Pseudocompact spaces with regular G-delta diagonals

This post complements two results discussed in two previous blog posts concerning G_\delta-diagonal. One result is that any compact space with a G_\delta-diagonal is metrizable (see here). The other result is that the compactness in the first result can be relaxed to countably compactness. Thus any countably compact space with a G_\delta-diagonal is metrizable (see here). The countably compactness in the second result cannot be relaxed to pseudocompactness. The Mrowka space is a pseudocompact space with a G_\delta-diagonal that is not submetrizable, hence not metrizable (see here). However, if we strengthen the G_\delta-diagonal to a regular G_\delta-diagonal while keeping pseudocompactness fixed, then we have a theorem. We prove the following theorem.

Theorem 1
If the space X is pseudocompact and has a regular G_\delta-diagonal, then X is metrizable.

All spaces are assumed to be Hausdorff and completely regular. The assumption of completely regular is crucial. The proof of Theorem 1 relies on two lemmas concerning pseudocompact spaces (one proved in a previous post and one proved here). These two lemmas work only for completely regular spaces.

The proof of Theorem 1 uses a metrization theorem. The best metrization to use in this case is Moore metrization theorem (stated below). The result in Theorem 1 is found in [2].

First some basics. Let X be a space. The diagonal of the space X is the set \Delta=\{ (x,x): x \in X \}. When the diagonal \Delta, as a subset of X \times X, is a G_\delta-set, i.e. \Delta is the intersection of countably many open subsets of X \times X, the space X is said to have a G_\delta-diagonal.

The space X is said to have a regular G_\delta-diagonal if the diagonal \Delta is a regular G_\delta-set in X \times X, i.e. \Delta=\bigcap_{n=1}^\infty \overline{U_n} where each U_n is an open subset of X \times X with \Delta \subset U_n. If \Delta=\bigcap_{n=1}^\infty \overline{U_n}, then \Delta=\bigcap_{n=1}^\infty \overline{U_n}=\bigcap_{n=1}^\infty U_n. Thus if a space has a regular G_\delta-diagonal, it has a G_\delta-diagonal. We will see that there exists a space with a G_\delta-diagonal that fails to be a regular G_\delta-diagonal.

The space X is a pseudocompact space if for every continuous function f:X \rightarrow \mathbb{R}, the image f(X) is a bounded set in the real line \mathbb{R}. Pseudocompact spaces are discussed in considerable details in this previous post. We will rely on results from this previous post to prove Theorem 1.

The following lemma is used in proving Theorem 1.

Lemma 2
Let X be a pseudocompact space. Suppose that O_1,O_2,O_2,\cdots is a decreasing sequence of non-empty open subsets of X such that \bigcap_{n=1}^\infty O_n=\bigcap_{n=1}^\infty \overline{O_n}=\{ x \} for some point x \in X. Then \{ O_n \} is a local base at the point x.

Proof of Lemma 2
Let O_1,O_2,O_2,\cdots be a decreasing sequence of open subsets of X such that \bigcap_{n=1}^\infty O_n=\bigcap_{n=1}^\infty \overline{O_n}=\{ x \}. Let U be open in X with x \in U. If O_n \subset U for some n, then we are done. Suppose that O_n \not \subset U for each n.

Choose open V with x \in V \subset \overline{V} \subset U. Consider the sequence \{ O_n \cap (X-\overline{V}) \}. This is a decreasing sequence of non-empty open subsets of X. By Theorem 2 in this previous post, \bigcap \overline{O_n \cap (X-\overline{V})} \ne \varnothing. Let y be a point in this non-empty set. Note that y \in \bigcap_{n=1}^\infty \overline{O_n}. This means that y=x. Since x \in \overline{O_n \cap (X-\overline{V})} for each n, any open set containing x would contain a point not in \overline{V}. This is a contradiction since x \in V. Thus it must be the case that x \in O_n \subset U for some n. \square

The following metrization theorem is useful in proving Theorem 1.

Theorem 3 (Moore Metrization Theorem)
Let X be a space. Then X is metrizable if and only if the following condition holds.

