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Today a student ask me the following question regarding topological groups in the tutorial centre.

Let $H$ be a subgroup of a topological group $G$. Suppose that $U$ is an open containing the identity such that $\overline{U} \cap H$ is closed. Prove that $H$ is closed.

I am a bit rusty on point set topology so I couldn't figure out a proper solution. I tried a straightforward approach without success:

Let $x \in \overline{H}$, we want to show that $x \in H$. Since $e \in U$ then $x \in xU$, and we can take an element $y \in xU \cap H$, therefore $y = xa = h$ for some $a \in U$, $h\in H$, so $x = ha^{-1}$

Trying to prove that $a^{-1} \in H$, I want to use somehow that $\overline{U} \cap H$ is closed; but I'm stuck at this point. I wish I could dedicate more time to solve the problem and review the basic properties of topological groups, but unfortunely I am overhelmed with my own stuff.

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  • $\begingroup$ Can you let $U = H - $ the limit points? $\endgroup$ – Larry B. Oct 31 '17 at 2:30
  • $\begingroup$ What do exactly you mean by $H$ - limit points? $\endgroup$ – C. Zhihao Oct 31 '17 at 2:45
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Suppose $x\in\overline{H}$. Let $V$ be a neighborhood of $1$ such that $VV^{-1}\subseteq U$. Since $Vx$ is an open neighborhood of $X$, $x\in\overline{H\cap Vx}$. Now pick some $y\in H\cap Vx$ and multiply everything by $y^{-1}$ on the right. We conclude that $xy^{-1}\in\overline{Hy^{-1}\cap Vxy^{-1}}$. Since $H$ is a subgroup and $y\in H$, $Hy^{-1}=H$. Also, since $y\in Vx$, $xy^{-1}\in V^{-1}$ so $Vxy^{-1}\subseteq VV^{-1}\subseteq U$. Thus $$xy^{-1}\in\overline{H\cap U}\subseteq \overline{H\cap\overline{U}}=H\cap \overline{U}.$$ In particular, $xy^{-1}\in H$, so since $y\in H$ and $H$ is a subgroup, $x\in H$.

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First, a bit of point set topology: $\bar U \cap H$ closed is equivalent to $U \cap H$ is closed in $U$.

Second, by homogeneity, for every $h\in H$, there exists $U_h$ neighborhood of $h$ so that $U_h \cap H$ is closed in $U_h$ ( take $U_h = h\cdot U$)

The above means that $H$ is locally closed in $G$. This is equivalent to : $H$ is the intersection between a closed set and an open set, or, what we are really after: $H$ is open in its closure $\bar H$.

Now, $\bar H$ is a topological group ( with the induced topology. Let's now recall: an open subgroup of a topological group is also closed. Indeed, any coset will be open, so any union of cosets. Therefore $H$ is also closed in $\bar H$, and so equal to $\bar H$, closed.

Take home message: a locally closed subgroup is closed.

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Let $(x_n, n\in I)$ be a net where $A$ is a directed set. Suppose $x=lim_nx_n, x_n\in H$, write $y_n=x^{-1}x_n$, $lim_ny_n=e$, there exists $N$ such that $n>N$ implies that $y_n\in U$. Let $n_0>N$ $z_n=x_n^{-1}x_{n_0}$, $lim_nz_n=x^{-1}x_{n_0}\in U$ implies that there exists $M$ such that $n>M$ implies that $x_n^{-1}x_{n_0}\in U\cap H\subset\bar U\cap H$ we deduce that, $x^{-1}x_{n_0}=lim_{n\geq M}x_n^{-1}x_{n_0}\in \bar U\cap H$ since $\bar U\cap H$ is closed. This implies that $x^{-1}x_{n_0}\in H$ and $x\in H$.

https://en.wikipedia.org/wiki/Net_(mathematics)#Properties

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    $\begingroup$ I think you need first countability to deal with sequences; which I think in general is not true for any topological group. $\endgroup$ – C. Zhihao Oct 31 '17 at 0:12
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    $\begingroup$ Even though you added the condition of being Hausdorff (to asure uniqueness in the limits), being First countable is necessary to say that the closure of H is exactly the set of limit points of sequences with values in H. Here the statement is for general Topological groups and none of this properties should be assumed. $\endgroup$ – C. Zhihao Oct 31 '17 at 1:23
  • $\begingroup$ @C.Zhihao: The exact same argument works in general if you just use nets instead of sequences. The argument also doesn't actually assume limits are unique anywhere and so does not require $G$ to be Hausdorff. $\endgroup$ – Eric Wofsey Oct 31 '17 at 5:25
  • $\begingroup$ @EricWofsey I think first countability is still needed to use that $\lim f(x_n) = f(x)$ under a continuous function $f$. However, your solution is correct and I will take it as the solution to the OP. $\endgroup$ – C. Zhihao Oct 31 '17 at 16:43

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