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Can you please help me with this question?

Let's $X$ be a topological space. Show that these two following conditions are equivalent :

  1. $X$ is hausdorff

  2. for all $x\in X$ intersection of all closed sets containing the neighborhoods of $x$ it's $\{x\}$.

Thanks a lot!

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    @William: I’m reasonably certain that Lilly means that the intersection of all closed nbhds of $x$ is $\{x\}$, ‘closed sets containing the neighborhoods of $x$’ being ‘closed sets having $x$ in their interiors’.2012-06-13

2 Answers 2

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Recall the definition of Hausdorff:

$X$ is a Hausdorff space if for every two distinct $x,y\in X$ there are disjoint open sets $U,V$ such that $x\in U$ and $y\in V$.

Suppose that $X$ is Hausdorff and $x\in X$. Suppose $y\neq x$. We have $U,V$ as in the definition. So $x\in U$ and $y\in V$ and $U\cap V=\varnothing$. Suppose that $F$ is a closed set containing an open neighborhood of $x$, intersecting this open set with $U$ yields an open neighborhood of $x$ which is a subset of $F$ so without loss of generality $U\subseteq F$. $F'=F\cap(X\setminus V)$ is closed and does not contain $y$. Furthermore $U$ itself is a subset of this closed set and $y\notin F'$. Therefore when intersecting all closed subsets which contain an open environment of $x$ we remove every other $y$, so the result is $\{x\}$.


On the other hand, suppose that for every $x\in X$ this intersection is $\{x\}$, by a similar process as above deduce that $X$ is Hausdorff as in the definition above.

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    I'd be very happy to hear about the downvote... (although "I just don't like you because of ..." is probably not a good explanation...)2017-10-16
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HINTS:

  1. If $x$ and $y$ are distinct points in a Hausdorff space, they have disjoint open nbhds $V_x$ and $V_y$, and $X\setminus V_y$ is a closed set containing $V_x$.

  2. If $F$ is a closed set containing an open nbhd $V$ of $x$, then $V$ and $X\setminus F$ are disjoint open sets.

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    How do you know that the closed set $F$ contains a nbhd $V$ of $x$?2017-09-18