Products and h-homogeneity

TL;DR

This paper investigates the properties of h-homogeneity in zero-dimensional spaces, demonstrating its productivity, existence of non-trivial products, and conditions under which infinite powers are h-homogeneous, thus advancing the understanding of zero-dimensional topology.

Contribution

It generalizes previous results on h-homogeneity, proves its productivity in zero-dimensional spaces, and relates key open questions to new partial answers.

Findings

h-homogeneity is productive in zero-dimensional spaces

Existence of non-empty zero-dimensional Y such that X×Y is h-homogeneous

Infinite powers of spaces with dense isolated points are h-homogeneous

Abstract

Building on work of Terada, we prove that h-homogeneity is productive in the class of zero-dimensional spaces. Then, by generalizing a result of Motorov, we show that for every non-empty zero-dimensional space $X$ there exists a non-empty zero-dimensional space $Y$ such that $X\times Y$ is h-homogeneous. Also, we simultaneously generalize results of Motorov and Terada by showing that if $X$ is a space such that the isolated points are dense then $X^\kappa$ is h-homogeneous for every infinite cardinal $\kappa$. Finally, we show that a question of Terada (whether $X^\omega$ is h-homogeneous for every zero-dimensional first-countable $X$) is equivalent to a question of Motorov (whether such an infinite power is always divisible by 2) and give some partial answers.

Equations

$$\mathcal{B}=\{\mathrm{Ex}(U)\times \mathrm{Ex}(V):U\text{ is open in }X\text{ and }V\text{ is open in }Y\}$$

$$C$$

$$Y\times C$$

$$X^{\omega }\cong (X^{\omega }{)}^{\omega }\cong (F\times Y{)}^{\omega }\cong F^{\omega }\times Y^{\omega }.$$

$$U_n=\{f\in X:f(i)=0\text{ for all }i<n\}.$$

$$X\cong (Y\oplus 1)\times X\cong (Y\times X)\oplus X,$$

$$Y^{\omega }\times (Y\oplus 1{)}^{\omega }\cong {\left(Y^{\omega }\times (Y\oplus 1)\right)}^{\omega }\cong (Y^{\omega }\oplus Y^{\omega }{)}^{\omega }\cong 2^{\omega }\times Y^{\omega },$$

$$X^{\omega }$$

Full text

All spaces in this paper are assumed to be Tychonoff. It is easy to see that every zero-dimensional space is Tychonoff. For all undefined topological notions, see [7]. For all undefined Boolean algebraic notions, see [9]. Recall that a subset of a space is clopen if it is closed and open.

The Cantor set, the rationals and the irrationals are examples of h-homogeneous spaces. Every connected space is h-homogeneous. A finite space is h-homogeneous if and only if it has size at most $1$. The concept of h-homogeneity has been studied (mostly in the zero-dimensional case) by several authors: see [10] for an extensive list of references.

We will denote by $\operatorname{\textrm{Clop}}(X)$ the Boolean algebra of the clopen subsets of $X$. Recall that a Boolean algebra $\mathcal{A}$ is homogeneous if $\mathcal{A}\upharpoonright a$ is isomorphic to $\mathcal{A}$ for every non-zero $a\in\mathcal{A}$, where $\mathcal{A}\upharpoonright a$ denotes the relative algebra $\{x\in\mathcal{A}:x\leq a\}$. If $X$ is h-homogeneous then $\operatorname{\textrm{Clop}}(X)$ is homogeneous; the converse holds if $X$ is compact and zero-dimensional (see the remarks following Definition 9.12 in [9]).

1. The productivity of h-homogeneity

In [20], the productivity of h-homogeneity is stated as an open problem (see also [10] and [11]), and it is shown that the product of zero-dimensional h-homogeneous spaces is h-homogeneous provided it is compact or non-pseudocompact (see Theorem 3.3 in [20]). The following theorem, proved by Terada under the additional assumption that $X$ is zero-dimensional (see Theorem 2.4 in [20]), is the key ingredient in the proof. Recall that a collection $\mathcal{B}$ consisting of non-empty open subsets of a space $X$ is a $\pi$-base if for every non-empty open subset $U$ of $X$ there exists $V\in\mathcal{B}$ such that $V\subseteq U$.

Theorem 2** (Terada).**

Assume that $X$ is non-pseudocompact. If $X$ has a $\pi$-base consisting of clopen sets that are homeomorphic to $X$ then $X$ is h-homogeneous.

