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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!

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  • $\begingroup$ Have you proved that (a) and (c) are equivalent? $\endgroup$ Jul 28 '20 at 17:55
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$\newcommand{\cl}{\operatorname{cl}}$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 $x\in\bigcap_{n\in\Bbb N}\cl U_n$.

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  • $\begingroup$ 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? $\endgroup$
    – Ishan Deo
    Jul 28 '20 at 18:25
  • $\begingroup$ @IshanDeo: I just went ahead and gave a proof that (a) implies (b). $\endgroup$ Jul 28 '20 at 19:03

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