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Please help me to prove the following statement:

In a first countable $T_1$ space, every point is a $G_\delta$ set.

A space is first countable if it has a countable base. Also in a $T_1$ space every singleton is closed. So how can I use these to prove the above statement?

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    You might also take a look at [$G_\delta$ singletons in compact Hausdorff and first countability](http://math.stackexchange.com/questions/240472/g-delta-singletons-in-compact-hausdorff-and-first-countability) for sort of the converse statement.2012-11-22

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First remember that a space is first countable if it has a countable base locally, that is, for any point $x$, there is a countable collection of open sets such that any open set containing $x$ contains one of these open sets. Edit: These open sets are neighborhoods of $x$, not just any open sets. Thanks to Stefan H. for pointing this out.

A $G_\delta$ set is a set that is a countable intersection of open sets. Suppose you're given a single point $x$ in a first-countable $T_1$ space. What is a natural countable collection of open sets to try to intersect to get $\{x\}$? Why does this give you only $\{x\}$? (Hint: you haven't used the $T_1$ property yet).

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    @StefanH.: That's right, my mistake.2012-11-22
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Let $X$ first-countable and $T_1$ and $x\in X$. Suppose that $\{V_1,V_2,... \}$ be a countable basis at point $x$. We claim that $\bigcap V_i=\{x\}$.

If $y\in \bigcap V_i$ where $x\not=y$, then there is an open nbhood $U$ of $x$ which does not contain $y$. But then some $V_i$ is contained in $U$ and also does not contain $y$. It follows that $y=x$. Hence $\{x\}$ is a $G_\delta$-set.

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A first countable space is a space which has countable basis at each point, not necessarily globally.

Hint: Choose any point $x_0$, and a countable basis $U_n$ of neighbourhoods of $x_0$. What is $\bigcap_n U_n$?