# Decreasing Sequence of Open Sets in Tychonoff Pseudocompact Spaces has Nonempty Intersection of Closures

Let $$(U_n)$$ be a decreasing sequence of non-empty open sets in a Tychonoff pseudocompact space $$X$$. Then, show that $$\cap \overline U_n \neq \phi$$

This was part of a problem in Willard. I was able to do the rest of the parts, but this one has still eluded me. Any help would be appreciated!

• Have you proved that (a) and (c) are equivalent? Jul 28 '20 at 17:55

$$\newcommand{\cl}{\operatorname{cl}}$$From what you said, I’m assuming that you’ve proved the equivalence of $$2$$(a) and $$2$$(c) in Exercise $$17$$J.

For $$n\in\Bbb N$$ let $$V_n=X\setminus\cl U_n$$, and let $$\mathscr{V}=\{V_n:n\in\Bbb N\}$$. If $$\bigcap_n\cl U_n=\varnothing$$, then $$\mathscr{V}$$ is an open cover of $$X$$, so there is a finite $$\mathscr{V}_0\subseteq\mathscr{V}$$ such that $$\bigcup\{\cl V_n:V_n\in\mathscr{V}_0\}=X$$. Moreover, since the sets $$U_n$$ are decreasing, the sets $$V_n$$ are increasing, so in fact there is an $$n\in\Bbb N$$ such that $$\cl V_n=X$$. But then $$U_n\subseteq X\setminus\cl V_n=\varnothing$$, which is impossible.

Added: Here’s one way to prove (b) directly from (a), which will let you complete your circle of implications. If there is an $$n_0\in\Bbb N$$ such that $$U_n=U_{n_0}$$ for all $$n\ge n_0$$, then clearly $$\bigcap_n\cl U_n\ne\varnothing$$, so we may as well assume that $$U_n\supsetneqq U_{n+1}$$ for each $$n\in\Bbb N$$, so that for each $$n\in\Bbb N$$ there is an $$x_n\in U_n\setminus U_{n+1}$$. $$X$$ is completely regular, so for each $$n\in\Bbb N$$ there is a continuous $$f_n:X\to[0,n]$$ such that $$f(x_n)=n$$, and $$f_n[X\setminus U_n]=\{0\}$$.

Suppose that for each $$x\in X$$ there are an open set $$V_x$$ and a $$k(x)\in\Bbb N$$ such that $$x\in V_x$$, and $$V_x\cap U_k=\varnothing$$ whenever $$k>k(x)$$. Then we can define a function

$$f:X\to\Bbb R:x\mapsto\sum_{n\in\Bbb N}f_n(x)\;,$$

since $$f_n(x)=0$$ for all $$k\ge k(x)$$. I’ll leave it to you to check that the function $$f$$ is continuous; use the fact that every $$x\in X$$ has an open nbhd $$V_x$$ that intersects only finitely many of the sets $$U_n$$.

This is impossible, since $$X$$ is pseudocompact, and $$f$$ is clearly unbounded. Thus, there must be some $$x\in X$$ such that every open nbhd of $$x$$ intersects infinitely many of the sets $$U_n$$. And since the sets $$U_n$$ are nested, this means that every open nbhd of $$x$$ intersects every $$U_n$$ and hence that $$x\in\cl U_n$$ for each $$n\in\Bbb N$$, so that $$x\in\bigcap_{n\in\Bbb N}\cl U_n$$.

• No, I haven't done a) $\implies$ c). I figured that we would have to prove the $3$ statements in a circular fashion, that is a) $\implies$ b) $\implies$ c) $\implies$ a), so I did b) $\implies$ c) $\implies$ a). Could you help me with a) $\implies$ c) too? Jul 28 '20 at 18:25
• @IshanDeo: I just went ahead and gave a proof that (a) implies (b). Jul 28 '20 at 19:03