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Embedding into free topological vector spaces on compact metrizable spaces
Topology and its Applications 233 (2018) 33–43
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Topology and its Applications www.elsevier.com/locate/topol
Embedding into free topological vector spaces on compact metrizable spaces Saak S. Gabriyelyan a,∗ , Sidney A. Morris b,c , a
Department of Mathematics, Ben-Gurion University of the Negev, P.O. 653, Beer-Sheva, Israel Faculty of Science and Technology, Federation University Australia, P.O. Box 663, Ballarat, Victoria, 3353, Australia c Department of Mathematics and Statistics, La Trobe University, Melbourne, Victoria, 3086, Australia b
a r t i c l e
i n f o
Article history: Received 9 April 2017 Received in revised form 12 September 2017 Accepted 13 September 2017 Available online 14 September 2017 Dedicated to the memory of W. Wistar Comfort MSC: 46A03 54A25 54D50
a b s t r a c t For a Tychonoff space X, let V(X) be the free topological vector space over X. Denote by I, G, Q and Sk the closed unit interval, the Cantor space, the Hilbert cube Q = IN and the k-dimensional unit sphere for k ∈ N, respectively. The main result is that V(R) can be embedded as a topological vector space in V(I). It is also shown that for a compact Hausdorff space K: (1) V(K) can be embedded in V(G) if and only if K is zero-dimensional and metrizable; (2) V(K) can be embedded in V(Q) if and only if K is metrizable; (3) V(Sk ) can be embedded in V(Ik ); (4) V(K) can be embedded in V(I) implies that K is finite-dimensional and metrizable. © 2017 Elsevier B.V. All rights reserved.
Keywords: Free topological vector space Free locally convex space Embedding Finite-dimensional Zero-dimensional Compact Cantor space Hilbert cube
1. Introduction In [6] we introduced the study of free topological vector spaces and showed that, in some ways, the topological structure of free topological vector spaces is nicer and better understood than that of free locally convex spaces; in particular, the free topological vector space on a kω -space is a kω -space and
* Corresponding author. E-mail addresses: [email protected] (S.S. Gabriyelyan), [email protected] (S.A. Morris). http://dx.doi.org/10.1016/j.topol.2017.09.008 0166-8641/© 2017 Elsevier B.V. All rights reserved.
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therefore complete. By contrast the free locally convex space on an infinite space X is complete if and only if X is Dieudonné complete and has no infinite compact subsets, see [14]. For many years, [1,7,8,10,12–14], the following question has been investigated: for what Tychonoff spaces X and Y can the free (free abelian) topological group F (Y ) (respectively, A(Y )) on Y be embedded as a topological subgroup of the free (respectively, free abelian) topological group F (X) (respectively, A(X)) on X? An analogous question for free locally convex spaces L(X) and L(Y ) over X and Y respectively was considered in [11]. We note that, in general, the question for free abelian topological groups is harder than that for free topological groups and the question for free locally convex spaces is harder again. In this paper we investigate the following: Question. For what Tychonoff spaces X and Y can the free topological vector space V(Y ) be embedded as a topological vector subspace of the free topological vector space V(X)? The free topological vector space V(X) over a Tychonoff space X is a pair consisting of a topological vector space V(X) and a continuous map i = iX : X → V(X) such that every continuous map f from X to a topological vector space E gives rise to a unique continuous linear operator f¯ : V(X) → E with f = f¯ ◦ i. Theorem 2.3 of [6] shows that for all Tychonoff spaces X, V(X) exists, is unique up to isomorphism of topological vector spaces, is Hausdorff and the mapping i is a homeomorphism of the topological space X onto its image in V(X). For a subspace Y of X, let V(Y, X) be the vector subspace of V(X) spanned by Y . Concerning the question raised above, Proposition 3.12 of [6] shows that, if K is a compact subspace of the Tychonoff space X, then the vector subspace V(K, X) of V(X) is V(K). Further, Corollary 3.11 of [6] says that if Y is a closed subspace of the kω -space X, then the vector subspace V(Y, X) of V(X) is V(Y ). It is known [9] that the free topological group F (R) on R can be embedded as a topological group in the free topological group F (I) on the unit interval I. It is also known [8] that the free abelian topological group A(R) on R can be embedded as a topological group in the free abelian topological group A(I) on the unit interval I, but this is significantly harder to prove. We hasten to point out that while R can be embedded as a subspace of I, no subgroup of F (I) or A(I) generated by a subspace Y of I is isomorphic as a topological group to F (R) or A(R). It is shown in Theorem 4.3 of [11] that the free locally convex space L(R) on R cannot be embedded as a topological vector subspace of the free locally convex space L(I) on I. The main theorem in this paper is that the free topological vector space V(R) on R can indeed be embedded as a topological vector space in the free topological vector space V(I) on I. It is also shown that for a compact Hausdorff space K: (1) V(K) can be embedded in V(G), where G is the Cantor space, if and only if K is zero-dimensional and metrizable; (2) V(K) can be embedded in V(Q), where Q denotes the Hilbert cube Q = IN , if and only if K is metrizable; (3) V(Sk ) can be embedded in V (Ik ), where Sk denotes the k-dimensional sphere, k ∈ N; (4) V(K) can be embedded in V(I) implies that K is finite-dimensional and metrizable. 2. Universal spaces In this section we study free topological vector spaces generated by the universal spaces: (1) the Hilbert cube Q = IN ; it is well-known that every metrizable compact space embeds into Q; (2) the Cantor compact space G; it is known that every zero-dimensional compact metrizable space embeds into G, see Example 3.1.28 and Theorem 6.2.16 of [3]. In so doing, we obtain new results not only for free topological vector spaces, but also for free locally convex spaces on these universal spaces. In what follows we use the following notation. Set N := {1, 2, . . . }. For a subset A of a vector space E and a natural number n ∈ N we denote by spn (A) the following subset of E
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spn (A) := {λ1 x1 + · · · + λn xn : λi ∈ [−n, n], xi ∈ A, ∀i = 1, . . . , n}, and set sp(A) := by
n∈N
spn (A), the span of A in E; we also define the continuous map πn : [−n, n]n ×An → E πn (λ1 , . . . , λn ) × (x1 , . . . , xn ) := λ1 x1 + · · · + λn xn ,
where “+” denotes the vector space addition in E, so πn [−n, n]n × An = spn (A). For every k, m ∈ N define Cmk := {(λ1 , . . . , λk ) ∈ Rk : 1/m ≤ |λi | ≤ m ∀i = 1, . . . , k}.
(2.1)
For a subspace Y of a Tychonoff space X, let V(Y, X) be the vector subspace of V(X) generated algebraically by Y . In what follows we shall essentially use the following results from [6], see Corollary 3.4, Proposition 3.9 and Proposition 3.12, respectively. Corollary 2.1 ([6]). Let X be a Tychonoff space. Then for every compact subset K of V(X) there is an n ∈ N such that K ⊆ spn (X). Proposition 2.2 ([6]). Let X = n∈N Cn be a kω -space and let Y be a subset of V(X) such that Y is a free vector space basis for the subspace, sp(Y ), that it generates. Assume that K1 , K2 , . . . is a sequence of compact subsets of Y such that Y = n∈N Kn is a kω -decomposition of Y inducing the same topology on Y that Y inherits as a subset of V(X). If for every n ∈ N there is a natural number m such that sp(Y ) ∩ spn (Cn ) ⊆ spm (Km ), then sp(Y ) is V(Y ), and both sp(Y ) and Y are closed subsets of V(X). Proposition 2.3 ([6]). If K is a compact subspace of a Tychonoff space X, then V(K, X) is V(K). To obtain a general assertion about embedding of free topological vector spaces over compact spaces, we consider the following three conditions on a class K of compact Hausdorff spaces: (a) K is closed under taking finite products, closed subspaces and Hausdorff quotients; (b) K contains I and has a universal (by inclusion) object, i.e., there exists Ku ∈ K such that every compact space C ∈ K embeds into K; (c) if a compact Hausdorff space K is the union of a countable family of compact spaces from K, then K ∈ K. A typical example of K satisfying (a)–(c) is the following. Example 2.4. Let Kκ be the class of all compact Hausdorff spaces of weight ≤ κ for an infinite cardinal κ. Then Kκ satisfies (a)–(c) with the universal object Iκ (see, Theorems 3.1.19 and 3.2.5 of [3]). Theorem 2.5. Let K be a class of compact Hausdorff spaces and let Z be a compact Hausdorff space. Then: (i) if K satisfies (a)–(b), then V(Z) embeds into V(Ku ) as a topological vector subspace if and only if Z ∈ K; (ii) if K satisfies (a)–(c), then L(Z) embeds into L(Ku ) as a locally convex vector subspace if and only if Z ∈ K. Proof. (i) Assume that V(Z) embeds into V(Ku ). By Corollary 2.1, Z is a subspace of spn (Ku ) for some n ∈ N. Therefore, by (a) and (b), Z belongs to K. The converse assertion follows from the universality of Ku and Proposition 2.3.
