5. Axiom F. Every normal function has a regular fixed point.
(Since the normal functions are proper classes, this axiom can be stated only as a scheme in the language of ZFC, with one instance for each formula, saying that if that formula defines a normal function, then it has a fixed point.) To see that this is much stronger than the axiom of inaccessibles, first note that we can enumerate the beth numbers greater than a given u by a normal function, and this must have a regular fixed point which must therefore be a strong inaccessible >u. So the axiom of inaccessibles follows. But then we can enumerate the class of strong inaccessibles by a function which is continuous at limits (i.e., we enumerate the closure of the class of strong inaccessibles), and this will be a normal function whose regular fixed points will be hyperinaccessibles. So (since we can start the enumeration of strong inaccessibles above any point we choose) we can deduce the existence of a proper class of hyperinaccessibles, and repeat the process to get hyperhyperinaccessibles, etc. This axiom does not seem to have received a name in the literature, probably because stronger axioms were proposed at the same time. We shall derive another form of axiom F in $4; meanwhile to give the stronger axioms, we need to relativize the notion of a normal function to a set, which in practice will be a cardinal. 2.6. Definition. For ordinal u, a functionf :6 + u is a normal function on u iff is increasing, continuous, and its range is unbounded in u, i.e., URange(f) = u. We shall really only be interested in this notion when u is a regular cardinal, thus we have: 2.7. Theorem (ZFC). The following are equivalent: (i) Every normal function on u has afixedpoint. (ii) u is regular and u > w. Proof. If u is regular and f :6 + u is unbounded in u, then we must have 6 3 u ; iff is normal on u, then in fact we must have 6 = u.
116
INACCESSIBLE AND MAHLO CARDINALS
[CH.4,
$3
Then if u > m, we can construct a fixed point off exactly as in the proof of 2.3. To prove the converse, note that w certainly has normal functions without fixed points (any increasingf: m --f m will be counted as a normal function on m). If u is not a limit, then no function can be unbounded in u, and if cf(u) < u, say cf(u) = 6, then we can find f : 6 + a which is unbounded, increasing and continuous, and such that f(q) > 6 for all 11 < 6; this f cannot have a fixed point. 0 We set out some of the properties of normal functions on regular cardinals in the exercises; in the next section we use them t o define Mahlo cardinals. 2.8. Exercise
(1) (a) Given a normal function f (on a regular cardinal K > cc) or on On), let f' be the derived function off. Then define f" by f o = f, f"" = (f*)',and if 1is a limit ordinal, f is the function enumerating all ordinals which are fixed points of each f' for 6 < 1. Show that iff is a normal function on On, thenf" is a normal function for each a ; iff is a normal function on K , then so isf" for ci < K. (b) We can also define the function g by
d.1
=f"(O) Show that g is also a normal function (in the same sense as f, i.e. on On or on K ) . (c) We can also define the diagonal function f A as the function which enumerates all ordinals u such that cc = f J ( a ) for all 6 < a. Show thatfA is also a normal function (in the same sense as f ) .
$3. Mahlo cardinals The definitions which Mahlo gave are effectively equivalent to the following: 3.1. Definition. K is weakly Muhlo iff K is an ordinal such that every normal function on K has a regular fixed point. K is strongly Muhlo iff every normal function on K has a strongly inaccessible fixed point. In view of 2.7, it is clear that weakly or strongly Mahlo ordinals must in fact be cardinals, since they must be regular, and they must be > m . We shall use Muhlo for strongly Mahlo in accordance with the
CH.4,
$31
117
MAHLO CARDINALS
general practice (though originally Mahlo himself considered only weakly Mahlo cardinals). First we show that these cardinals must themselves be inaccessible: 3.2. Lemma. If K is weakly Mahlo, then K is weakly inaccessible; i f K is strongly Mahlo, then K is strongly inaccessible. Proof. We have seen that K is regular. Suppose that K is not a limit cardinal, then K is 6 + (i.e. N(6)) for some 6, and f : K + K given by f(u) = 6 u is a normal function on K whose fixed points cannot be regular. So K cannot be weakly Mahlo. If K is not a strong limit cardinal, then 26 2 K for some 6 < K , and again f : K + K given by f(u) = 6 u is a normal function; in this case its fixed points cannot be strong limit cardinals since they are between 6 and 2d, so they are not strongly inaccessible and K is not strongly Mahlo. 0
