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Suppose that $X$ has countable basis, then every open covering of $X$ contains a countable subcollection covering $X$.

Proof

{$U_i;i\in I$} is open cover for ($X,\tau$) and $B$ = {$V_n;n\in N$} is countable base for topology $\tau$.We have that $X=\cup_i U_i$.For every point $x$ in $X$ let's choose any $U_i$ s.t. $x \in U_i$.Denote that $U_i$ as $U_x$.There exists $V_x \in B$ s.t. $x\in V_x\subset U_x$.Collection of $V_x$ is countable.For every $V_x$ choose $U_x$ s.t. $V_x\subset U_x$.Because $X=\cup_xV_x$ follows that $X=\cup U_x$ and number of $U_x$-s are countable.

I understood proof only last part I can't understand.Why we need that number of $U_x$ to be countable?

We have already open cover for $X$ and found {$V_x$} which is countable subcollection covering $X$.

unit 1991
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  • My answer in the duplicate shows how we use (only) countable choice instead of full choice as in your proof. The basic $V_x$ are not the subcover, but you need to pick supersets from the original cover you start with. – Henno Brandsma Nov 20 '21 at 23:14

1 Answers1

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{$U_i;i\in I$} is open cover for ($X,\tau$) and $B$ = {$V_n;n\in N$} is countable base for topology $\tau$.We have that $X=\cup_i U_i$.For every point $x$ in $X$ let's choose any $U_i$ s.t. $x \in U_i$.Denote that $U_i$ as $U_x$.There exists $V_x \in B$ s.t. $x\in V_x\subset U_x$.

Now as these $\{V_{x}\}$ which correspond to the $U_{x}$'s is a subcollection of $B$. Hence is countable as a subset of countable set is countable. Now it's just a matter of indexing. Explicitly we were looking at $U_{0}=\{U_{x}\in \{U_{i}\}_{i\in I} : x\in V_{x}\subset U_{x}\}$. correspondingly we have $B_{0}=\{V_{x}: V_{x}\subset U_{x}\}$. These sets $B_{0}$ and $U_{0}$ are in bijection and $B_{0}$ being a subset of countable set is countable.

Write out all such $V_{x}$'s as a sequence say $\{V_{1},V_{2},.....\}$. Correspondingly we have $\{U_{1},U_{2},....\}$.

Thus $\{U_{k}\}_{k=1}^{\infty}$ is a countable subcover of $\{U_{i}\}_{i\in I}$

Mr.Gandalf Sauron
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  • But {$V_x$} is also countable subcover of $U_i,i \in I$.Why we need another ${U_{k}}_{k=1}^{\infty}$? – unit 1991 Nov 20 '21 at 19:48
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    You have confused the definition of subcovers. It is not asked that whether we can find a countable subcover by basis elements. We need a countable subcover of $U_{i}$'s that is we need a countable subcollection of $U_{i}$. Recall what we do to prove compactness, we take an arbitrary cover $U_{\alpha}$ and show there exists $U_{\alpha_{1}},U_{\alpha_{2}},...U_{\alpha_{n}}$ such that they cover $X$. I cannot just take another basis and say that they do so. That would prove absolutely nothing – Mr.Gandalf Sauron Nov 20 '21 at 20:25
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    When you are asked to find a subcollection of some collection of sets you necessarily have to form that subcollection by members of that original collection itself. – Mr.Gandalf Sauron Nov 20 '21 at 20:27
  • It now makes sense,thank you! – unit 1991 Nov 20 '21 at 20:30