There exists a decreasing sequence \mathcal{B}_1,\mathcal{B}_2,\mathcal{B}_3,\cdots of open covers of X such that for each x \in X, the sequence \{ St(St(x,\mathcal{B}_n),\mathcal{B}_n):n=1,2,3,\cdots \} is a local base at the point x.

For any family \mathcal{U} of subsets of X, and for any A \subset X, the notation St(A,\mathcal{U}) refers to the set \cup \{U \in \mathcal{U}: U \cap A \ne \varnothing \}. In other words, it is the union of all sets in \mathcal{U} that contain points of A. The set St(A,\mathcal{U}) is also called the star of the set A with respect to the family \mathcal{U}. If A=\{ x \}, we write St(x,\mathcal{U}) instead of St(\{ x \},\mathcal{U}). The set St(St(x,\mathcal{B}_n),\mathcal{B}_n) indicated in Theorem 3 is the star of the set St(x,\mathcal{B}_n) with respect to the open cover \mathcal{B}_n.

Theorem 3 follows from Theorem 1.4 in [1], which states that for any T_0-space X, X is metrizable if and only if there exists a sequence \mathcal{G}_1, \mathcal{G}_2, \mathcal{G}_3,\cdots of open covers of X such that for each open U \subset X and for each x \in U, there exist an open V \subset X and an integer n such that x \in V and St(V,\mathcal{G}_n) \subset U.

Proof of Theorem 1

Suppose X is pseudocompact such that its diagonal \Delta=\bigcap_{n=1}^\infty \overline{U_n} where each U_n is an open subset of X \times X with \Delta \subset U_n. We can assume that U_1 \supset U_2 \supset \cdots. For each n \ge 1, define the following:

    \mathcal{U}_n=\{ U \subset X: U \text{ open in } X \text{ and } U \times U \subset U_n \}

Note that each \mathcal{U}_n is an open cover of X. Also note that \{ \mathcal{U}_n \} is a decreasing sequence since \{ U_n \} is a decreasing sequence of open sets. We show that \{ \mathcal{U}_n \} is a sequence of open covers of X that satisfies Theorem 3. We establish this by proving the following claims.

Claim 1. For each x \in X, \bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}.

To prove the claim, let x \ne y. There is an integer n such that (x,y) \notin \overline{U_n}. Choose open sets U and V such that (x,y) \in U \times V and (U \times V) \cap \overline{U_n}=\varnothing. Note that (x,y) \notin U_k and (U \times V) \cap U_n=\varnothing.

We want to show that V \cap St(x,\mathcal{U}_n)=\varnothing, which implies that y \notin \overline{St(x,\mathcal{U}_n)}. Suppose V \cap St(x,\mathcal{U}_n) \ne \varnothing. This means that V \cap W \ne \varnothing for some W \in \mathcal{U}_n with x \in W. Then (U \times V) \cap (W \times W) \ne \varnothing. Note that W \times W \subset U_n. This implies that (U \times V) \cap U_n \ne \varnothing, a contradiction. Thus V \cap St(x,\mathcal{U}_n)=\varnothing. Since y \in V, y \notin \overline{St(x,\mathcal{U}_n)}. We have established that for each x \in X, \bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}.

Claim 2. For each x \in X, \{ St(x,\mathcal{U}_n) \} is a local base at the point x.

Note that \{ St(x,\mathcal{U}_n) \} is a decreasing sequence of open sets such that \bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}. By Lemma 2, \{ St(x,\mathcal{U}_n) \} is a local base at the point x.

Claim 3. For each x \in X, \bigcap_{n=1}^\infty \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n)}=\{ x \}.

Let x \ne y. There is an integer n such that (x,y) \notin \overline{U_n}. Choose open sets U and V such that (x,y) \in U \times V and (U \times V) \cap \overline{U_n}=\varnothing. It follows that (U \times V) \cap \overline{U_t}=\varnothing for all t \ge n. Furthermore, (U \times V) \cap U_t=\varnothing for all t \ge n. By Claim 2, choose integers i and j such that St(x,\mathcal{U}_i) \subset U and St(y,\mathcal{U}_j) \subset V. Choose an integer k \ge \text{max}(n,i,j). It follows that (St(x,\mathcal{U}_i) \times St(y,\mathcal{U}_j)) \cap U_k=\varnothing. Since \mathcal{U}_k \subset \mathcal{U}_i and \mathcal{U}_k \subset \mathcal{U}_j, it follows that (St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap U_k=\varnothing.

