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Let $X$ be a mertic space, and let $\{V_{\alpha}\}$ be a collection open subsets of $X$ such that, for every $x \in X$ and for every open set $G \subset X$ with $x\in G$, there is some $V_\alpha$ such that $x \in V_\alpha \subset G$. Then the collection $\{V_\alpha\}$ is said to be a base for $X$.

Now suppose $X$ has a countable base $\{V_1, V_2, V_3, \ldots\}$, and let $\{O_\beta\}$ be an open cover of $X$; that is, let $\{O_\beta\}$ be some collection of open sets such that $X \subset \cup_{\beta} O_\beta$.

Then how to show that some countable subcollection of $\{O_\beta\}$ also covers $X$?

Of course, every element $x \in X$ is in some $O_\beta$, which in turn is a union of some subcollection of the countable collection $\{V_1, V_2, V_3, \ldots\}$. What next? How to obtain a countable subcollection of $\{O_\beta\}$?

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well you need the axiom of choice. since $\{V_\alpha\}$ is countable, and is a basis, for every $V_i$ choose $O_i \in \{O_{\beta}\}$ satisfying $V_i \subset O_i$. we may write $O_i = \psi(V_i)$

every $x \in X$ is covered by at least one member of $\{O_{\beta}\}$. since this is open it must contain some $V_x$ with $x \in V_x$. now take the corresponding $O_x = \psi(V_x)$

the collection $\{O_x\}_{x \in X}$ is countable and covers $X$

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    $\begingroup$ You don’t actually need the full strength of the axiom of choice; the result is actually equivalent to $CC(\Bbb R)$, the axiom of countable choice for subsets of $\Bbb R$, which says that if $\{A_n:n\in\Bbb N\}$ is a family of non-empty subsets of $\Bbb R$, then $\prod_{n\in\Bbb N}S_n\ne\varnothing$. More important, it’s not true that every member of the base is contained in some member of the open cover. For each $i$ let $\mathcal{O}_i=\{O\in\mathcal{O}:V_i\subseteq O\}$, and then choose one member of each non-empty $\mathscr{O}_i$. $\endgroup$ – Brian M. Scott Oct 11 '14 at 5:15
  • $\begingroup$ thank you Brian. i was hoping someone might clarify what weaker assumption would suffice. and thank you also for picking up on the empty set loose end in my argument. i did have a feeling of unease, but being still somewhat of a tyro, i ignored it and blundered on $\endgroup$ – David Holden Oct 11 '14 at 5:22
  • $\begingroup$ You’re welcome. $\endgroup$ – Brian M. Scott Oct 11 '14 at 5:28
  • $\begingroup$ David Holden, how do we know that for every $V_i$, there does exist a \textit{single} $O_\beta$ that covers this $V_i$? $\endgroup$ – Saaqib Mahmood Oct 15 '14 at 0:03
  • $\begingroup$ good point. Brian spotted my error concerning that unjustified assumption and has given the correct treatment in his comment above. $\endgroup$ – David Holden Oct 15 '14 at 0:14
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What you are asking is to prove that every second countable space is Lindelof (in more common notation).

So, let's prove $\text{Second Countable}\implies\text{Lindelof}$.

Let $X$ be second countable with countable basis $\mathscr{B}$, and let $\Omega=\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an open cover for $X$. By assumption, for each $\alpha\in\mathcal{A}$ we can cover $U_\alpha$ with some collection $B_\alpha$ of elements of $\mathscr{B}$. Note then that $\displaystyle \Sigma=\bigcup_{\alpha\in\mathcal{A}}B_\alpha$ is a countable open cover for $X$. So, for each element $O$ of $\Sigma$ choose an element $U$ of $\Omega$ containing it. Then, this subset, call it $\Gamma$, of $\Omega$ is an open cover of $X$ (since its union contains the union over all the elements of $\Sigma$ which is $X$) and is countable since there is a surjection $\Sigma\to\Gamma$ and $\Sigma$ is countable. Thus, $\Gamma$ is our desired countable subcover of $\Omega$.

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  • $\begingroup$ Paul, thank you for your answer, but are you sure you haven't made a mistake with the symbols and notation you've used? I'm afraid I'm not being able to fully make sense of your answer. $\endgroup$ – Saaqib Mahmood Oct 15 '14 at 0:01

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