# Topology-first countable axiom

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 for sort of the converse statement. – Stefan Hamcke Nov 22 '12 at 17:08

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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There is a minor flaw in your formulation. The countable collection of open sets has to be a countable collection of neighborhoods of $x$. I guess that is what you meant. For example in the quotient space $\mathbb {R/Z}$ the intervalls $(n,\ n+\frac1k),\ n,k\in\mathbb Z-\{0\}$ form a countable family of open sets for the point $\{\mathbb Z\}$ with that property. But $\{\mathbb Z\}$ does not have a countable local base. – Stefan Hamcke Nov 22 '12 at 17:42
Stefan H. is right (in principle, I'm not sure about the example given...). For reference, such a collection is called a local $\pi$-base, and the corresponding property is countable $\pi$-character (denoted $\pi\chi(X)\leq \aleph_0$ using the notation of the Handbook of set-theoretic topology). – tomasz Nov 22 '12 at 18:04
@StefanH.: Correct me if I'm wrong, but doesn't the collection you specified fail to have the property by $U=(\bigcup (k-\frac{1}{\lvert k \rvert},k+\frac{1}{\lvert k\rvert})/_\sim$? – tomasz Nov 22 '12 at 18:15
For a more accessible (to me, at least) example, consider an uncountable set $X'$ and the space $X=\{x_0\}\sqcup X'$ with the topology given by basis of all cofinite subsets of $X$ containing $\{x_0\}$ and $\{x_0\}$ itself. Then for any $x\neq x_0$ we have that $\{x_0\}$ is in every neigbhourhood of $x$ (indeed, it is in any open set), but the intersection of any countable family of neighbourhoods of $x$ is cocountable, so no such family forms a basis of neighbourhoods of $x$. – tomasz Nov 22 '12 at 18:31
@StefanH.: Thanks for the correction! – Ben Nov 22 '12 at 18:31

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 $\bigcap_n U_n$?

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