# Prove every countable open cover of $X$ has a finite collection whose closures cover $X$, directly.

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.

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 $$X$$.