The proof of Theorem 2 uses the fact that a zero-dimensional non-pseudocompact space can be written as the disjoint union of infinitely many of its non-empty clopen subsets (the converse is also true, trivially). However, that is the only consequence of zero-dimensionality that is actually used (see Appendix A). Therefore such assumption is redundant by the following lemma, whose proof we leave to the reader.

Lemma 3**.**

Assume that $X$ is non-pseudocompact. If $X$ has a $\pi$-base consisting of clopen sets then $X$ can be written as the disjoint union of infinitely many of its non-empty clopen subsets.

Using Theorem 2 one can easily prove the following.

Theorem 4** (Terada).**

Assume that $X=\prod_{i\in I}X_{i}$ is non-pseudocompact. If $X_{i}$ is h-homogeneous and it has a $\pi$-base consisting of clopen sets for every $i\in I$ then $X$ is h-homogeneous.

For proofs of the following basic facts about $\beta X$, see Section 11 in [12]. Given any open subset $U$ of $X$, define $\operatorname{Ex}(U)=\beta X\setminus\operatorname{cl}_{\beta X}(X\setminus U)$. It is easy to see that $\operatorname{Ex}(U)$ is the biggest open subset of $\beta X$ such that its intersection with $X$ is $U$. If $C$ is a clopen subset of $X$ then $\operatorname{Ex}(C)=\operatorname{cl}_{\beta X}(C)$, hence $\operatorname{Ex}(C)$ is clopen in $\beta X$. Furthermore, the collection $\{\operatorname{Ex}(U):U\textrm{ is open in }X\}$ is a base for $\beta X$.

Remark*.*

It is not true that $\beta X$ is zero-dimensional whenever $X$ is zero-dimensional (see Example 6.2.20 in [7] or Example 3.39 in [22]). If $\beta X$ is zero-dimensional then $X$ is called strongly zero-dimensional.

We will need the following theorem (see Theorem 8.25 in [22]); see also Exercise 3.12.20(d) in [7]. Recall that a subspace $Y$ of $X$ is $C^{\ast}$-embedded in $X$ if every bounded continuous function $f:Y\longrightarrow\mathbb{R}$ admits a continuous extension to $X$.

Theorem 5** (Glicksberg).**

Assume that $X=\prod_{i\in I}X_{i}$ is pseudocompact. Then $X$ is $C^{\ast}$-embedded in $\prod_{i\in I}\beta X_{i}$.

Remark*.*

The reverse implication is also true, under the additional assumption that $\prod_{j\neq i}X_{j}$ is infinite for every $i\in I$. Such assumption is clearly not needed in the above statement (see Proposition 8.2 in [22]).

Proposition 6**.**

Assume that $X\times Y$ is pseudocompact. If $C$ is a clopen subset of $X\times Y$ then $C$ can be written as the union of finitely many open rectangles.

Proof.

It follows from Theorem 5 that $X\times Y$ is $C^{\ast}$-embedded in $\beta X\times\beta Y$. By the universal property of the Čech-Stone compactification (see Corollary 3.6.3 in [7]), there exists a homeomorphism $h:\beta X\times\beta Y\longrightarrow\beta(X\times Y)$ such that $h(x,y)=(x,y)$ whenever $(x,y)\in X\times Y$.

Let $C$ be a clopen subset of $X\times Y$. Since $\{\operatorname{Ex}(U):U\textrm{ is open in }X\}$ is a base for $\beta X$ and $\{\operatorname{Ex}(V):V\textrm{ is open in }Y\}$ is a base for $\beta Y$, the collection

[TABLE]

is a base for $\beta X\times\beta Y$. Therefore $\{h[B]:B\in\mathcal{B}\}$ is a base for $\beta(X\times Y)$, hence we can write $\operatorname{Ex}(C)=h[B_{1}]\cup\cdots\cup h[B_{n}]$ for some $B_{1},\ldots,B_{n}\in\mathcal{B}$ by compactness.

Finally, if we let $B_{i}=\operatorname{Ex}(U_{i})\times\operatorname{Ex}(V_{i})$ for each $i$, we get

[TABLE]

that concludes the proof. ∎

Lemma 7**.**

Assume that $C$ is a clopen subset of $X\times Y$ that can be written as the union of finitely many rectangles. Then $C$ can be written as the union of finitely many pairwise disjoint clopen rectangles.