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(ii) Assume that L(Z) embeds into L(Ku ). Then Z = n∈N Z ∩ spn (Ku ) . Hence, by (a)-(c), Z ∈ K. Conversely, assume that Z ∈ K. Then, by the universality of Ku , Z embeds into Ku . Thus L(Z) embeds into L(Ku ) by Lemma 3.3 in [5]. 2 Recalling that Q = IN is the Hilbert cube, Example 2.4 and Theorem 2.5 immediately imply Theorem 2.6. For a compact Hausdorff space Z the following assertions are equivalent: (i) V(Z) embeds into V(Q) as a topological vector space; (ii) L(Z) embeds into L(Q) as a locally convex space; (iii) Z is metrizable. In what follows we use the universality of the (compact) Cantor space G in the class of zero-dimensional compact metrizable spaces. Theorem 2.7. For a compact Hausdorff space Z the following assertions are equivalent: (i) V(Z) embeds into V(G) as a topological vector space; (ii) L(Z) embeds into L(G) as a locally convex space; (iii) Z is metrizable and zero-dimensional. In particular, V(I) does not embed into V(G) as a topological vector space, and L(I) does not embed into L(G) as a locally convex vector space. Proof. (i) ⇒ (iii) Assume that V(Z) embeds into V(G). By Corollary 2.1, there is n ∈ N such that Z ⊆ spn (G). Since spn (G) is a continuous image of the metrizable compact space [−n, n]n × Gn , we obtain that Z is metrizable. Let us show that Z is zero-dimensional. For every n ∈ N, set sp∗n (G) := λ1 x1 + · · · + λn xn : λ1 , . . . , λn ∈ R, x1 , . . . , xn ∈ G = {0} ∪
n
λ1 x1 + · · · + λk xk : 0 ≤ x1 < · · · < xk ≤ 1, xi ∈ G, λi ∈ R \ {0} ∀1 ≤ i ≤ k .
k=1
For every 1 ≤ k ≤ n and each natural number m > k, set Kmk
1 2 k k := x = (x1 , . . . , xk ) ∈ G : 0 ≤ x1 , x1 − 2 ≤ x2 − 2 ≤ · · · ≤ xk − 2 . m m m
Then (the sets Cmk are defined in (2.1)) sp∗n (G) = {0} ∪
n
∞
πk Cmk × Kmk ,
(2.2)
k=1 m=k+1
and the restriction πmk of the canonical map πk onto Cmk × Kmk is one-to-one and hence πmk is a homeomorphism. For every 1 ≤ k ≤ n and m > k, set Zmk := Z ∩ πk Cmk × Kmk and Tmk := πk−1 (Zmk ). So Tmk is a compact subset of Cmk × Kmk homeomorphic to Zmk . Note that Zmk is linearly independent since Z is a vector basis for V(Z). Denote by pk the projection of Cmk × Kmk onto Kmk .
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Claim. We show that |Tmk ∩ p−1 k (x)| ≤ k for every x = (x1 , . . . , xk ) ∈ Kmk . Indeed, supposing the converse we find (x1 , . . . , xk ) ∈ Kmk and (λi1 , . . . , λik ) ∈ Cmk , 1 ≤ i ≤ k + 1, such that vi := λi1 x1 + · · · + λik xk ∈ Zmk ,
1 ≤ i ≤ k + 1.
Therefore the vectors v1 , . . . , vk+1 from Zmk are dependent which is impossible. Since pk is closed, the claim and the Hurewicz theorem [2, Problem 1.12.A] imply that Tmk and Zmk are zero-dimensional. Now (2.2) and Theorem 1.3.1 of [2] imply that the metrizable compact space ∞ Z = {0} ∩ Z ∪
∞
Zmk ,
where Zmk := Z ∩ πk Cmk × Kmk ,
(2.3)
k=1 m=k+1
is zero-dimensional as well. (ii) ⇒ (iii) Assume that L(Z) embeds into L(G). Note that L(G) = n∈N spn (G) and Z has representation (2.3). As we showed above, all the Zmk in (2.3) are zero-dimensional and metrizable. So the compact set Z is the union of a sequence of metrizable compact spaces. Therefore Z has countable network, and hence Z is metrizable by Theorem 3.1.19 of [3]. Now Theorem 1.3.1 of [2] implies that Z is also zero-dimensional. (iii) ⇒ (i), (ii) Assume that Z is a zero-dimensional compact metrizable space. By the universal property of G, the space Z embeds into G. Thus L(Z) embeds into L(G) as a locally convex space by Lemma 3.3 in [5], and V(Z) embeds into V(G) as a topological vector space by Proposition 2.3. 2 We complement Theorem 2.7 by the following assertion using an analogous idea to that in the proof of Theorem 2.7. Proposition 2.8. Let Z be a compact Hausdorff space and let K be a compact countable metrizable space. (i) If V(Z) embeds into V(K) as a topological vector space, then Z is countable and metrizable; (ii) if L(Z) embeds into L(K) as a locally convex space, then Z is countable and metrizable. Proof. Since K is countable it is zero-dimensional. Therefore we can consider K as a closed subspace of the Cantor space G. By Proposition 2.3, V(K) (L(Z)) embeds into V(G) (respectively, L(G)), and therefore V(Z) (L(Z)) embeds into V(G) (respectively, L(G)) as well. Thus Z is metrizable by Theorem 2.7. To prove that Z is countable we proceed as in the proof of Theorem 2.7. For every n ∈ N, set sp∗n (K) := λ1 x1 + · · · + λn xn : λ1 , . . . , λn ∈ R, x1 , . . . , xn ∈ K = {0} ∪
n
λ1 x1 + · · · + λk xk : 0 ≤ x1 < · · · < xk ≤ 1, xi ∈ K, λi ∈ R \ {0} ∀1 ≤ i ≤ k .
k=1
For every 1 ≤ k ≤ n and each natural number m > k, set Kmk :=
1 2 k x = (x1 , . . . , xk ) ∈ K : 0 ≤ x1 , x1 − 2 ≤ x2 − 2 ≤ · · · ≤ xk − 2 m m m k
so Kmk is a countable compact set. Then (the sets Cmk are defined in (2.1))
,
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sp∗n (K) = {0} ∪
n
∞
πk Cmk × Kmk ,
(2.4)
k=1 m=k+1
and the restriction πmk of the canonical map πk onto Cmk × Kmk is one-to-one and hence πmk is a homeomorphism. For every 1 ≤ k ≤ n and m > k, set Zmk := Z ∩ πk Cmk × Kmk and Tmk := πk−1 (Zmk ). So Tmk is a compact subset of Cmk × Kmk homeomorphic to Zmk . Denote by pk the projection of Cmk × Kmk onto Kmk . As we proved in the claim of Theorem 2.7, |Tmk ∩ p−1 k (x)| ≤ k for every x = (x1 , . . . , xk ) ∈ Kmk . Since all the Kmk are countable, we obtain that all the Tmk are countable, and hence Z ⊆ {0} ∪
n
∞
πk Tmk
k=1 m=k+1
is also countable. 2 3. Embedding of V(X) into V(I) First we prove the following assertion. Proposition 3.1. For every n ∈ N, the subset spn (I) of V(I) and L(I) is compact metrizable of dimension 2n, that is, dim spn (I) = 2n. Proof. For every n ∈ N, we have spn (I) = πn [−n, n]n × In and hence spn (I) is compact and metrizable. For every n ∈ N, set sp∗n (I) := λ1 x1 + · · · + λn xn : λ1 , . . . , λn ∈ R, x1 , . . . , xn ∈ I = {0} ∪
n
λ1 x1 + · · · + λk xk : 0 ≤ x1 < · · · < xk ≤ 1, λ1 , . . . , λn ∈ R \ {0} .
