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How can I see that singleton sets are closed in Hausdorff space? That is, why is $X\setminus \{x\}$ open?

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    $\begingroup$ Exercise. Prove the stronger theorem that every singleton of a T1 space is closed. $\endgroup$ – William Elliot Oct 30 '17 at 2:59
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For every point $a$ distinct from $x$, there is an open set containing $a$ that does not contain $x$. Call this open set $U_a$.

Then, $\displaystyle \bigcup_{a \in X \setminus \{x\}} U_a = X \setminus \{x\}$, making $X \setminus \{x\}$ open.

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  • $\begingroup$ Thank you for explaining! $\endgroup$ – HBHSU Oct 30 '17 at 0:56
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To show $X-\{x\}$ is open, let $y \in X -\{x\}$ be some arbitrary element. We want to find some open set $W$ so that $y \in W \subseteq X-\{x\}$.

But $y \in X -\{x\}$ implies $y\neq x$. By the Hausdorff property, there are open, disjoint $U,V$ so that $x \in U$ and $y\in V$. Since they are disjoint, $x\not\in V$, so we have $y\in V \subseteq X-\{x\}$, proving $X -\{x\}$ is open.

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Observe that if a$\in X-{x}$ then this means that $a\neq x$ and so you can find disjoint open sets $U_1,U_2$ of $a,x$ respectively. Hence $U_1$ $\cap$ $\{$ x $\}$ is empty which means that $U_1$ is contained in the complement of the singleton set consisting of the element x.

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Different proof, not requiring a complement of the singleton.

$y \in X, \ x \in cl_\underline{X}(\{y\}) \Rightarrow \forall U \in U(x): y \in U$.

In $T2$ (as well as in $T1$) right-hand-side of the implication is true only for $x = y$.

Therefore, $cl_\underline{X}(\{y\}) = \{y\}$ and thus $\{y\}$ is closed.

PS. Here $U(x)$ is a neighbourhood filter of the point $x$.

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Every singleton is compact. Thus, a more interesting challenge is:

Theorem   Every compact subspace of an arbitrary Hausdorff space is closed in that space.

The main stepping stone:  show that for every point of the space that doesn't belong to the said compact subspace, there exists an open subset of the space which includes the given point, and which is disjoint with the subspace.

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Let $F$ be the family of all open sets that do not contain $x.$ Every $y\in X \setminus \{x\}$ belongs to at least one member of $F$ while $x$ belongs to no member of $F.$ So the $open$ set $\cup F$ is equal to $X\setminus \{x\}.$

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