+
+
We can now go further using the following: 3.3. Definition. K is 0-weakly hyperinaccessible iff K is weakly inacces1-weakly hyperinaccessible iff K is regular and there are sible. K is u K u-weakly hyperinaccessible cardinals below K . For limit A, K is Aweakly hyperinaccessible iff K is u-weakly hyperinaccessible for all u < A. Similarly for u-strongly hyperinaccessible. Now we can show:
+
3.4. Theorem (ZFC). If
K is weakly Mahlo, then K is u-weakly hyperinaccessible for all u < K ; if K is strongly Mahlo, then K is u-strongly hyperinaccessible for all u < K . Proof. The proof is identical for the strong or weak case, and we shall write u-hyp for either u-weakly or a-strongly hyperinaccessible according to which case is being considered. First we note that if K is weakly Mahlo, then any normal function on K must have K regular fixed points. For supposef:K -,K is normal on K . Then for any fixed p < K , if g(u) = f(p a), then g : --f~K is also normal on K , and so g has a regular fixed point 6. But then 6 = = f(p + 6), and so since f(p 6) 2 p 6, we have p 6 = 6 = = f(p + a), i.e., 6 is a regular fixed point off greater than p. Since K is regular, this means that there must be K such regular fixed points. This gives the induction step from u 1 to u 2 immediately. Suppose that K is a + 1-hyp, then K is a regular fixed point of fa, wherefx is the normal function enumerating the closure of the class of u-hyp cardinals. Hence
+
+
+
+
+
+
118
r
[CH.4,
INACCESSIBLE AND MAHLO CARDINALS
$3
K is a normal function on K , and so has K regular fixed points. Each of these is then a 1-hyp and less than K , i.e., K is a 2-hyp. The induction step for A-hyp is trivial for limit A, since we have defined A-hyp as: a-hyp for all a < I . It remains to show that if K is A-hyp, 1-hyp. where A is a limit < K , then K is I So suppose that K is A-hyp, and I < K . Then for 6 < I , let X , be the closure of {a < K [ a is 8-hyp). Then since K is 8 1-hyp, card(Xd) = K for each 6 < I , and all regular members of X , are d-hyp. We need to show that X, = X , has cardinal K also. Then, since X , is also closed, the function enumerating X , will be normal on K and so will 1-hyp. have K regular fixed points; these will all be I-hyp, and so K is A So suppose that p < K , and set
fa
+
+
+
+
nd<,
+
a, = Inf(X, - p) for 6 < A,
u6<,
and let y = ad. Since K is regular and I < K, we have y < K . But if 6 < 6' < A, then ad,E X,,and so y is a limit of members of A', and since X,is closed, we have y E X , for each 8 < A, i.e., y E X,. Hence X,is cofinal in K , and since K is regular, it must have cardinal K , as required. 0 This theorem is not the strongest we can prove in this connection; indeed, much of the interest in Mahlo cardinals arises because there seems to be no strongest statement which can be made, and the Mahlo cardinals are larger than any which can be given, using methods like those of $3.3. Of course this is an imprecise statement; a little more justification for it is given in Exercise 3.7(2). Another way of looking at Mahlo cardinals is in terms of stationary sets:
3.5. Definition. A subset x of K is called stationary in K (sometimes Mahlo in K ) if, for every closed, unbounded subset y of K , x n y # 8. For cardinals cofinal with w , this notion is not of much interest; a subset will be stationary only if it contains a final segment. But if cf(K) > o,then the closed unbounded subsets of K generate a filter on K ; the intersection of any two closed unbounded subsets of K is again closed and unbounded in K . (To see this, if x and y are both closed, unbounded, in K , and a < K , put a. = Inf(x - a ) ; a1 = Inf(y - ao), as = Inf(x - al), etc., and a , = a,. Then amE X n y since it is a limit point of both, and a 6 a, < K since cf(K) > w . ) So for
u,,,
~13.4,$31
MAHU)
CARDINALS
119
cf(K) > o,any set of this filter (i.e., any set which contains a closed unbounded subset of K ) will be stationary, and the stationary sets are those which meet every member of this fdter. Now the definition of Mahlo cardinals can be rephrased as: K is weakly Mahlo if the regular cardinals less than K form a stationary subset of K ; K is strongly Mahlo if the strongly inaccessible cardinals less than K form a stationary subset of K.