k=1
For every 1 ≤ k ≤ n and each natural number m > k, set Kmk
1 2 k k := x = (x1 , . . . , xk ) ∈ I : 0 ≤ x1 , x1 − 2 ≤ x2 − 2 ≤ · · · ≤ xk − 2 . m m m
Then (the sets Cmk are defined in (2.1)) sp∗n (I) = {0} ∪
n
∞
πk Cmk × Kmk ,
(3.1)
k=1 m=k+1
and the restriction πmk of the canonical map πk onto Cmk × Kmk is one-to-one and hence πmk is a homeomorphism. As subspaces of R2k with nonempty interior the compact sets Cmk × Kmk have dimension 2k by Theorem 3.1.4 of [2], and hence all the πk Cmk × Kmk have dimension 2k. Now Theorems 3.1.4 and 3.1.8 of [2] and (3.1) show that sp∗n (I) has dimension 2n. Thus the compact subspace spn (I) of sp∗n (I) has dimension ≤ 2n. As 2spn (I) is homeomorphic to spn (I) and contains πn Cn+1,n × Kn+1,n , the dimension of spn (I) is bigger than 2n. Thus the dimension of spn (I) is equal to 2n. 2 Recall that a normal space X is said to be countable dimensional if it is a countable union of normal finite dimensional subspaces; otherwise, X is called uncountable dimensional, see Definition 5.1.1 of [4]. The Hilbert cube Q is an example of an uncountable dimensional space.
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Corollary 3.2. The Hilbert cube Q does not embed into V(I). Indeed, if X is any normal space contained in V(I), then X is countable dimensional. Proof. By Theorem 3.1 of [6], V(I) is a kω -space and V(I) = n∈N spn I is a kω -decomposition of V(I). So X = n∈N X ∩ spn (I) . Since for every n ∈ N, the space X ∩ spn (I) is metrizable and has dimension ≤ 2n by Proposition 3.1, the space X is countable dimensional. 2 Theorem 3.3. Let K be a compact Hausdorff space. If V(K) embeds into V(I) as a topological vector space, then K is finite-dimensional and metrizable. Proof. Assume that V(K) embeds into V(I). By Corollary 2.1, there is an n ∈ N such that K ⊆ spn (I). Since spn (I) is a continuous image of the metrizable compact space [−n, n]n × In , we obtain that K is metrizable, and K is finite-dimensional by Proposition 3.1. 2 Corollary 3.4. Let X be a Tychonoff space such that V(X) embeds into V(I) as a topological vector space. Then every compact subspace K of X is metrizable and finite-dimensional. In particular, if X = n∈N Kn is a kω -space, then all the Kn are finite-dimensional metrizable compact spaces. Proof. By the assumption and Proposition 2.3, V(K) embeds into V(I) and Theorem 3.3 applies. 2 We do not know whether the converse assertion in Theorem 3.3 is true. Open question. Let K be a finite-dimensional metrizable compact space. Does V(K) embed into V(I) as a topological vector space? In [11] it is proved that if for a Tychonoff space X the free locally convex space L(X) embeds into L(I) as a locally convex subspace, then X is a metrizable countable-dimensional compact. In particular, L(R) does not embed into L(I) as a locally convex subspace. But for free topological vector spaces the situation different as the following theorem shows. Theorem 3.5. V(R) embeds into V(I) as a topological vector space. Proof. We prove the theorem in four steps. Step 1. The basic construction. Take two sequences {ak }k∈Z , {bk }k∈Z ⊂ I such that 0 < a0 < b0 < a1 < b1 < a−1 < b−1 < a2 < b2 < a−2 < b−2 < · · · < 1, and set Ik = [ak , bk ] for every k ∈ Z. For k = 0, define the homeomorphism g0,0 : [0, 1] → I0 by g0,0 (x) := a0 + (b0 − a0 )x. For every k ∈ Z \ {0}, set Sk := 8(T1 + · · · + T|k| ) and Ak :=
1 Sk , 2
where Tn := 1 + · · · + n.
For every k ∈ Z \ {0} and i ∈ N such that 1 ≤ i ≤ Ak , we define closed pairwise disjoint intervals by Ii,k
bk − ak bk − ak := ak + (2i − 1), ak + 2i ⊂ Ik , Sk Sk
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and define the homeomorphisms gi,k : [k, k + 1] → Ii,k by gi,k (x) := ak +
bk − ak 2i − 1 + (x − k) . Sk
For every k ∈ Z, define the maps Hk : [k, k + 1] → V(I) by H0 (x) := g0,0 (x),
if k = 0, and
Hk (x) := g1,k (x) + g2,k (x) + · · · + gAk ,k (x), if k = 0,
(3.2)
where “+” denotes the vector space addition in V(I). Now we define the map χ : R → V(I) inductively as follows: if x ∈ [0, 1], set χ(x) := H0 x ; if k ∈ N and x ∈ [k, k + 1], put k Hi (i) − Hi−1 (i) , χ(x) := Hk x − Hk k + χ k = Hk x − i=1
and if −k ∈ N and x ∈ [k, k + 1], we set −1 χ(x) := Hk x − Hk k + 1 + χ k + 1 = Hk x − Hi (i + 1) − Hi+1 (i + 1) . i=k
Clearly, χ is well-defined and continuous. Since all the intervals Ii,k are disjoint and the functions gi,k are bijective, the map χ is one-to-one. For every n ∈ N, set Yn := χ([−n, n]) and put Y := χ(R). Step 2. For every s ∈ N there is M (s) ∈ N such that sp(Y ) ∩ sps (I) ⊆ spM (s) (YM (s) ). Indeed, fix t ∈ sp(Y ) ∩ sps (I). So there are distinct x1 , . . . , xs ∈ I, distinct y1 , . . . , ym ∈ Y , nonzero real numbers a1 , . . . , am and nonzero numbers λ1 , . . . , λs ∈ [−s, s] such that t = a1 y1 + · · · + am ym = λ1 x1 + · · · + λs xs .
(3.3)
By construction, there are r ∈ N, integer numbers n1 < · · · < nr , natural numbers q1 , . . . , qr with q1 + · · · + qr = m, and pairwise distinct elements zj,i ∈ R, where 1 ≤ j ≤ qi for every 1 ≤ i ≤ r, such that ⎧ ⎪ [0, 1], if ni = 0, ⎪ ⎨ (1) z1,i , . . . , zqi ,i ∈ (ni , ni + 1], if ni > 0, ⎪ ⎪ ⎩ [ni , ni + 1), if ni < 0, (2) for every y ∈ {y1 , . . . , ym } there is unique (j, i) such that y = χ(zj,i ). So we can uniquely represent t in the form t=
qi r
aj,i χ(zj,i ) = λ1 x1 + · · · + λs xs .
(3.4)
i=1 j=1
Since all the intervals Ii,k are disjoint and the functions gi,k are bijective, the construction of the map χ and (3.4) imply the following: if zj,i ∈ (ni , ni + 1), then aj,i ∈ {λ1 , . . . , λs } and χ(zj,i ) has at least Ani
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distinct summands from the basis I of V(I) which do not appear in another summand in the middle sum of (3.4). Therefore (q1 − 1) + · · · + (qr − 1) ≤ s.
(3.5)
Assume that nr > 0, then χ(zqr ,r ) has at least Anr distinct basic elements from I which do not appear in other summands in the middle sum of (3.4). Therefore, (3.4) implies that Anr ≤ s and |aqr ,r | ≤ s. If n1 < 0, the same argument shows that An1 ≤ s and |aq1 ,1 | ≤ s. Since Ak ≥ 4|k|, this implies in particular that r ≤ s, and therefore (3.5) yields m = q1 + · · · + qr ≤ 2s. Now if nr > 0, let c ∈ N be the least index such that nc ≥ 0. By the definition of χ and (3.4), for every qr qr 1 ≤ i ≤ nr the coefficient of Hi (i) in the sum j=1 aj,r χ(zj,r ) is − j=1 aj,r , and hence qr aj,r ≤ qr · s ≤ 2s · s. j=1
(3.6)
Therefore, if c < r and znr−1 ,r−1 = nr−1 , by (3.4), the coefficient anr−1 ,r−1 satisfies an
r−1 ,r−1
≤ s + qr · s = (qr + 1) · s ≤ 2s2 .