We claim that St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing. Suppose not. Choose w \in St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k). It follows that w \in B for some B \in \mathcal{U}_k such that B \cap St(x,\mathcal{U}_k) \ne \varnothing and B \cap St(y,\mathcal{U}_k) \ne \varnothing. Furthermore (St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap (B \times B)=\varnothing. Note that B \times B \subset U_k. This means that (St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap U_k \ne \varnothing, contradicting the fact observed in the preceding paragraph. It must be the case that St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing.

Because there is an open set containing y, namely St(y,\mathcal{U}_k), that contains no points of St(St(x,\mathcal{U}_k), \mathcal{U}_k), y \notin \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n)}. Thus Claim 3 is established.

Claim 4. For each x \in X, \{ St(St(x,\mathcal{U}_n),\mathcal{U}_n)) \} is a local base at the point x.

Note that \{ St(St(x,\mathcal{U}_n),\mathcal{U}_n) \} is a decreasing sequence of open sets such that \bigcap_{n=1}^\infty \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n))}=\{ x \}. By Lemma 2, \{ St(St(x,\mathcal{U}_n),\mathcal{U}_n) \} is a local base at the point x.

In conclusion, the sequence \mathcal{U}_1,\mathcal{U}_2,\mathcal{U}_3,\cdots of open covers satisfies the properties in Theorem 3. Thus any pseudocompact space with a regular G_\delta-diagonal is metrizable. \square

Example

Any submetrizable space has a G_\delta-diagonal. The converse is not true. A classic example of a non-submetrizable space with a G_\delta-diagonal is the Mrowka space (discussed here). The Mrowka space is also called the psi-space since it is sometimes denoted by \Psi(\mathcal{A}) where \mathcal{A} is a maximal family of almost disjoint subsets of \omega. Actually \Psi(\mathcal{A}) would be a family of spaces since \mathcal{A} is any maximal almost disjoint family. For any maximal \mathcal{A}, \Psi(\mathcal{A}) is a pseudocompact non-submetrizable space that has a G_\delta-diagonal. This example shows that the requirement of a regular G_\delta-diagonal in Theorem 1 cannot be weakened to a G_\delta-diagonal. See here for a more detailed discussion of this example.

Reference

  1. Gruenhage, G., Generalized Metric Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 423-501, 1984.
  2. McArthur W. G., G_\delta-Diagonals and Metrization Theorems, Pacific Journal of Mathematics, Vol. 44, No. 2, 613-317, 1973.

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Ψ-Spaces – spaces from almost disjoint families

As the title suggests the \Psi-spaces are defined using almost disjoint families, in our case, of subsets of \omega. This is a classic example of a pseudocompact space that is not countably compact. This example is due to Mrowka ([3]) and Isbell (credited in [2]), It is sometimes called the Mrowka space in the literature. This is another example that is a useful counterexample and set-theoretic construction. This being an introduction, I prove that the \Psi-space, when it is defined using a maximal almost disjoint family of subsets of \omega, is pseudocompact and not countably compact. On the other hand, I show that if a normal space is pseudocompact space, then it is countably compact. All spaces in this note is at least Hausdorff.

A space X is countably compact if every countable open cover of X has a finite subcover. According to Theorem 3.10.3 in [1], a space X is countably compact if and only if every infinite subset of X has an accumulation point. A space X is pseudocompact if every real-valued continuous function defined on X is bounded. It is clear that if X is a countably compact space, then it is pseudocompact. We show that the converse does not hold by using the example of \Psi-space. We also show that the converse does hold for normal spaces.

Let \mathcal{A} be a family of infinite subsets of \omega. The family \mathcal{A} is said to be an almost disjoint family if for each two distinct A,B \in \mathcal{A}, A \cap B is finite. The almost disjoint family \mathcal{A} is said to be a maximal almost disjoint family if B is an infinite subset of \omega such that B \notin \mathcal{A}, then B \cap A is infinite for some A \in \mathcal{A}.

