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.
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.