(3.7)
Analogously, assume that c < r − 1 and znr−2 ,r−2 = nr−2 . Then the coefficient of Hnr−2 (nr−2 ) in the middle sum of (3.4) is qr−1 −1
anr−2 ,r−2 −
j=1
aj,r−1 − anr−1 ,r−1 −
qr
aj,r .
j=1
This and (3.4)–(3.7) imply an
r−2 ,r−2
≤ s + (qr−1 − 1) · s + (qr + 1) · s + qr · s = s(2qr + qr−1 + 1) < s · 4s.
Continuing this process we obtain that there is M+ > 0 such that |ani ,i | ≤ M+ for every i > c. In a similar way one can show that if n1 < 0, then there is M− > 0 such that |ani ,i | ≤ M− for every i such that ni > 0. Now, if nc = 0, the boundedness of aj,i corresponding to i = c and (3.4) easily imply that there exists an M such that |aj,i | ≤ M,
for every 1 ≤ i ≤ r and 1 ≤ j ≤ qi .
Then M (s) := max{2s, M } is as desired. Step 3. Let us show that if a1 y1 + · · · + am ym = 0, then a1 = · · · = am = 0. Indeed, we can represent 0 in the form (3.4). Now, as above, if nr > 0, then χ(zqr ,r ) has at least Anr distinct basic elements from I which do not appear in other summands in the middle sum of (3.4). So aqr ,r = 0. Analogously, if n1 < 0 we obtain that aq1 ,1 = 0. Therefore r = 1 and n1 = 0. In this case we also easily obtain that aj,1 = 0 for every 1 ≤ j ≤ q1 . Thus a1 = · · · = am = 0 as desired. Step 4. We claim that Y is a closed subset of V(I). First we show that Y ∩ sps (I) = Ys ∩ sps (I) for every s ∈ N. Indeed, let y := χ(x) = λ1 x1 + · · · + λs xs ∈ Y ∩ sps (I).
(3.8)
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If x ∈ [0, 1], then y ∈ Y1 and we are done. Suppose that x ∈ (k, k + 1] for some k ≥ 0, or x ∈ [k, k + 1) for some k < 0. If y ∈ / Ys , then either k ≥ s or k + 1 ≤ −s. In both cases y has at least Ak ≥ 4|k| > s distinct basic summands from I which contradicts (3.8). Hence y ∈ Ys . Thus Y ∩ sps (I) ⊆ Ys ∩ sps (I). The converse inclusion is clear. Fix now a closed subset F of R. Then, for every s ∈ N, we have χ(F ) ∩ sps (I) = χ(F ) ∩ Y ∩ sps (I) = χ(F ) ∩ Ys ∩ sps (I) = χ(F ) ∩ χ([−s, s]) ∩ sps (I) = χ F ∩ [−s, s] ∩ sps (I), and hence χ(F ) ∩ sps (I) is a closed subset of sps (I). Since V(I) = s∈N sps (I) is a kω -space by Theorem 3.1 of [6], we obtain that χ(F ) is closed in V(I). Therefore χ is a closed map. Thus χ is a homeomorphism of R onto Y . Finally, Steps 2–4 show that we can apply Proposition 2.2 to get that V(R) is linearly isomorphic to the closed linear subspace sp(Y ) of V(I). 2 Corollary 3.6 ([8]). A(R) embeds into A(I) as a topological group. Proof. The definition of the map χ in Step 1 of the proof of Theorem 3.5 (see also (3.2)) shows that A(R) is an algebraic subgroup of A(I). As A(X) is a closed subgroup of V(X) for every Tychonoff space X by Proposition 5.1 of [6], Theorem 3.5 implies that A(R) embeds into A(I) as a topological group. 2 We end the paper with the following result. In [10] it was shown that A(Sk ) embeds into A(Ik ), where Sk is the k-dimensional unit sphere. The proof of this fact is rather complicated. Essentially using the vector space properties below we give an explicit direct and simple construction of an embedding of V(Sk ) into V(Ik ). Proposition 3.7. For every k ∈ N, the space V(Sk ) embeds into V(Ik ) as a closed vector subspace. Proof. For a Tychonoff space X and x ∈ X, denote by [x] the element x considered as an element of the Hamel basis X of V(X). It will be more convenient to work with the closed symmetric unit interval I∗ := [−1, 1] than with I. Denote by Bk the closed unit ball in Rk and let x be the standard norm of x ∈ Bk . Clearly, Bk ⊆ Ik∗ and Sk−1 is the boundary of Bk . Set 3 := (3, . . . , 3), 8 := (8, . . . , 8) ∈ Rk and define injective continuous functions a(x), b(x) : Bk → V(Ik∗ ) as follows a(x) :=
x x 3+x 8+x + 1 − x and b(x) := + 1 − x . 10 10 10 10
It is easy to see that a(x) = b(y) if and only if x = y ∈ Sk−1 . Set K := A ∪ B, where A := {a(x) : x ∈ Bk } and B := {b(x) : x ∈ Bk }. By the above we obtain that K is homeomorphic to Sk . We shall show that the vector subspace sp(K) of V(Ik∗ ) generated by K is V(Sk ). By Proposition 2.2 it suffices to show that (i) K is a free basis for sp(K); (ii) for every n ∈ N there is a natural number m such that sp(K) ∩ spn (Ik∗ ) ⊆ spm (K). Let x1 , . . . , xl be distinct elements of Sk−1 , y1 , . . . , ys be distinct elements of Bk \ Sk−1 , z1 , . . . , zq be distinct elements of Bk \ Sk−1 and let λx1 , . . . , λxl , λy1 , . . . , λys , λz1 , . . . , λzq ∈ R \ {0}.
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Set T :=
l i=1
λxi a(xi )
+
s j=1
λyj a(yj )
+
q
λzt b(zt ).
t=1
By the construction of a(x) and b(x), it is easy to see that the number of nonzero basic element from Ik∗ in the decomposition of T is bigger or equal than l + s + q. The same holds also in the cases s = 0 and q = 0. So, (1) T = 0 that proves (i), and (2) sp(K) ∩ spn (Ik∗ ) ⊆ spn (K) that proves (ii). 2 Acknowledgements The second author thanks Ben Gurion University of the Negev for hospitality. References [1] R. Brown, J.P.L. Hardy, Subgroups of free topological groups and free topological products of topological groups, J. Lond. Math. Soc. 10 (1975) 431–440. [2] R. Engelking, Dimension Theory, PWN, Warszawa, 1978. [3] R. Engelking, General Topology, Heldermann Verlag, Berlin, 1989. [4] R. Engelking, Theory of Dimensions. Finite and Infinite, Heldermann Verlag, 1995. [5] S. Gabriyelyan, The k-space property for free locally convex spaces, Can. Math. Bull. 57 (2014) 803–809. [6] S.S. Gabriyelyan, S.A. Morris, Free topological vector spaces, Topol. Appl. 223 (2017) 30–49. [7] E. Katz, S.A. Morris, Free abelian topological groups on countable CW -complexes, Bull. Aust. Math. Soc. 41 (1990) 451–456. [8] E. Katz, S.A. Morris, P. Nickolas, A free subgroup of the free abelian topological group on the unit interval, Bull. Lond. Math. Soc. 14 (1982) 399–402. [9] E. Katz, S.A. Morris, P. Nickolas, Characterization of bases of subgroups of free topological groups, J. Lond. Math. Soc. (2) 27 (1983) 421–426. [10] E. Katz, S.A. Morris, P. Nickolas, Free abelian topological groups on spheres, Quart. J. Math. Oxford 35 (1984) 173–181. [11] A. Leiderman, S.A. Morris, V. Pestov, The free abelian topological group and the free locally convex space on the unit interval, J. Lond. Math. Soc. 56 (1997) 529–538. [12] P. Nickolas, Subgroups of the free topological group on [0, 1], J. Lond. Math. Soc. 12 (1976) 199–205. [13] V.G. Pestov, Some properties of free topological groups, Mosc. Univ. Math. Bull. 37 (1982) 46–49. [14] V.V. Uspenski˘ı, Free topological groups of metrizable spaces, Math. USSR, Izv. 37 (1991) 657–680.
