8
$\begingroup$

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?

$\endgroup$
  • 10
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
10
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
11
$\begingroup$

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 $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.