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In a Tychonoff space X we have:

a)If $U_1\supset U_2 \supset ...$ is a decreasing sequence of nonempty open sets in $X$, then $\cap \overline{U_n} \neq \emptyset$.

b)Every countable open cover of $X$ has a finite collection whose closures cover $X$.

Problem Prove, directly, a) implies b).

My attempt. Let $\mathbf{O}=\{O_n: n\in \mathbb{N}\}$ a countable open cover of $X$. Let $U_1= \cup_{n=1}^{\infty}{O_n}, U_2=U_1- O_1, U_3=U_2-O_2,...$ We see $U_1 \supset U_2 \supset ...$ Exists $N \in \mathbb{N}$ such that $\forall m \geq N $, $U_m \subset X$ and $U_m \supset U_{m+1} \supset..$, then thoses open sets are subsets of $X$ and a decreasing sequence, then $\cap \overline{U_m} \neq \emptyset$, but I don't see how to have a subcollection of that $U_m$ cush that $X \subset \cup_{i=1}^{n_0} \overline{U_{m_i}}$. Could you help me?, please.

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Unfortunately, $U_1\setminus O_1$ need not be open, so you won’t be able to apply (a) to your sets $U_n$.

For $n\in\Bbb Z^+$ let

$$U_n=X\setminus\bigcup_{k=1}^n\operatorname{cl}O_k\;;$$

then each $U_n$ is open, and $U_n\supseteq U_{n+1}$ for each $n\in\Bbb Z^+$. There are two possibilities. If some $U_n=\varnothing$, then $\bigcup_{k=1}^n\operatorname{cl}O_k=X$, and we’re done.

Otherwise we can apply (a) to conclude that

$$\begin{align*} \varnothing&\ne\bigcap_{n\ge 1}U_n\\ &=\bigcap_{n\ge 1}\left(X\setminus\bigcup_{k=1}^n\operatorname{cl}O_k\right)\\ &=X\setminus\bigcup_{n\ge 1}\left(\bigcup_{k=1}^n\operatorname{cl}O_k\right)\\ &=X\setminus\bigcup_{n\ge 1}\operatorname{cl}O_n\\ &\subseteq X\setminus\bigcup\mathbf{O}\\ &=\varnothing\;; \end{align*}$$

this is impossible, so some $U_n$ must be empty, and $\mathbf{O}$ must have a finite subfamily whose closures cover $X$.

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