Proof.

For every $x\in X$, let $C_{x}=\{y\in Y:(x,y)\in C\}$ be the corresponding vertical cross-section. For every $y\in Y$, let $C^{y}=\{x\in X:(x,y)\in C\}$ be the corresponding horizontal cross-section. Since $C$ is clopen, each cross-section is clopen.

Let $\mathcal{A}$ be the Boolean subalgebra of $\operatorname{\textrm{Clop}}(X)$ generated by $\{C^{y}:y\in Y\}$. Since $\mathcal{A}$ is finite, it must be atomic. Let $P_{1},\ldots,P_{m}$ be the atoms of $\mathcal{A}$. Similarly, let $\mathcal{B}$ be the Boolean subalgebra of $\operatorname{\textrm{Clop}}(Y)$ generated by $\{C_{x}:x\in X\}$, and let $Q_{1},\ldots,Q_{n}$ be the atoms of $\mathcal{B}$.

Observe that the rectangles $P_{i}\times Q_{j}$ are clopen and pairwise disjoint. Furthermore, given any $i,j$, either $P_{i}\times Q_{j}\subseteq C$ or $(P_{i}\times Q_{j})\cap C=\varnothing$. Hence $C$ is the union of a (finite) collection of such rectangles. ∎

Proposition 8**.**

Assume that $X\times Y$ is pseudocompact. If $X$ is h-homogeneous and $Y$ is h-homogeneous then $X\times Y$ is h-homogeneous.

Proof.

Assume that $X$ and $Y$ are h-homogeneous. If $X$ and $Y$ are both connected then $X\times Y$ is connected. So assume without loss of generality that $X$ is not connected. Since $X$ is also h-homogeneous, it follows that $X\cong n\times X$ whenever $1\leq n<\omega$. Therefore $X\times Y\cong n\times X\times Y$ whenever $1\leq n<\omega$.

Now let $C$ be a non-empty clopen subset of $X\times Y$. By Proposition 6 and Lemma 7, we can write $C$ as the disjoint union of finitely many, say $n$, non-empty clopen rectangles. By the h-homogeneity of $X$ and $Y$, each such rectangle is homeomorphic to $X\times Y$. Therefore $C\cong n\times X\times Y\cong X\times Y$. ∎

Corollary 9**.**

Assume that $X=X_{1}\times\cdots\times X_{n}$ is pseudocompact. If each $X_{i}$ is h-homogeneous then $X$ is h-homogeneous.

An obvious modification of the proof of Proposition 6 yields the following.

Proposition 10**.**

Assume that $X=\prod_{i\in I}X_{i}$ is pseudocompact. If $C$ is a clopen subset of $X$ then $C$ can be written as the union of finitely many open rectangles.

Corollary 11**.**

Assume that $X=\prod_{i\in I}X_{i}$ is pseudocompact. If $C$ is a clopen subset of $X$ then $C$ depends on finitely many coordinates.

Remark*.*

The zero-dimensional case of Corollary 11 is a trivial consequence of a result by Broverman (see Theorem 2.6 in [2]).

Theorem 12**.**

Assume that $X=\prod_{i\in I}X_{i}$ is pseudocompact. If $X_{i}$ is h-homogeneous for every $i\in I$ then $X$ is h-homogeneous.

Proof.

Assume that each $X_{i}$ is h-homogeneous. Let $C$ be a non-empty clopen subset of $X$. By Corollary 11, there exists a finite subset $F$ of $I$ such that $C$ is homeomorphic to $D\times\prod_{i\in I\setminus F}X_{i}$, where $D$ is a non-empty clopen subset of $\prod_{i\in F}X_{i}$. But $\prod_{i\in F}X_{i}$ is h-homogeneous by Corollary 9, so $D\cong\prod_{i\in F}X_{i}$. Hence $C\cong X$. ∎

Theorem 13**.**

If $X_{i}$ is h-homogeneous and it has a $\pi$-base consisting of clopen sets for every $i\in I$ then $X=\prod_{i\in I}X_{i}$ is h-homogeneous.

Proof.

If $X$ is pseudocompact, apply Theorem 12; if $X$ is non-pseudocompact, apply Theorem 4. ∎

Corollary 14**.**

If $X_{i}$ is h-homogeneous and zero-dimensional for every $i\in I$ then $\prod_{i\in I}X_{i}$ is h-homogeneous.

