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Let $G$ be a topological group equipped with a metric $d$ and $H < G$ a subgroup of G.

Now I am dealing with the proof that $H$ is closed in $G$.

Proof Let's assume that $H$ is not closed. Therefore $G \setminus H$ is not open and in particular $H \neq G$. That means that there exists a $g \in G \setminus H$ so that for every $\epsilon > 0$ the intersection $B_g(\epsilon) \not\subset (G \setminus H)$ where $B_g(\epsilon) := \{ g' \in G | d(x,x')<\epsilon\}$. This must also mean that $B_g(\epsilon) \cap H \neq \emptyset$.

So let's take for each $n \in \mathbb{N}$ an $g_n \in B_g(\frac{1}{n}) \cap H$. By construction we get $g_n \longrightarrow g$ for $n \longrightarrow \infty$.

Since $G$ is a topological group, meaning that the group multiplication map $G \times G \to G$ and the inverse map $G \to G$ are continuous, one can conclude $g_{n+1}^{-1} g_n \to g^{-1} g = 1$ for $n \to \infty$.

And now my problem: In our proof it is said that this sequence can't stabilize, meaning $\exists N \in \mathbb{N} \forall m \geq N: g_{m+1}^{-1}g_m = g_{N+1}^{-1} g_N $. I understand why in a discrete metric space every sequence has to stabilize, so I don't need to think about this anymore. But I don't really get why this particular sequence does not stabilize.

Could someone explain this to me? Thank you in advance.

Diglett
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2 Answers2

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The point is that the sequence $g_{m+1}^{-1}g_m$ would have to stabilise as $1$, since that's the limit of that sequence. From that we can show that the sequence $g_m$ stabilises, since we keep getting $g_{m+1}=g_m$ past that point. But $g_m$ can't stabilise, because in $G$ it's tending to $g$ which is not a member of $H$ and therefore not a member of the sequence.

Chessanator
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  • To be honest, I'd have written the proof to avoid proof by contradiction, starting by taking a sequence in $H$ that converges in $G$ to $g$, then using the fact that $g_{m+1}^{-1}g_m$ is a convergent sequence in H to prove that it stabilises, using that to prove that $g_m$ stabilises, and concluding that the limit of a sequence in H lies in H. I think that way around is a lot neater and more intuitive. – Chessanator Nov 07 '17 at 00:54
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But $e\in H$ since $H$ is discrete there exists $c>0$ such that $B(e,c)\cap H=e$, there exists $N\geq 0$ such that $g_{n+1}^{-1}g_n\in B(e,c)\cap H=e$ implies $g_{n+1}=g_n, n\geq N$. We deduce that $lim_ng_n=g_N=g\in H$ contradiction.

Tsemo Aristide
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