3.6. Mahlo's operation In terms of stationary sets, we can set out the process which Mahlo used in passing from the inaccessible cardinals to the Mahlo cardinals: Definition. Suppose X is any class of cardinals. Then H ( X ) = {a E X [ X n a is stationary in a}. Then H is Mahlo's operation: If X is the class of regular cardinals, then H ( X ) is the class of weakly Mahlo cardinals; if X is the class of strongly inaccessible cardinals, then H ( X ) is the class of strongly Mahlo cardinals. We can now iterate the operation, and define H a for ordinals a by:
HO(X) = X , Ha"(X) = H(H"(x)), H A ( X )= na,,Ha(X) if Lim(A). If X is the class of strongly inaccessible cardinals, then H"(X) is the class of strongly Mahlo cardinals of kind a. Thus the strongly Mahlo cardinals of the second kind will be just those cardinals K such that every normal function on K has a fixed point which is strongly Mahlo of the first kind. We can then diagonalize this operation, and define a E H*(X) iff Vg < a ( a E P ( X ) ) .
Then if X is the class of strongly inaccessible cardinals, H A ( X )is the class of (strongly) hyper-Mahlo cardinals, (HA)A(X)is the class of hyper-hyper-Mahlo cardinals, etc. (Then K is hyper-Mahlo iff it is Mahlo of kind K . ) It is clear that there is no end to the diagonalizations which can be performed on the operation H, to give higher and higher classes of Mahlo cardinals.
I20
[CH.4,
INACCESSIBLE AND MAHLO CARDINALS
$3
3.7. Exercises (1) Show that the filter generated by the closed unbounded subsets of K is cf(K)-closed; that is, the intersection of a set of < c f ( ~ ) sets of the filter is again a set of the filter. [Use the argument given in 3.5 to show that a set of < c f ( ~ )closed unbounded subsets of K has an intersection which is unbounded in K . ] (2) Show that theorem 3.4 can be strengthened. Show that if K is Mahlo, then K is not the first cardinal u such that u is u-hyperinaccessible. Show that there must be K cardinals u below K such that a is u-hyperinaccessible. [For cc < K , let f(u) be the first ordinal /3 such that there are a cardinals below fi which are cr-hyperinaccessible; show that fis a normal function on K . ]
+
(3) Show that if K is Mahlo, then V, is a model of ZFC axiom F. Show that the converse of this is not true. [Since axiom F is first-order, arguments such as used in Exercise 1.6(4) will show this.] However, show that if a second-order form of axiom F holds in V,, then K is Mahlo. [Compare Exercise 1.6(6).J (4) Regressive functions and stationary sets. If X c K and f:X --f K , then f is called regressive on X if f(u) < u for all u in X , a # 0. Establish the following characterization of stationary sets : If Cf(K) > w , then X c K is stationary in K iff for every function f regressive on X , there is some ,!? < K such that { % € X I f(a) < p} is unbounded in K . [If X is not stationary in K, there is a B c K - X which is closed, unbounded in K . Set f(a) = U ( B n u) for u E X , and show that f is regressive on X and for B < K , ( a E X < p} is bounded by Inf(B - (fi 1)). To prove the converse, set D = f”X n ( K - X ) and consider two cases : First if D is bounded below K, take an upper bound 8, of D and note that if we put f ‘((a) = a,f”f+’(u)= fcf“(u)) iff”(.) E X,f””(a) = 0 otherwise, then for each u there is a first n(u) such that f”‘”’(a) = 0. Take the least n such that {f”(u) I u E X , n(u) = n l} = Z is unbounded in K and show that f ” (2)is bounded by b. If D is unbounded in K , find a contradiction from the assumption: (i) for each u E X , there is a 4(a) E K such that /I> +(a) impliesf(p) > a.
+
).(fI
+
CH.4,
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REFLECTION PRINCIPLES FOR MAHLO CARDINALS
121
Set X' = f - ' ( D ) and defme a cofinal sequence in X', (at 16 < T ) , such that: (ii) t < T,I < T implies $(ae) < a,,. Then Y = { f ( a t ) I 5' < T } is unbounded in K and a subset of K - X; since X is stationary, X n Cl(Y) # 0 (where Cl(Y) is the closure of Y). Then if a* E X n C1( Y ) , f ( a * ) < a*, and this contradicts (i) and (ii).] As a corollary of this, deduce the following theorem, due to Fodor: Suppose that K is regular, > o,Xis stationary on K , andfis regressive on X.Then for some a < ~ , f - ~ (isa stationary ) on K . [Show first: (iii) if { Y, 1 a < T } is a sequence of pairwise disjoint, non-empty, non-stationary subsets of K , with T < K , such that if U = = {y, I a < T } , where y, = Y,, then U is not stationary also, then Y= < Y, is not stationary. Prove this by using the characterization of stationary sets above; show that a sequence of regressive functions for each Y, - {y,}, and one for U, can be fitted together to give a regressive function on Y. Now set Y, = { E E X I f ( t ) = a> and use (iii) on this to show that some Y, must be stationary.]