Question*.*

Can the zero-dimensionality requirement be dropped in Corollary 14?

2. Some applications

The compact case of the following result was essentially proved by Motorov (see Theorem 0.2(9) in [16] and Theorem 2 in [15]).

Theorem 15**.**

Assume that $X$ has a $\pi$-base $\mathcal{B}$ consisting of clopen sets. Then $Y=(X\times 2\times\prod\mathcal{B})^{\kappa}$ is h-homogeneous for every infinite cardinal $\kappa$.

Proof.

One can easily check that $Y$ has a $\pi$-base consisting of clopen sets that are homeomorphic to $Y$. Therefore, if $Y$ is non-pseudocompact, the result follows from Theorem 2.

On the other hand, an analysis of Motorov’s proof shows that the only consequence of the compactness of $Y$ that is used is the fact that clopen sets in $Y$ depend on finitely many coordinates. Therefore the same proof works if $Y$ is pseudocompact by Corollary 11. We reproduce such proof for the convenience of the reader.

Assume that $Y$ is pseudocompact and let $C$ be a non-empty clopen subset of $Y$. The fact that $C$ depends on finitely many coordinates implies that $C\cong Y\times C$. So it will be enough to show that $Y\times C\cong Y$.

Let $B$ be a clopen subset of $C$ that is homeomorphic to $Y$. Let $D=C\setminus B$ and $E=(Y\setminus C)\oplus B$. Observe that $Y\cong Y^{2}\cong(Y\times D)\oplus(Y\times E)$ and that $Y\oplus Y\cong 2\times Y\cong Y$. So

[TABLE]

that concludes the proof. ∎

Remark*.*

In [15] and [16], Motorov asked whether the $2$ can be dropped in the definition of $Y$. This is certainly true if $Y$ is non-pseudocompact, but we do not know the answer in general. Observe that if the answer were ‘yes’ then Theorem 18 would become an immediate corollary of Theorem 15.

Corollary 16**.**

For every non-empty zero-dimensional space $X$ there exists a non-empty zero-dimensional space $Y$ such that $X\times Y$ is h-homogeneous. Furthermore, if $X$ is compact, then $Y$ can be chosen to be compact.

Question*.*

Is it true that for every non-empty space $X$ there exists a non-empty space $Y$ such that $X\times Y$ is h-homogeneous? If $X$ is compact, can $Y$ be chosen to be compact?

Remark*.*

In [21], using a very brief and elegant argument, Uspenskiĭ proved that for every non-empty space $X$ there exists a non-empty space $Y$ such that $X\times Y$ is homogeneous (in the sense of Definition 19). However, it is not true that $Y$ can be chosen to be compact whenever $X$ is compact: Motorov proved that the closure in the plane of $\{(x,\sin(1/x)):x\in(0,1]\}$ is not the retract of any compact homogeneous space (see Section 3 in [1] for a proof).

The following was proved by Matveev (see Proposition 3 in [10]) under the additional assumption that $X$ is zero-dimensional, even though such assumption is not actually used in the proof (see Appendix A). Recall that a sequence $\langle A_{n}:n\in\omega\rangle$ of subsets of a space $X$ converges to a point $x$ if for every neighborhood $U$ of $x$ there exists $N\in\omega$ such that $A_{n}\subseteq U$ for each $n\geq N$.

Proposition 17** (Matveev).**

Assume that $X$ has a $\pi$-base consisting of clopen sets that are homeomorphic to $X$. If there exists a sequence $\langle U_{n}:n\in\omega\rangle$ of non-empty open subsets of $X$ that converges to a point then $X$ is h-homogeneous.

The case $\kappa=\omega$ of the following result is an easy consequence of Proposition 17. Motorov first proved it under the additional assumption that $X$ is a zero-dimensional first-countable compact space (see Theorem 0.2(2) in [16] and Theorem 1 in [15]). Terada proved it for an arbitrary infinite $\kappa$, under the additional assumption that $X$ is zero-dimensional and non-pseudocompact (see Corollary 3.2 in [20]).

Theorem 18**.**

Assume that $X$ is a space such that the isolated points are dense. Then $X^{\kappa}$ is h-homogeneous for every infinite cardinal $\kappa$.

Proof.

