If a topological space is Hausdorff, then every point is closed. Is the converse true?

Edited: Let $G$ be a topological group and $H$ the intersection of all neighborhoods of zero. Since every coset of $H$ is closed, every point of $G/H$ will be closed. Why does that make $G/H$ Hausdorff?

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    $\begingroup$ No, take $\mathbb{R}$ with the cofinite topology. $\endgroup$ – user29743 Sep 3 '12 at 20:31
  • $\begingroup$ In fact, the part you just erased from your question is very important: look at Tomasz's answer below. $\endgroup$ – M Turgeon Sep 3 '12 at 20:44
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    $\begingroup$ @MTurgeon: I added it back, thanks :-) $\endgroup$ – Manos Sep 3 '12 at 20:48
  • $\begingroup$ @Manos Well, I think it fits very well into this question. It shows that being a topological group imposes some constraints on the topology. $\endgroup$ – M Turgeon Sep 3 '12 at 20:48

Notice that $H$ is a closed normal subgroup of $G$, for that see e.g. this for proof that it is a closed subgroup (equal to $\operatorname{cl} \{e\}$), and for normality just notice that conjugation preserves the neighbourhoods of identity (as a set), so it does preserve intersection as well.

From that we see that $G/H$ is a topological group.

It is a known fact that for topological groups, $T_0$ implies completely regular Hausdorff. Every point being closed is equivalent to $T_1$, from which $T_{3\frac {1}{2}}$, so in particular $T_2$, follows.

A proof can be found in many places, e.g. Engelking's General Topology iirc.

A short one for closed $\{e\}\implies T_2$: notice that Hausdorffness is equivalent to the diagonal being closed. But the diagonal is the preimage of identity by the map $(x,y)\mapsto xy^{-1}$.

In general, we do not have the implication, as shown by e.g. the cofinite topology on an infinite space.

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    $\begingroup$ I like this proof much better than the usual one involving symmetric neighborhoods and $VV \subseteq U$ and so on. [Not that those are hard facts, but I find it hard to remember to use them.] Very nice! $\endgroup$ – Dylan Moreland Sep 3 '12 at 20:50
  • $\begingroup$ @tomasz: Since $H$ need not be normal in $G$, $G/H$ is not necessarily a group. Thus we don't have an identity element in general. How does then the Hausdorffness of $G/H$ follow? $\endgroup$ – Manos Sep 3 '12 at 20:53
  • $\begingroup$ @DylanMoreland: I don't know much about it, but I think the proof you refer to is much more general, since it works for arbitrary uniform spaces. $\endgroup$ – tomasz Sep 3 '12 at 20:54
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    $\begingroup$ I believe this $H$ is the closure of $\{e\}$ (see this question), which is normal: a different conjugate $H'$ of $H$ would yield a smaller (check this) closed subgroup $H' \cap H \ni e$. $\endgroup$ – Dylan Moreland Sep 3 '12 at 20:57
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    $\begingroup$ It's a general fact that the closure of a normal subgroup group is normal: $gN = Ng$ implies that $g\overline{N} = \overline{gN} = \overline{Ng} = \overline{N}g$. $\endgroup$ – t.b. Sep 3 '12 at 21:10

A space $X$ has the property that all singletons are closed if and only if it is $T_1$, meaning that whenever $x,y\in X$ and $x\ne y$, there is an open set $U$ such that $x\in U$ and $y\notin U$. The definition is symmetric, so there is also an open set $V$ such that $y\in V$ and $x\notin V$, but there is no guarantee that $U$ and $V$ can be chosen to be disjoint. For example, if $X=\Bbb N$, and the open sets are $\varnothing$ and the sets whose complements in $\Bbb N$ are finite, then $X$ is $T_1$ but not Hausdorff: in this space $U\cap V\ne\varnothing$ whenever $U$ and $V$ are non-empty open sets, but for any distinct $m,n\in X$, $X\setminus\{n\}$ is an open nbhd of $m$ that does not contain $n$.


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