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why is one point set in a first countable T1 space a Gδ? st in first countable hausdorff space every one point set is Gδ set? are they same I am asking they have a same proof

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    You just stated the same thing twice – 2017-02-14
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    One doubt what is T2 – 2017-02-14
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    $\mathsf{T1}$ means if $x,y$ are distinct points, then there are respective neighborhoods $U,V$ of $x,y$ such that $y \notin U$ and $x \notin V$. However $U,V$ don't have to be entirely disjoint from each other. On the other hand, $\mathsf{T2}$ means that if $x,y$ are distinct points, then there are respective $disjoint$ neighborhoods $U,V$ such that $y \notin U$ and $x \notin V$. Note that $\mathsf{T2}$ just means Hausdorff. – 2017-02-14
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    so In the above the property there u took T1 why? – 2017-02-14
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    In the following proof I have one doubt how will u take T1 because in my assignment i have been asked to prove for hausdorff – 2017-02-14
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    $\mathsf{T1}$ and $\mathsf{T2}$ are properties that a topological space $X$ may or may not have. If $X$ has the property $\mathsf{T1}$ we call $X$ a $\mathsf{T1}$-space, and if $X$ has the property $\mathsf{T2}$, we call $X$ a $\mathsf{T2}$-space. Now, you can easily see by definition, if $X$ has the property $\mathsf{T2}$, then it also has the property $\mathsf{T1}$ property. Thus every $\mathsf{T2}$-space is necessarily a $\mathsf{T1}$-space and anything proved for an arbitrary $\mathsf{T1}$ space must therefore be true for an arbitrary $\mathsf{T2}$ space as well – 2017-02-14
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    Thnx I understood quite completely – 2017-02-14

2 Answers 2

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Let $X$ be a first countable $\mathsf{T1}$ space. Fix an element $x \in X$. There exists a countable collection $\mathbf{U}$ of open sets about $x$ with the property that any open set $V \ni x$ is refined by some $U \in \mathbf{U}$ in the sense that $x \in U \subset V$.

Now let $E = \bigcap_{U \in \mathbf{U}}U$. Then clearly $E$ is a $G\delta$ set. Can you use the $\mathsf{T1}$ axiom to show that $E = \{x\}$?

Hint: say $y \in X \backslash \{x\}$. By the $\mathsf{T1}$ axiom there is some open set $V$ containing $x$ but not containing $y$. Refine $V$ accordingly.

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    Thanks now I got understood I was bit confused of how to take the neighbourhood – 2017-02-14
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Let $X$ be a first countable $T_1$ space. Let $x \in X$. Then $x$ has a countable local base $U_n$, which are all open and contain $x$. Claim $\{x\} = \cap_n U_n$, which shows that the set $\{x\}$ is a $G_\delta$.

The left to right inclusion is immediate. So assume we have some $y\in \cap_n U_n$ and suppose that $y\neq x$. By the $T_1$ property, there is some open set $O$ such that $x \in O, y \notin O$. As the $U_n$ form a local base for $x$, for some $m$ we have $U_m \subset O$. But then $y \notin U_m$ (as $y \notin O$) contrary to $y \in \cap_n U_n$. So the assumption that $y\neq x$ was wrong so $y= x$ and the other inclusion has been shown.