We will show that $X^{\omega}$ is h-homogeneous and it has a $\pi$-base consisting of clopen sets. Since $X^{\kappa}\cong(X^{\omega})^{\kappa}$ for every infinite cardinal $\kappa$, an application of Theorem 13 will conclude the proof.

Let $D$ be the set of isolated points of $X$ and let $\operatorname{Fn}(\omega,D)$ be the set of finite partial functions from $\omega$ to $D$. Given $s\in\operatorname{Fn}(\omega,D)$, define $U_{s}=\{f\in X^{\omega}:f\supseteq s\}$. Now fix $d\in D$ and let $g\in X^{\omega}$ be the constant function with value $d$. It is easy to see that $\langle U_{g\upharpoonright n}:n\in\omega\rangle$ is a sequence of open sets in $X^{\omega}$ that converges to $g$. Furthermore $\mathcal{B}=\{U_{s}:s\in\operatorname{Fn}(\omega,D)\}$ is a $\pi$-base for $X^{\omega}$ consisting of clopen sets that are homeomorphic to $X^{\omega}$. So $X^{\omega}$ is h-homogeneous by Proposition 17. ∎

3. Infinite powers of zero-dimensional first-countable spaces

Definition 19**.**

A space $X$ is homogeneous if for every $x,y\in X$ there exists a homeomorphism $f:X\longrightarrow X$ such that $f(x)=y$.

It is well-known (and easy to prove) that every zero-dimensional first-countable h-homogeneous space is homogeneous. As announced by Motorov (see Theorem 0.1 in [16]), the converse holds for zero-dimensional first-countable compact spaces of uncountable cellularity (see Theorem 2.5 in [8] for a proof). In [3], Van Douwen constructed a zero-dimensional first-countable compact homogeneous space $X$ that is not h-homogeneous (actually, $X$ has no proper subspaces that are homeomorphic to $X$). In [17], using similar techniques, Motorov constructed a zero-dimensional first-countable compact homogeneous space that is not divisible by $2$ (in the sense of Definition 21); see also Theorem 7.7 in [13].

In [20], Terada asked whether $X^{\omega}$ is h-homogeneous for every zero-dimensional first-countable $X$. In [4], the following remarkable theorem is proved.

Theorem 20** (Dow and Pearl).**

If $X$ is a zero-dimensional first-countable space then $X^{\omega}$ is homogeneous.

However, Terada’s question remains open. In [15] and [16], Motorov asks whether such an infinite power is always divisible by $2$. Using Theorem 20, we will show that the two questions are equivalent: actually even weaker conditions suffice (see Proposition 24).

Definition 21**.**

A space $F$ is a factor of $X$ (or $X$ is divisible by $F$) if there exists $Y$ such that $F\times Y\cong X$. If $F\times X\cong X$ then $F$ is a strong factor of $X$ (or $X$ is strongly divisible by $F$).

We will use the following lemma freely in the rest of this section.

Lemma 22**.**

The following are equivalent.

1. (1)

$F$* is a factor of $X^{\omega}$.* 2. (2)

$F$* is a strong factor of $X^{\omega}$.* 3. (3)

$F^{\omega}$* is a strong factor of $X^{\omega}$.*

Proof.

The implications $\ref{strongfactor}\rightarrow\ref{factor}$ and $\ref{infinitestrongfactor}\rightarrow\ref{factor}$ are clear.

Assume 1. Then there exists $Y$ such that $F\times Y\cong X^{\omega}$, hence

[TABLE]

Since multiplication by $F$ or by $F^{\omega}$ does not change the right-hand side, it follows that 2 and 3 hold. ∎

Lemma 23**.**

Assume that $Y$ is a non-empty zero-dimensional first-countable space. Then $X=(Y\oplus 1)^{\omega}$ is h-homogeneous and $X\cong Y^{\omega}\times(Y\oplus 1)^{\omega}\cong 2^{\omega}\times Y^{\omega}$.

Proof.

Recall that $1=\{0\}$ and let $g\in X$ be the constant function with value [math]. For each $n\in\omega$, define

[TABLE]

Observe that $\mathcal{B}=\{U_{n}:n\in\omega\}$ is a local base for $X$ at $g$ consisting of clopen sets that are homeomorphic to $X$. But $X$ is homogeneous by Theorem 20, therefore it has such a local base at every point. In conclusion $X$ has a base (hence a $\pi$-base) consisting of clopen sets that are homeomorphic to $X$. It follows from Proposition 17 that $X$ is h-homogeneous.