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
Embedding into free topological vector spaces on compact metrizable spaces Saak S. Gabriyelyan a,∗ , Sidney A. Morris b,c , a
Department of Mathematics, Ben-Gurion University of the Negev, P.O. 653, Beer-Sheva, Israel Faculty of Science and Technology, Federation University Australia, P.O. Box 663, Ballarat, Victoria, 3353, Australia c Department of Mathematics and Statistics, La Trobe University, Melbourne, Victoria, 3086, Australia b
a r t i c l e
i n f o
Article history: Received 9 April 2017 Received in revised form 12 September 2017 Accepted 13 September 2017 Available online 14 September 2017 Dedicated to the memory of W. Wistar Comfort MSC: 46A03 54A25 54D50
a b s t r a c t For a Tychonoff space X, let V(X) be the free topological vector space over X. Denote by I, G, Q and Sk the closed unit interval, the Cantor space, the Hilbert cube Q = IN and the k-dimensional unit sphere for k ∈ N, respectively. The main result is that V(R) can be embedded as a topological vector space in V(I). It is also shown that for a compact Hausdorff space K: (1) V(K) can be embedded in V(G) if and only if K is zero-dimensional and metrizable; (2) V(K) can be embedded in V(Q) if and only if K is metrizable; (3) V(Sk ) can be embedded in V(Ik ); (4) V(K) can be embedded in V(I) implies that K is finite-dimensional and metrizable. © 2017 Elsevier B.V. All rights reserved.
Keywords: Free topological vector space Free locally convex space Embedding Finite-dimensional Zero-dimensional Compact Cantor space Hilbert cube
1. Introduction In [6] we introduced the study of free topological vector spaces and showed that, in some ways, the topological structure of free topological vector spaces is nicer and better understood than that of free locally convex spaces; in particular, the free topological vector space on a kω -space is a kω -space and
* Corresponding author. E-mail addresses: [email protected] (S.S. Gabriyelyan), [email protected] (S.A. Morris). http://dx.doi.org/10.1016/j.topol.2017.09.008 0166-8641/© 2017 Elsevier B.V. All rights reserved.
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therefore complete. By contrast the free locally convex space on an infinite space X is complete if and only if X is Dieudonné complete and has no infinite compact subsets, see [14]. For many years, [1,7,8,10,12–14], the following question has been investigated: for what Tychonoff spaces X and Y can the free (free abelian) topological group F (Y ) (respectively, A(Y )) on Y be embedded as a topological subgroup of the free (respectively, free abelian) topological group F (X) (respectively, A(X)) on X? An analogous question for free locally convex spaces L(X) and L(Y ) over X and Y respectively was considered in [11]. We note that, in general, the question for free abelian topological groups is harder than that for free topological groups and the question for free locally convex spaces is harder again. In this paper we investigate the following: Question. For what Tychonoff spaces X and Y can the free topological vector space V(Y ) be embedded as a topological vector subspace of the free topological vector space V(X)? The free topological vector space V(X) over a Tychonoff space X is a pair consisting of a topological vector space V(X) and a continuous map i = iX : X → V(X) such that every continuous map f from X to a topological vector space E gives rise to a unique continuous linear operator f¯ : V(X) → E with f = f¯ ◦ i. Theorem 2.3 of [6] shows that for all Tychonoff spaces X, V(X) exists, is unique up to isomorphism of topological vector spaces, is Hausdorff and the mapping i is a homeomorphism of the topological space X onto its image in V(X). For a subspace Y of X, let V(Y, X) be the vector subspace of V(X) spanned by Y . Concerning the question raised above, Proposition 3.12 of [6] shows that, if K is a compact subspace of the Tychonoff space X, then the vector subspace V(K, X) of V(X) is V(K). Further, Corollary 3.11 of [6] says that if Y is a closed subspace of the kω -space X, then the vector subspace V(Y, X) of V(X) is V(Y ). It is known [9] that the free topological group F (R) on R can be embedded as a topological group in the free topological group F (I) on the unit interval I. It is also known [8] that the free abelian topological group A(R) on R can be embedded as a topological group in the free abelian topological group A(I) on the unit interval I, but this is significantly harder to prove. We hasten to point out that while R can be embedded as a subspace of I, no subgroup of F (I) or A(I) generated by a subspace Y of I is isomorphic as a topological group to F (R) or A(R). It is shown in Theorem 4.3 of [11] that the free locally convex space L(R) on R cannot be embedded as a topological vector subspace of the free locally convex space L(I) on I. The main theorem in this paper is that the free topological vector space V(R) on R can indeed be embedded as a topological vector space in the free topological vector space V(I) on I. It is also shown that for a compact Hausdorff space K: (1) V(K) can be embedded in V(G), where G is the Cantor space, if and only if K is zero-dimensional and metrizable; (2) V(K) can be embedded in V(Q), where Q denotes the Hilbert cube Q = IN , if and only if K is metrizable; (3) V(Sk ) can be embedded in V (Ik ), where Sk denotes the k-dimensional sphere, k ∈ N; (4) V(K) can be embedded in V(I) implies that K is finite-dimensional and metrizable. 2. Universal spaces In this section we study free topological vector spaces generated by the universal spaces: (1) the Hilbert cube Q = IN ; it is well-known that every metrizable compact space embeds into Q; (2) the Cantor compact space G; it is known that every zero-dimensional compact metrizable space embeds into G, see Example 3.1.28 and Theorem 6.2.16 of [3]. In so doing, we obtain new results not only for free topological vector spaces, but also for free locally convex spaces on these universal spaces. In what follows we use the following notation. Set N := {1, 2, . . . }. For a subset A of a vector space E and a natural number n ∈ N we denote by spn (A) the following subset of E
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spn (A) := {λ1 x1 + · · · + λn xn : λi ∈ [−n, n], xi ∈ A, ∀i = 1, . . . , n}, and set sp(A) := by
n∈N
spn (A), the span of A in E; we also define the continuous map πn : [−n, n]n ×An → E πn (λ1 , . . . , λn ) × (x1 , . . . , xn ) := λ1 x1 + · · · + λn xn ,
where “+” denotes the vector space addition in E, so πn [−n, n]n × An = spn (A). For every k, m ∈ N define Cmk := {(λ1 , . . . , λk ) ∈ Rk : 1/m ≤ |λi | ≤ m ∀i = 1, . . . , k}.
(2.1)
For a subspace Y of a Tychonoff space X, let V(Y, X) be the vector subspace of V(X) generated algebraically by Y . In what follows we shall essentially use the following results from [6], see Corollary 3.4, Proposition 3.9 and Proposition 3.12, respectively. Corollary 2.1 ([6]). Let X be a Tychonoff space. Then for every compact subset K of V(X) there is an n ∈ N such that K ⊆ spn (X). Proposition 2.2 ([6]). Let X = n∈N Cn be a kω -space and let Y be a subset of V(X) such that Y is a free vector space basis for the subspace, sp(Y ), that it generates. Assume that K1 , K2 , . . . is a sequence of compact subsets of Y such that Y = n∈N Kn is a kω -decomposition of Y inducing the same topology on Y that Y inherits as a subset of V(X). If for every n ∈ N there is a natural number m such that sp(Y ) ∩ spn (Cn ) ⊆ spm (Km ), then sp(Y ) is V(Y ), and both sp(Y ) and Y are closed subsets of V(X). Proposition 2.3 ([6]). If K is a compact subspace of a Tychonoff space X, then V(K, X) is V(K). To obtain a general assertion about embedding of free topological vector spaces over compact spaces, we consider the following three conditions on a class K of compact Hausdorff spaces: (a) K is closed under taking finite products, closed subspaces and Hausdorff quotients; (b) K contains I and has a universal (by inclusion) object, i.e., there exists Ku ∈ K such that every compact space C ∈ K embeds into K; (c) if a compact Hausdorff space K is the union of a countable family of compact spaces from K, then K ∈ K. A typical example of K satisfying (a)–(c) is the following. Example 2.4. Let Kκ be the class of all compact Hausdorff spaces of weight ≤ κ for an infinite cardinal κ. Then Kκ satisfies (a)–(c) with the universal object Iκ (see, Theorems 3.1.19 and 3.2.5 of [3]). Theorem 2.5. Let K be a class of compact Hausdorff spaces and let Z be a compact Hausdorff space. Then: (i) if K satisfies (a)–(b), then V(Z) embeds into V(Ku ) as a topological vector subspace if and only if Z ∈ K; (ii) if K satisfies (a)–(c), then L(Z) embeds into L(Ku ) as a locally convex vector subspace if and only if Z ∈ K. Proof. (i) Assume that V(Z) embeds into V(Ku ). By Corollary 2.1, Z is a subspace of spn (Ku ) for some n ∈ N. Therefore, by (a) and (b), Z belongs to K. The converse assertion follows from the universality of Ku and Proposition 2.3.