There is an almost disjoint family \mathcal{A} of subsets of \omega such that \lvert \mathcal{A} \lvert=\text{continuum}. To see this, identify \omega (the set of all natural numbers) with \mathbb{Q}=\lbrace{r_0,r_1,r_2,...}\rbrace (the set of all rational numbers). For each real number x, choose a subsequence of \mathbb{Q} consisting of distinct elements that converges to x. Then the family of all such sequences of rational numbers would be an almost disjoint family. By a Zorn’s Lemma argument, this almost disjoint family is contained within a maximal almost disjoint family. Thus we also have a maximal almost disjoint family of cardinality continuum. On the other hand, there is no countably infinite maximal almost disjoint family of subsets of \omega. See comment below Theorem 2.

Let \mathcal{A} be an infinite almost disjoint family of subsets of \omega. Let’s define the space \Psi(\mathcal{A}). The underlying set is \Psi(\mathcal{A})=\mathcal{A} \cup \omega. Points in \omega are isolated. For A \in \mathcal{A}, a basic open set of of the form \lbrace{A}\rbrace \cup (A-F) where F \subset \omega is finite. It is straightforward to verify that \Psi(\mathcal{A}) is Hausdorff, first countable and locally compact. It has a countable dense set of isolated points. Note that \mathcal{A} is an infinite discrete and closed set in the space \Psi(\mathcal{A}). Thus \Psi(\mathcal{A}) is not countably compact. We have the following theorems.

Theorem 1. Let \mathcal{A} be an infinite maximal almost disjoint family of subsets of \omega. Then \Psi(\mathcal{A}) is pseudocompact.

Proof. Suppose we have a continuous f:\Psi(\mathcal{A}) \rightarrow \mathbb{R} that is unbounded. We can find an infinite B \subset \omega such that f is unbounded on B. If B \in \mathcal{A}, then we have a contradiction since \lbrace{f(n):n \in B}\rbrace is a sequence that does not converge to f(B). So we have B \notin \mathcal{A}. By the maximality of \mathcal{A}, C=B \cap A is infinite for some A \in \mathcal{A}. Then \lbrace{f(n):n \in C}\rbrace is a sequence that does not converge to f(A), another contradiction. So \Psi(\mathcal{A}) is pseudocompact.

Theorem 2. For a normal space X, if X is pseudocompact, then X is countably compact.

Proof. Suppose X is not countably compact. Then we have an infinite closed and discrete set A=\lbrace{a_0,a_1,a_2,...}\rbrace in X. Define f:A \rightarrow \mathbb{R} by f(a_n)=n. According to the Tietze-Urysohn Theorem, in a normal space, any continuous function defined on a closed subset of X can be extended to a continuous function defined on all of X. Then f:A \rightarrow \mathbb{R} can be extended to a continuous f^*:X \rightarrow \mathbb{R}, making X not pseudocompact.

Comment
If there is a countably infinite maximal almost disjoint family \mathcal{B} of subsets of \omega, then \Psi(\mathcal{B}) is a countable first countable space and is thus has a countable base (hence is normal). By Theorem 1, it is pseudocompact. By Theorem 2, it is countably compact. Yet \mathcal{B} is an infinite closed and discrete subset of \Psi(\mathcal{B}), contradicting that it is countably compact. Thus there is no countably infinite maximal almost disjoint family \mathcal{B} of subsets of \omega. In fact, we have the following corollary.

Corollary. If \mathcal{A} is an infinite maximal almost disjoint family of subsets of \omega, then \Psi(\mathcal{A}) cannot be normal.

Reference

  1. Engelking, R., General Topology, Revised and Completed Edition, 1988, Heldermann Verlag Berlin.
  2. Gillman, L., and Jerison, M., Rings of Continuous Functions, 1960, Van Nostrand, Princeton, NJ.
  3. Mrowka, S., On completely regular spaces, Fund. Math., 41, (1954) 105-106.