To prove the second statement, observe that

[TABLE]

hence $X\cong Y\times X$ by h-homogeneity. It follows that $X\cong Y^{\omega}\times(Y\oplus 1)^{\omega}$. Finally,

[TABLE]

that concludes the proof. ∎

Proposition 24**.**

Assume that $X$ is a zero-dimensional first-countable space containing at least two points. Then the following are equivalent.

1. (1)

$X^{\omega}\cong(X\oplus 1)^{\omega}$. 2. (2)

$X^{\omega}\cong Y^{\omega}$* for some space $Y$ with at least one isolated point.* 3. (3)

$X^{\omega}$* is h-homogeneous.* 4. (4)

$X^{\omega}$* has a non-empty clopen subset that is strongly divisible by $2$.* 5. (5)

$X^{\omega}$* has a proper clopen subset that is homeomorphic to $X^{\omega}$.* 6. (6)

$X^{\omega}$* has a proper clopen subset that is a factor of $X^{\omega}$.*

Proof.

The implication $\ref{xoplus1}\rightarrow\ref{yisolated}$ is trivial; the implication $\ref{yisolated}\rightarrow\ref{hhomogeneous}$ follows from Lemma 23; the implications $\ref{hhomogeneous}\rightarrow\ref{divisiblebytwo}\rightarrow\ref{properclopen}\rightarrow\ref{properclopenfactor}$ are trivial.

Assume that 6 holds. Let $C$ be a proper clopen subset of $X^{\omega}$ that is a factor of $X^{\omega}$ and let $D=X^{\omega}\setminus C$. Then

[TABLE]

hence $X^{\omega}\cong(1\oplus D)^{\omega}\times X^{\omega}$. Since $(1\oplus D)^{\omega}\cong 2^{\omega}\times D^{\omega}$ by Lemma 23, it follows that $2^{\omega}$ is a factor of $X^{\omega}$. So $2^{\omega}$ is a strong factor of $X^{\omega}$. Therefore 1 holds by Lemma 23. ∎

The next two propositions show that in the pseudocompact case we can say something more.

Proposition 25**.**

Assume that $X$ is a zero-dimensional first-countable space such that $X^{\omega}$ is pseudocompact. Then $C^{\omega}\cong(X\oplus 1)^{\omega}$ for every non-empty proper clopen subset $C$ of $X^{\omega}$.

Proof.

Let $C$ be a non-empty proper clopen subset of $X^{\omega}$. It follows from Corollary 11 that $C\cong C\times X^{\omega}$, hence $C^{\omega}\cong C^{\omega}\times X^{\omega}$. Since $C^{\omega}\times X^{\omega}$ clearly has a proper clopen subset that is homeomorphic to $C^{\omega}\times X^{\omega}$, Proposition 24 implies that $C^{\omega}$ is h-homogeneous, hence strongly divisible by $2$. So $C^{\omega}\cong 2^{\omega}\times C^{\omega}\cong 2^{\omega}\times C^{\omega}\times X^{\omega}$. Since $2^{\omega}\times X^{\omega}\cong(X\oplus 1)^{\omega}$ by Lemma 23, it follows that $C^{\omega}\cong C^{\omega}\times(X\oplus 1)^{\omega}$.

On the other hand, $(X\oplus 1)^{\omega}\cong X^{\omega}\times(X\oplus 1)^{\omega}$ by Lemma 23. Hence $(X\oplus 1)^{\omega}$ has a clopen subset homeomorphic to $C\times(X\oplus 1)^{\omega}$. But Lemma 23 shows that $(X\oplus 1)^{\omega}$ is h-homogeneous, so $C\times(X\oplus 1)^{\omega}\cong(X\oplus 1)^{\omega}$. Therefore $C^{\omega}\times(X\oplus 1)^{\omega}\cong(X\oplus 1)^{\omega}$, that concludes the proof. ∎

Proposition 26**.**

In addition to the hypotheses of Proposition 24, assume that $X^{\omega}$ is pseudocompact. Then the following can be added to the list of equivalent conditions.

1. (7)

$X^{\omega}$* has a non-empty proper clopen subset that is homeomorphic to $Y^{\omega}$ for some space $Y$.*

Proof.

The implication $5\rightarrow 7$ is trivial.