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(ii) Assume that L(Z) embeds into L(Ku ). Then Z = n∈N Z ∩ spn (Ku ) . Hence, by (a)-(c), Z ∈ K. Conversely, assume that Z ∈ K. Then, by the universality of Ku , Z embeds into Ku . Thus L(Z) embeds into L(Ku ) by Lemma 3.3 in [5]. 2 Recalling that Q = IN is the Hilbert cube, Example 2.4 and Theorem 2.5 immediately imply Theorem 2.6. For a compact Hausdorff space Z the following assertions are equivalent: (i) V(Z) embeds into V(Q) as a topological vector space; (ii) L(Z) embeds into L(Q) as a locally convex space; (iii) Z is metrizable. In what follows we use the universality of the (compact) Cantor space G in the class of zero-dimensional compact metrizable spaces. Theorem 2.7. For a compact Hausdorff space Z the following assertions are equivalent: (i) V(Z) embeds into V(G) as a topological vector space; (ii) L(Z) embeds into L(G) as a locally convex space; (iii) Z is metrizable and zero-dimensional. In particular, V(I) does not embed into V(G) as a topological vector space, and L(I) does not embed into L(G) as a locally convex vector space. Proof. (i) ⇒ (iii) Assume that V(Z) embeds into V(G). By Corollary 2.1, there is n ∈ N such that Z ⊆ spn (G). Since spn (G) is a continuous image of the metrizable compact space [−n, n]n × Gn , we obtain that Z is metrizable. Let us show that Z is zero-dimensional. For every n ∈ N, set sp∗n (G) := λ1 x1 + · · · + λn xn : λ1 , . . . , λn ∈ R, x1 , . . . , xn ∈ G = {0} ∪
n
λ1 x1 + · · · + λk xk : 0 ≤ x1 < · · · < xk ≤ 1, xi ∈ G, λi ∈ R \ {0} ∀1 ≤ i ≤ k .
k=1
For every 1 ≤ k ≤ n and each natural number m > k, set Kmk
1 2 k k := x = (x1 , . . . , xk ) ∈ G : 0 ≤ x1 , x1 − 2 ≤ x2 − 2 ≤ · · · ≤ xk − 2 . m m m
Then (the sets Cmk are defined in (2.1)) sp∗n (G) = {0} ∪
n
∞
πk Cmk × Kmk ,
(2.2)
k=1 m=k+1
and the restriction πmk of the canonical map πk onto Cmk × Kmk is one-to-one and hence πmk is a homeomorphism. For every 1 ≤ k ≤ n and m > k, set Zmk := Z ∩ πk Cmk × Kmk and Tmk := πk−1 (Zmk ). So Tmk is a compact subset of Cmk × Kmk homeomorphic to Zmk . Note that Zmk is linearly independent since Z is a vector basis for V(Z). Denote by pk the projection of Cmk × Kmk onto Kmk .
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Claim. We show that |Tmk ∩ p−1 k (x)| ≤ k for every x = (x1 , . . . , xk ) ∈ Kmk . Indeed, supposing the converse we find (x1 , . . . , xk ) ∈ Kmk and (λi1 , . . . , λik ) ∈ Cmk , 1 ≤ i ≤ k + 1, such that vi := λi1 x1 + · · · + λik xk ∈ Zmk ,
1 ≤ i ≤ k + 1.
Therefore the vectors v1 , . . . , vk+1 from Zmk are dependent which is impossible. Since pk is closed, the claim and the Hurewicz theorem [2, Problem 1.12.A] imply that Tmk and Zmk are zero-dimensional. Now (2.2) and Theorem 1.3.1 of [2] imply that the metrizable compact space ∞ Z = {0} ∩ Z ∪
∞
Zmk ,
where Zmk := Z ∩ πk Cmk × Kmk ,
(2.3)
k=1 m=k+1
is zero-dimensional as well. (ii) ⇒ (iii) Assume that L(Z) embeds into L(G). Note that L(G) = n∈N spn (G) and Z has representation (2.3). As we showed above, all the Zmk in (2.3) are zero-dimensional and metrizable. So the compact set Z is the union of a sequence of metrizable compact spaces. Therefore Z has countable network, and hence Z is metrizable by Theorem 3.1.19 of [3]. Now Theorem 1.3.1 of [2] implies that Z is also zero-dimensional. (iii) ⇒ (i), (ii) Assume that Z is a zero-dimensional compact metrizable space. By the universal property of G, the space Z embeds into G. Thus L(Z) embeds into L(G) as a locally convex space by Lemma 3.3 in [5], and V(Z) embeds into V(G) as a topological vector space by Proposition 2.3. 2 We complement Theorem 2.7 by the following assertion using an analogous idea to that in the proof of Theorem 2.7. Proposition 2.8. Let Z be a compact Hausdorff space and let K be a compact countable metrizable space. (i) If V(Z) embeds into V(K) as a topological vector space, then Z is countable and metrizable; (ii) if L(Z) embeds into L(K) as a locally convex space, then Z is countable and metrizable. Proof. Since K is countable it is zero-dimensional. Therefore we can consider K as a closed subspace of the Cantor space G. By Proposition 2.3, V(K) (L(Z)) embeds into V(G) (respectively, L(G)), and therefore V(Z) (L(Z)) embeds into V(G) (respectively, L(G)) as well. Thus Z is metrizable by Theorem 2.7. To prove that Z is countable we proceed as in the proof of Theorem 2.7. For every n ∈ N, set sp∗n (K) := λ1 x1 + · · · + λn xn : λ1 , . . . , λn ∈ R, x1 , . . . , xn ∈ K = {0} ∪
n
λ1 x1 + · · · + λk xk : 0 ≤ x1 < · · · < xk ≤ 1, xi ∈ K, λi ∈ R \ {0} ∀1 ≤ i ≤ k .
k=1
For every 1 ≤ k ≤ n and each natural number m > k, set Kmk :=
1 2 k x = (x1 , . . . , xk ) ∈ K : 0 ≤ x1 , x1 − 2 ≤ x2 − 2 ≤ · · · ≤ xk − 2 m m m k
so Kmk is a countable compact set. Then (the sets Cmk are defined in (2.1))
,
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sp∗n (K) = {0} ∪
n
∞
πk Cmk × Kmk ,
(2.4)
k=1 m=k+1
and the restriction πmk of the canonical map πk onto Cmk × Kmk is one-to-one and hence πmk is a homeomorphism. For every 1 ≤ k ≤ n and m > k, set Zmk := Z ∩ πk Cmk × Kmk and Tmk := πk−1 (Zmk ). So Tmk is a compact subset of Cmk × Kmk homeomorphic to Zmk . Denote by pk the projection of Cmk × Kmk onto Kmk . As we proved in the claim of Theorem 2.7, |Tmk ∩ p−1 k (x)| ≤ k for every x = (x1 , . . . , xk ) ∈ Kmk . Since all the Kmk are countable, we obtain that all the Tmk are countable, and hence Z ⊆ {0} ∪
n
∞
πk Tmk
k=1 m=k+1
is also countable. 2 3. Embedding of V(X) into V(I) First we prove the following assertion. Proposition 3.1. For every n ∈ N, the subset spn (I) of V(I) and L(I) is compact metrizable of dimension 2n, that is, dim spn (I) = 2n. Proof. For every n ∈ N, we have spn (I) = πn [−n, n]n × In and hence spn (I) is compact and metrizable. For every n ∈ N, set sp∗n (I) := λ1 x1 + · · · + λn xn : λ1 , . . . , λn ∈ R, x1 , . . . , xn ∈ I = {0} ∪
n
λ1 x1 + · · · + λk xk : 0 ≤ x1 < · · · < xk ≤ 1, λ1 , . . . , λn ∈ R \ {0} .
