This is question from my assignment i have tried to do some proof on it Let $\{x\}$ be one point set in $X$ since $X$ is countable then there exists a countable basis for $x$ since $X$ is hausdorff for any $x,y\in X$ with $x$ not equal to $y$ there is a open set $U$ containing $x$ and $V$ containing $y$ im not able to do the proof further can u help me with this proof?
show that in first countable hausdorff space every open set is G δ set?
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general-topology
self-learning
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4What, exactly, are you trying to prove? Your title indicates that you want to prove that in any first-countable Hausdorff space every open set is G-δ (which is trivially true for _any_ space if you know what G-δ means). But the body of your question seems different. Please [edit] your question so that the title and body match. – 2017-02-13
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0http://math.stackexchange.com/q/2139379/4280 is probably what you mean? – 2017-02-13
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0A countable space need not have a countable local base at any point, so that's where your suggestion fails. – 2017-02-13
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0"since $X$ is countable".... First-countable means every point in $X$ has a countable local base. It does not mean the space is countable. The Q in the title is unprovable because there are uncountable first-countable spaces in which not every open set is $G_{\delta}.$ – 2017-02-13
1 Answers
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Borrowing my own answer from here:
Suppose $X = \{x_n: n \in \mathbb{N}\}$ to make it explicitly countable. As $X$ is $T_1$ in particular, finite sets are closed, so all sets $U_n = X\setminus\{x_n\}$ are open. Now $$\{x_m\} =\cap \{U_n: n \neq m\}$$ for every $m$, making all singletons a $G_\delta$.
Note that Hausdorff is slightly overkill, only $T_1$ is used.