n
u,
$4. Reflection principles for Mahlo cardinals
Ltvy and Bernays have both given strengthenings of the reflection principle R,, and the first of these turns out to be equivalent to axiom F. For any formula with at most x,, . . ., x , free, we get an instance of the reflection principle V a 38 > a (Inac(B) A Vx,, . . ., x, E Vs (4 tf 4'0)). R1 4.1. Theorem (ZF). R1 is equivalent to axiom F. Proof. First we derive F from R1.We shall use the form of R1which reflects two formulas at once; this will be equivalent to R1exactly as in ch. 3 $6.4 for R,. Suppose that f is a definable normal function. We must show that f has an inaccessible fixed point. We use R1on the formulas
f(Y) = 6, VY 36 CfW = 4-
Then we get an inaccessible B such that, for y,6 < /?,
f(r) = f3wf '(74 = 6
(so the definition off will be absolute for V8),and
VY < B 36 < B CfV4r)= d),
122
INACCESSIBLE AND MAHLO CARDINALS
[CH.4,
$4
(since V y 38 ( f ( y ) = 6) holds in Y by hypothesis). But together these say VY < B 36 < B (fb) = 61,
and so since ,8 is a limit ordinal,f(/?) = Uy<,f(y) < /?,i.e., B is an inaccessible fixed point off. The proof in the opposite direction, F implies R,, is an extension of the proof of R, given in ch. 3 $6.3. In that proof we defined a function f on all ordinals, for a given formula 4, such that f was increasing and had the property that if X = V,,,, for any <, then X reflects 4. The f as defined was not necessarily continuous, so let g be the result of making f continuous, i.e., g enumerates the closure of Range(f). (Then for IZ < co, g(n) = f ( n ) , for a > co, g(a 1) = f ( a ) , and for limits A, g(4 =
+
Umf(a).>
Then in fact g will have the same property as$ We need to check that for limit 1, X = V,,,, reflects 4, but this follows just as for f, since if xl,. . ., x,, y,, . . ., yr E X, then x,, . . ., x,, y,, . . ., y r E V&,) for some a < 1. So g,(xl,. . ., x,, yl,. . ., y,.) < f ( a 1) < g(A), and so the induction on r will hold as before and so X will reflect 4. But now g is a definable normal function (with parameter a only) and so by F it will have an inaccessible fixed point. This will be a /I> cc such that /? is inaccessible and V, reflects 4, hence R, holds. 0
+
We can now strengthen R1 to: R2
V a 38 > a (/I is Mahlo
A
Vx,,
. . ., x, E
V,
(4 ++ CvB)),
which will be equivalent to a scheme asserting that every normal function has a Mahlo fixed point, and then again using hyper Mahlo to get RB,etc. LCvy has given a way of writing these reflection principles using the notion of a standard complete model of a theory (which will be Z or ZF or a theory extending ZF). The idea is to write the schemes of the theory in second-order form, so that we get the natural models of the theories. We then cannot put this in the form x k ZF, whichwas used in ch. 3 55.9, but we can write them directly: 4.2. Definition. SCMZ(x)for
Trans(x) A Vy E x Vz ( z c y + z E x) A x k Z “x is a standard complete model of Z ’
CH.4, $41
REFLECTION PRINCIPLES FOR MAHLO CARDINALS
123
SCMZF(x)for SCMZ(x) A V f v y ~ ~ ( C f : x - - , x ) ~ f " y ~ x ) . For this definition we shall have: 4.3. Theorem (ZF). SCMZF(x)c-) 3cr (Inac(or) A x Proof. Exercise.
Now we can re-write Ro and R1as
=
V,).