Assume that 7 holds. Let $C$ be a non-empty proper clopen subset of $X^{\omega}$ that is homeomorphic to $Y^{\omega}$ for some space $Y$. Then clearly $C^{\omega}\cong C$. Therefore $C\cong(X\oplus 1)^{\omega}$ by Proposition 25. Hence $C$ is strongly divisible by $2$ by Lemma 23, showing that 4 holds. ∎

Finally, we point out that Proposition 24 can be used to give a positive answer to Terada’s question for a certain class of spaces. We will need the following definition.

Definition 27**.**

A space $X$ is ultraparacompact if every open cover of $X$ has a refinement consisting of pairwise disjoint clopen sets.

It is easy to see that every ultraparacompact space is zero-dimensional. As noted by Nyikos in [18], a space is ultraparacompact if and only if it is paracompact and strongly zero-dimensional (this is proved like Proposition 1.2 in [5]). A metric space $X$ is ultraparacompact if and only if $\dim X=0$ (see Theorem 7.2.4 in [7]); see also Theorem 7.3.3 in [7]. For such a metric space $X$, Van Engelen proved that $X^{\omega}$ is h-homogeneous if $X$ is of the first category in itself or $X$ has a completely metrizable dense subset (see Theorem 4.2 and Theorem 4.4 in [6]). It follows that $X^{\omega}$ is h-homogeneous if $X$ is analytic (see Corollary 31). For related results, see also Theorem 8 and Theorem 9 in [19].

Proposition 28**.**

Assume that $X$ is a (zero-dimensional) first-countable space. If $X^{\omega}$ is ultraparacompact and non-Lindelöf then $X^{\omega}$ is h-homogeneous.

Proof.

Let $\mathcal{U}$ be an open cover of $X^{\omega}$ with no countable subcovers. By ultraparacompactness, there exists a refinement $\mathcal{V}$ of $\mathcal{U}$ consisting of pairwise disjoint non-empty clopen sets. Let $\mathcal{V}=\{C_{\alpha}:\alpha\in\kappa\}$ be an enumeration without repetitions, where $\kappa$ is an uncountable cardinal.

Now fix $x\in X^{\omega}$ and a local base $\{U_{n}:n\in\omega\}$ at $x$ consisting of clopen sets. Since $X^{\omega}$ is homogeneous by Theorem 20, for each $\alpha<\kappa$ we can find $n(\alpha)\in\omega$ such that a homeomorphic clopen copy $D_{\alpha}$ of $U_{n(\alpha)}$ is contained in $C_{\alpha}$. Since $\kappa$ is uncountable, there exists an infinite $S\subseteq\kappa$ such that $n(\alpha)=n(\beta)$ for every $\alpha,\beta\in S$. It is easy to check that $\bigcup_{\alpha\in S}D_{\alpha}$ is a non-empty clopen subset of $X^{\omega}$ that is strongly divisible by $2$. Therefore $X^{\omega}$ is h-homogeneous by Proposition 24. ∎

An application of Corollary 4.1.16, Theorem 7.3.2 and Theorem 7.3.16 in [7] immediately yields the following.

Corollary 29**.**

Assume that $X$ is a metric space such that $\dim X=0$. If $X$ is non-separable then $X^{\omega}$ is h-homogeneous.

Acknowledgements. The author thanks his advisor Ken Kunen and the anonymous referee for valuable comments on earlier versions of this paper.

Appendix A Proofs of the results by Terada and Matveev

In this section we will present a somewhat unified approach to the proofs of Theorem 2 and Proposition 17. Notice that zero-dimensionality is never needed.

Proof of Theorem 2.

Assume that $X$ has a $\pi$-base consisting of clopen sets that are homeomorphic to $X$. By Lemma 3, we can fix a collection $\{X_{n}:n\in\omega\}$ consisting of pairwise disjoint non-empty clopen subsets of $X$ such that $X=\bigcup_{n\in\omega}X_{n}$. Let $C$ be a non-empty clopen subset of $X$. Since $C$ contains a clopen subset that is homeomorphic to $X$, we can fix a collection $\{C_{n}:n\in\omega\}$ consisting of pairwise disjoint non-empty clopen subsets of $C$ such that $C=\bigcup_{n\in\omega}C_{n}$.