k=1
For every 1 ≤ k ≤ n and each natural number m > k, set Kmk
1 2 k k := x = (x1 , . . . , xk ) ∈ I : 0 ≤ x1 , x1 − 2 ≤ x2 − 2 ≤ · · · ≤ xk − 2 . m m m
Then (the sets Cmk are defined in (2.1)) sp∗n (I) = {0} ∪
n
∞
πk Cmk × Kmk ,
(3.1)
k=1 m=k+1
and the restriction πmk of the canonical map πk onto Cmk × Kmk is one-to-one and hence πmk is a homeomorphism. As subspaces of R2k with nonempty interior the compact sets Cmk × Kmk have dimension 2k by Theorem 3.1.4 of [2], and hence all the πk Cmk × Kmk have dimension 2k. Now Theorems 3.1.4 and 3.1.8 of [2] and (3.1) show that sp∗n (I) has dimension 2n. Thus the compact subspace spn (I) of sp∗n (I) has dimension ≤ 2n. As 2spn (I) is homeomorphic to spn (I) and contains πn Cn+1,n × Kn+1,n , the dimension of spn (I) is bigger than 2n. Thus the dimension of spn (I) is equal to 2n. 2 Recall that a normal space X is said to be countable dimensional if it is a countable union of normal finite dimensional subspaces; otherwise, X is called uncountable dimensional, see Definition 5.1.1 of [4]. The Hilbert cube Q is an example of an uncountable dimensional space.
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Corollary 3.2. The Hilbert cube Q does not embed into V(I). Indeed, if X is any normal space contained in V(I), then X is countable dimensional. Proof. By Theorem 3.1 of [6], V(I) is a kω -space and V(I) = n∈N spn I is a kω -decomposition of V(I). So X = n∈N X ∩ spn (I) . Since for every n ∈ N, the space X ∩ spn (I) is metrizable and has dimension ≤ 2n by Proposition 3.1, the space X is countable dimensional. 2 Theorem 3.3. Let K be a compact Hausdorff space. If V(K) embeds into V(I) as a topological vector space, then K is finite-dimensional and metrizable. Proof. Assume that V(K) embeds into V(I). By Corollary 2.1, there is an n ∈ N such that K ⊆ spn (I). Since spn (I) is a continuous image of the metrizable compact space [−n, n]n × In , we obtain that K is metrizable, and K is finite-dimensional by Proposition 3.1. 2 Corollary 3.4. Let X be a Tychonoff space such that V(X) embeds into V(I) as a topological vector space. Then every compact subspace K of X is metrizable and finite-dimensional. In particular, if X = n∈N Kn is a kω -space, then all the Kn are finite-dimensional metrizable compact spaces. Proof. By the assumption and Proposition 2.3, V(K) embeds into V(I) and Theorem 3.3 applies. 2 We do not know whether the converse assertion in Theorem 3.3 is true. Open question. Let K be a finite-dimensional metrizable compact space. Does V(K) embed into V(I) as a topological vector space? In [11] it is proved that if for a Tychonoff space X the free locally convex space L(X) embeds into L(I) as a locally convex subspace, then X is a metrizable countable-dimensional compact. In particular, L(R) does not embed into L(I) as a locally convex subspace. But for free topological vector spaces the situation different as the following theorem shows. Theorem 3.5. V(R) embeds into V(I) as a topological vector space. Proof. We prove the theorem in four steps. Step 1. The basic construction. Take two sequences {ak }k∈Z , {bk }k∈Z ⊂ I such that 0 < a0 < b0 < a1 < b1 < a−1 < b−1 < a2 < b2 < a−2 < b−2 < · · · < 1, and set Ik = [ak , bk ] for every k ∈ Z. For k = 0, define the homeomorphism g0,0 : [0, 1] → I0 by g0,0 (x) := a0 + (b0 − a0 )x. For every k ∈ Z \ {0}, set Sk := 8(T1 + · · · + T|k| ) and Ak :=
1 Sk , 2
where Tn := 1 + · · · + n.
For every k ∈ Z \ {0} and i ∈ N such that 1 ≤ i ≤ Ak , we define closed pairwise disjoint intervals by Ii,k
bk − ak bk − ak := ak + (2i − 1), ak + 2i ⊂ Ik , Sk Sk
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and define the homeomorphisms gi,k : [k, k + 1] → Ii,k by gi,k (x) := ak +
bk − ak 2i − 1 + (x − k) . Sk
For every k ∈ Z, define the maps Hk : [k, k + 1] → V(I) by H0 (x) := g0,0 (x),
if k = 0, and
Hk (x) := g1,k (x) + g2,k (x) + · · · + gAk ,k (x), if k = 0,
(3.2)
where “+” denotes the vector space addition in V(I). Now we define the map χ : R → V(I) inductively as follows: if x ∈ [0, 1], set χ(x) := H0 x ; if k ∈ N and x ∈ [k, k + 1], put k Hi (i) − Hi−1 (i) , χ(x) := Hk x − Hk k + χ k = Hk x − i=1
and if −k ∈ N and x ∈ [k, k + 1], we set −1 χ(x) := Hk x − Hk k + 1 + χ k + 1 = Hk x − Hi (i + 1) − Hi+1 (i + 1) . i=k
Clearly, χ is well-defined and continuous. Since all the intervals Ii,k are disjoint and the functions gi,k are bijective, the map χ is one-to-one. For every n ∈ N, set Yn := χ([−n, n]) and put Y := χ(R). Step 2. For every s ∈ N there is M (s) ∈ N such that sp(Y ) ∩ sps (I) ⊆ spM (s) (YM (s) ). Indeed, fix t ∈ sp(Y ) ∩ sps (I). So there are distinct x1 , . . . , xs ∈ I, distinct y1 , . . . , ym ∈ Y , nonzero real numbers a1 , . . . , am and nonzero numbers λ1 , . . . , λs ∈ [−s, s] such that t = a1 y1 + · · · + am ym = λ1 x1 + · · · + λs xs .
(3.3)
By construction, there are r ∈ N, integer numbers n1 < · · · < nr , natural numbers q1 , . . . , qr with q1 + · · · + qr = m, and pairwise distinct elements zj,i ∈ R, where 1 ≤ j ≤ qi for every 1 ≤ i ≤ r, such that ⎧ ⎪ [0, 1], if ni = 0, ⎪ ⎨ (1) z1,i , . . . , zqi ,i ∈ (ni , ni + 1], if ni > 0, ⎪ ⎪ ⎩ [ni , ni + 1), if ni < 0, (2) for every y ∈ {y1 , . . . , ym } there is unique (j, i) such that y = χ(zj,i ). So we can uniquely represent t in the form t=
qi r
aj,i χ(zj,i ) = λ1 x1 + · · · + λs xs .
(3.4)
i=1 j=1
Since all the intervals Ii,k are disjoint and the functions gi,k are bijective, the construction of the map χ and (3.4) imply the following: if zj,i ∈ (ni , ni + 1), then aj,i ∈ {λ1 , . . . , λs } and χ(zj,i ) has at least Ani
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distinct summands from the basis I of V(I) which do not appear in another summand in the middle sum of (3.4). Therefore (q1 − 1) + · · · + (qr − 1) ≤ s.
(3.5)
Assume that nr > 0, then χ(zqr ,r ) has at least Anr distinct basic elements from I which do not appear in other summands in the middle sum of (3.4). Therefore, (3.4) implies that Anr ≤ s and |aqr ,r | ≤ s. If n1 < 0, the same argument shows that An1 ≤ s and |aq1 ,1 | ≤ s. Since Ak ≥ 4|k|, this implies in particular that r ≤ s, and therefore (3.5) yields m = q1 + · · · + qr ≤ 2s. Now if nr > 0, let c ∈ N be the least index such that nc ≥ 0. By the definition of χ and (3.4), for every qr qr 1 ≤ i ≤ nr the coefficient of Hi (i) in the sum j=1 aj,r χ(zj,r ) is − j=1 aj,r , and hence qr aj,r ≤ qr · s ≤ 2s · s. j=1
(3.6)
Therefore, if c < r and znr−1 ,r−1 = nr−1 , by (3.4), the coefficient anr−1 ,r−1 satisfies an
r−1 ,r−1
≤ s + qr · s = (qr + 1) · s ≤ 2s2 .