-
. . ., x , E y (4 p)); Vx 3y ( x € y A SCMzF(y) A Vx,, . . ., x, E Y (4 p)). RF If we then write ZF1 for ZF + R1,and define an appropriate notion R:
Vx 3y ( x E y
A
SCMz(y)
A
Vx,,
f-)
of SCMZF1,we can then write
vx 3y ( x E Y A SCMZF1(y)A Vx,, . . ., x , E Y (4 p)). R: We can then define ZF2 and repeat this, paralleling the applications of Mahlo's operation. We leave the details as an exercise. f+
4.4. A note on the plausibility of these cardinals Since we cannot prove in ZFC that there are strongly inaccessible cardinals, it is clear that we cannot prove that Mahlo cardinals exist; axioms would have to be added to say that they do exist, such as the schemes R1,R2,etc. How plausible are these axioms? There is no question of ever proving them consistent with ZFC, of course, since we cannot prove that ZFC is consistent, using methods available in ZFC. It may happen that they will one day be proved inconsistent with ZFC, but if they are in fact consistent, then we shall never have a proofwe shall never know by looking within ZFC. We can only hope for heuristic justification from outside. We look for justification for these axioms from the point of view of the cumulative type structure, where we want to say that the collection of levels, which is indexed by the ordinals, is a very rich structure with no conceivable end. An axiom saying that Mahlo cardinals of the first kind exist can then be regarded as saying that the class of strongly inaccessible cardinals is so rich that there are members K of the class such that no normal function on K can avoid this class; however we climb through K , provided we are continuous at limits (so that we are
124
INACCESSIBLE AND MAHLO CARDINALS
[CH.4
enumerating a closed subset of K ) , we shall eventually have to hit a strongly inaccessible. In this sense, axiom F can be regarded as the ultimate in saying that inaccessibles, hyperinaccessibles, etc., of all sorts exist, and the existence of Mahlo cardinals can be regarded as the next step after this-it always seems a plausible step, in view of the reflection principle, to take a property of the whole universe (such as axiom F), and postulate that it already holds at some level V,. (See Exercise 3.7(3) in this connection.) We shall find axioms in later chapters which imply that Mahlo cardinals of all sorts exist (and far more than that), but their justification from the point of view of the cumulative type structure is much more problematic; it cannot be given from the viewpoint of starting “from below” with simpler ideas, and saying as much as possible in that direction, then relativizing to a set and continuing the process. 4.5. Exercises
(1) Show that if Z F is consistent, then Z F is an essentially infinite extension of Z, that is, for any sentence (b such that ZF !- (by there is an axiom y of ZF such that Z #J Y y. [The method is to use R, to find an a such that V, I. Z 4, then if all axioms of ZF follow from Z -t (b, V, would be a model of ZF.] Show similarly that ZF, is an essentially infinite extension of ZF, and that ZF2 is an essentially infinite extension of ZF1, etc. (provided these theories are consistent.)
+
+
(2) Prove Theorem 4.3. However, show that SCMZ(y)as defined in 4.2 does not imply that y is V , for a limit ordinal il > o.(For example, the models given in Exercise 1.6(1) for Z will show this.) : and RT given in 4.3 are Show also that the forms of reflection R indeed equivalent to R, and R1,respectively. [These will be equivalent in the same sense as the other equivalent forms for Royas noted in ch. 3 §6.5(5).]
Notes to Chapter 4 Shepherdson [1951] gives most of the properties of V, as in $1. He uses the term complete for transitive, and supercomplete for transitive and closed under taking subsets (in connection with models of set
CH.~]
NOTES
125
theory). Other results are in Montague and Vaught [1959]. Mahlo’s work appeared in Mahlo [1911], [I9121 and [1913], though as noted only for weakly inaccessible cardinals. The reflection principles of $4 are in Bernays [1961] and LCvy [1960a]. The result of $1.6(6) is often taken as a better way of looking at classes than an impredicative class theory such as that of Morse [1965] (see ch. 1 55.3). If we consider that we have all classes to quantify over, then we have simply stopped forming sets at some strongly inaccessible cardinal (and we shouId have gone further). Note that the assumption of a collection of Skolem functions for V (as in $1.5) is not particularly easy to justify from the point of view of the cumulative type structure. Another argument for the existence of ordinals a such that V, < V is as follows. Take all the formulas of the language as a simple sequence 40, d2, . . .. Then using the reflection principle Ro, for each n find a normal function f, such that for each cc, VfnCa) reflects 4,,. (Note that ch. 3 51.6 does in fact prove that we can define such a normal function, for each formula 4.)Then if we can take the collection of allf, for n < w , we shall have common points of their ranges, and if a is in Range&) for all n < w , then V, < V. [We cannot of course define the collection {f,, 1 n < o} in ZFC, since the definitions of thef, will vary in length with n ( n is not a variable offn), but it seems a reasonable collection to assume, when each separatef, is definable. The proof can be carried out in the impredicative Bernays-Morse theory, but there it really proves the result asked for in 51.6(8). See ch. 5 $7.18(3) for more detail.]