We will recursively construct clopen sets $Y_{n}\subseteq X_{n}$ and $D_{n}\subseteq C_{n}$, together with partial homeomorphisms $h_{n}$ and $k_{n}$ for every $n\in\omega$. In the end, setting $h=\bigcup_{n\in\omega}(h_{n}\cup k_{n})$ will yield the desired homeomorphism. Start by setting $Y_{0}=\varnothing$ and $h_{0}=\varnothing$. Then, let $D_{0}\subseteq C_{0}$ be a clopen set that is homeomorphic to $X_{0}\setminus Y_{0}$ and fix a homeomorphism $k_{0}:X_{0}\setminus Y_{0}\longrightarrow D_{0}$. Now assume that clopen sets $D_{n}\subseteq C_{n}$ and $Y_{n}\subseteq X_{n}$ have been defined. Let $Y_{n+1}\subseteq X_{n+1}$ be a clopen set that is homeomorphic to $C_{n}\setminus D_{n}$ and fix a homeomorphism $h_{n+1}:Y_{n+1}\longrightarrow C_{n}\setminus D_{n}$. Then, let $D_{n+1}\subseteq C_{n+1}$ be a clopen set that is homeomorphic to $X_{n+1}\setminus Y_{n+1}$ and fix a homeomorphism $k_{n+1}:X_{n+1}\setminus Y_{n+1}\longrightarrow D_{n+1}$. ∎

Proof of Proposition 17.

Let $\langle U_{n}:n\in\omega\rangle$ be a sequence of non-empty open subsets of $X$ that converges to a point $x$. One can easily obtain a sequence $\langle X_{n}:n\in\omega\rangle$ of pairwise disjoint non-empty clopen sets that converges to $x$, such that $x\notin X_{n}$ for each $n\in\omega$. Let $C$ be a non-empty clopen subset of $X$. Let $B$ be a clopen subset of $C$ that is homeomorphic to $X$. Fix a homeomorphism $f:X\longrightarrow B$ and let $C_{n}=f[X_{n}]$ for each $n\in\omega$.

Now define clopen sets $Y_{n}\subseteq X_{n}$ and $D_{n}\subseteq C_{n}$ for each $n\in\omega$ and a (partial) homeomorphism $h$ as in the proof of Theorem 2, but start by choosing $Y_{0}$ homeomorphic to $C\setminus B$ and fixing a homeomorphism $h_{0}:Y_{0}\longrightarrow C\setminus B$. Finally, extend $h$ by setting $h(x)=f(x)$ for every $x\in X\setminus\bigcup_{n\in\omega}X_{n}$. It is easy to check that this yields the desired homeomorphism. ∎

Appendix B Some Descriptive Set Theory

The following results seem to be folklore, but we could not find satisfactory references. For the definitions of analytic and property of Baire, see [14].

Theorem 30**.**

Let $X$ be an analytic metric space. Then either $X$ has a completely metrizable dense subset or $X$ has a non-empty open subset of the first category.

Proof.

Let $\widetilde{X}$ be the completion of $X$. By Theorem A.13.13 in [14], $X$ has the property of Baire in $\widetilde{X}$. Therefore, by Proposition A.13.10 in [14], we can write $X=G\cup M$, where $G$ is a $G_{\delta}$ subset of $\widetilde{X}$ and $M$ is of the first category in $\widetilde{X}$.

Since $G$ is a $G_{\delta}$ subset of the complete metric space $\widetilde{X}$, it is completely metrizable (see Theorem A.6.3 in [14]). Since $X$ is dense in $\widetilde{X}$, the set $M$ is of the first category in $X$ as well (see Exercise A.13.7 in [14]). In conclusion, if $G$ is dense in $X$ then the first alternative in the statement of the theorem will hold, otherwise the second alternative will hold. ∎

Corollary 31**.**

Let $X$ be an analytic metric space. Then either $X^{\omega}$ has a completely metrizable dense subset or $X^{\omega}$ is of the first category in itself.

Proof.

If $X$ has a completely metrizable dense subset $D$ then $D^{\omega}$ is a completely metrizable dense subset of $X^{\omega}$ (see Lemma A.6.2 in [14]).

So assume that $X$ has a non-empty open subset $U$ of the first category. Observe that $M_{n}=\{f\in X^{\omega}:f(n)\in U\}$ is of the first category in $X^{\omega}$ for every $n\in\omega$. Also, it is clear that $(X\setminus U)^{\omega}$ is closed nowhere dense in $X^{\omega}$. It follows that $X^{\omega}$ is of the first category in itself. ∎