(3.7)
Analogously, assume that c < r − 1 and znr−2 ,r−2 = nr−2 . Then the coefficient of Hnr−2 (nr−2 ) in the middle sum of (3.4) is qr−1 −1
anr−2 ,r−2 −
j=1
aj,r−1 − anr−1 ,r−1 −
qr
aj,r .
j=1
This and (3.4)–(3.7) imply an
r−2 ,r−2
≤ s + (qr−1 − 1) · s + (qr + 1) · s + qr · s = s(2qr + qr−1 + 1) < s · 4s.
Continuing this process we obtain that there is M+ > 0 such that |ani ,i | ≤ M+ for every i > c. In a similar way one can show that if n1 < 0, then there is M− > 0 such that |ani ,i | ≤ M− for every i such that ni > 0. Now, if nc = 0, the boundedness of aj,i corresponding to i = c and (3.4) easily imply that there exists an M such that |aj,i | ≤ M,
for every 1 ≤ i ≤ r and 1 ≤ j ≤ qi .
Then M (s) := max{2s, M } is as desired. Step 3. Let us show that if a1 y1 + · · · + am ym = 0, then a1 = · · · = am = 0. Indeed, we can represent 0 in the form (3.4). Now, as above, if nr > 0, then χ(zqr ,r ) has at least Anr distinct basic elements from I which do not appear in other summands in the middle sum of (3.4). So aqr ,r = 0. Analogously, if n1 < 0 we obtain that aq1 ,1 = 0. Therefore r = 1 and n1 = 0. In this case we also easily obtain that aj,1 = 0 for every 1 ≤ j ≤ q1 . Thus a1 = · · · = am = 0 as desired. Step 4. We claim that Y is a closed subset of V(I). First we show that Y ∩ sps (I) = Ys ∩ sps (I) for every s ∈ N. Indeed, let y := χ(x) = λ1 x1 + · · · + λs xs ∈ Y ∩ sps (I).
(3.8)
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If x ∈ [0, 1], then y ∈ Y1 and we are done. Suppose that x ∈ (k, k + 1] for some k ≥ 0, or x ∈ [k, k + 1) for some k < 0. If y ∈ / Ys , then either k ≥ s or k + 1 ≤ −s. In both cases y has at least Ak ≥ 4|k| > s distinct basic summands from I which contradicts (3.8). Hence y ∈ Ys . Thus Y ∩ sps (I) ⊆ Ys ∩ sps (I). The converse inclusion is clear. Fix now a closed subset F of R. Then, for every s ∈ N, we have χ(F ) ∩ sps (I) = χ(F ) ∩ Y ∩ sps (I) = χ(F ) ∩ Ys ∩ sps (I) = χ(F ) ∩ χ([−s, s]) ∩ sps (I) = χ F ∩ [−s, s] ∩ sps (I), and hence χ(F ) ∩ sps (I) is a closed subset of sps (I). Since V(I) = s∈N sps (I) is a kω -space by Theorem 3.1 of [6], we obtain that χ(F ) is closed in V(I). Therefore χ is a closed map. Thus χ is a homeomorphism of R onto Y . Finally, Steps 2–4 show that we can apply Proposition 2.2 to get that V(R) is linearly isomorphic to the closed linear subspace sp(Y ) of V(I). 2 Corollary 3.6 ([8]). A(R) embeds into A(I) as a topological group. Proof. The definition of the map χ in Step 1 of the proof of Theorem 3.5 (see also (3.2)) shows that A(R) is an algebraic subgroup of A(I). As A(X) is a closed subgroup of V(X) for every Tychonoff space X by Proposition 5.1 of [6], Theorem 3.5 implies that A(R) embeds into A(I) as a topological group. 2 We end the paper with the following result. In [10] it was shown that A(Sk ) embeds into A(Ik ), where Sk is the k-dimensional unit sphere. The proof of this fact is rather complicated. Essentially using the vector space properties below we give an explicit direct and simple construction of an embedding of V(Sk ) into V(Ik ). Proposition 3.7. For every k ∈ N, the space V(Sk ) embeds into V(Ik ) as a closed vector subspace. Proof. For a Tychonoff space X and x ∈ X, denote by [x] the element x considered as an element of the Hamel basis X of V(X). It will be more convenient to work with the closed symmetric unit interval I∗ := [−1, 1] than with I. Denote by Bk the closed unit ball in Rk and let x be the standard norm of x ∈ Bk . Clearly, Bk ⊆ Ik∗ and Sk−1 is the boundary of Bk . Set 3 := (3, . . . , 3), 8 := (8, . . . , 8) ∈ Rk and define injective continuous functions a(x), b(x) : Bk → V(Ik∗ ) as follows a(x) :=
x x 3+x 8+x + 1 − x and b(x) := + 1 − x . 10 10 10 10
It is easy to see that a(x) = b(y) if and only if x = y ∈ Sk−1 . Set K := A ∪ B, where A := {a(x) : x ∈ Bk } and B := {b(x) : x ∈ Bk }. By the above we obtain that K is homeomorphic to Sk . We shall show that the vector subspace sp(K) of V(Ik∗ ) generated by K is V(Sk ). By Proposition 2.2 it suffices to show that (i) K is a free basis for sp(K); (ii) for every n ∈ N there is a natural number m such that sp(K) ∩ spn (Ik∗ ) ⊆ spm (K). Let x1 , . . . , xl be distinct elements of Sk−1 , y1 , . . . , ys be distinct elements of Bk \ Sk−1 , z1 , . . . , zq be distinct elements of Bk \ Sk−1 and let λx1 , . . . , λxl , λy1 , . . . , λys , λz1 , . . . , λzq ∈ R \ {0}.
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Set T :=
l i=1
λxi a(xi )
+
s j=1
λyj a(yj )
+
q
λzt b(zt ).
t=1
By the construction of a(x) and b(x), it is easy to see that the number of nonzero basic element from Ik∗ in the decomposition of T is bigger or equal than l + s + q. The same holds also in the cases s = 0 and q = 0. So, (1) T = 0 that proves (i), and (2) sp(K) ∩ spn (Ik∗ ) ⊆ spn (K) that proves (ii). 2 Acknowledgements The second author thanks Ben Gurion University of the Negev for hospitality. References [1] R. Brown, J.P.L. Hardy, Subgroups of free topological groups and free topological products of topological groups, J. Lond. Math. Soc. 10 (1975) 431–440. [2] R. Engelking, Dimension Theory, PWN, Warszawa, 1978. [3] R. Engelking, General Topology, Heldermann Verlag, Berlin, 1989. [4] R. Engelking, Theory of Dimensions. Finite and Infinite, Heldermann Verlag, 1995. [5] S. Gabriyelyan, The k-space property for free locally convex spaces, Can. Math. Bull. 57 (2014) 803–809. [6] S.S. Gabriyelyan, S.A. Morris, Free topological vector spaces, Topol. Appl. 223 (2017) 30–49. [7] E. Katz, S.A. Morris, Free abelian topological groups on countable CW -complexes, Bull. Aust. Math. Soc. 41 (1990) 451–456. [8] E. Katz, S.A. Morris, P. Nickolas, A free subgroup of the free abelian topological group on the unit interval, Bull. Lond. Math. Soc. 14 (1982) 399–402. [9] E. Katz, S.A. Morris, P. Nickolas, Characterization of bases of subgroups of free topological groups, J. Lond. Math. Soc. (2) 27 (1983) 421–426. [10] E. Katz, S.A. Morris, P. Nickolas, Free abelian topological groups on spheres, Quart. J. Math. Oxford 35 (1984) 173–181. [11] A. Leiderman, S.A. Morris, V. Pestov, The free abelian topological group and the free locally convex space on the unit interval, J. Lond. Math. Soc. 56 (1997) 529–538. [12] P. Nickolas, Subgroups of the free topological group on [0, 1], J. Lond. Math. Soc. 12 (1976) 199–205. [13] V.G. Pestov, Some properties of free topological groups, Mosc. Univ. Math. Bull. 37 (1982) 46–49. [14] V.V. Uspenski˘ı, Free topological groups of metrizable spaces, Math. USSR, Izv. 37 (1991) 657–680.