2

Here is a lemma of John Ratcliffe's book:

Lemma: If $G$ is a topological group with a metric topology, then every discrete subgroup of $G$ is closed in $G$.

Proof. Let $\Gamma$ be a discrete subgroup of $G$ and suppose that $G − \Gamma$ is not open. Then there is a $g$ in $G − \Gamma$ and $g_n$ in $B(g, 1/n) \cap \Gamma$ for each integer $n > 0$. As $g_n \to g$ in $G$, we have that $g_ng_{n+1}^{-1} \to 1$ in $\Gamma$. But $\{g_ng_{n+1}^{-1}\}$ is not eventually constant, which contradicts Lemma 2. Therefore, the set $G − \Gamma$ must be open, and so $\Gamma$ is closed in $G$.

Q1: Why is it clear that $\{g_ng_{n+1}^{-1}\}$ is not eventually constant?

Q2: Is metric topology important here? (I think being Hausdorff suffices.)

YCor
  • 17,878
C.F.G
  • 8,523

1 Answers1

2

If ${g_n g_{n+1}^{-1}}$ were eventually constant (let's say $g_n g_{n+1}^{-1} = g_m g_{m+1}^{-1}$ for all $n,m \geq N$, where $N$ is some fixed natural number), then the limit of the sequence (which is unique because metric topologies are Hausdorff) would be $g_N g_{N+1}^{-1}$. At the same time, the limit of the sequence is $1$, so $g_n g_{n+1}^{-1} = 1$ for all $n \geq N$. This means that $g_n = g_{n+1}$ for all $n \geq N$, and thus $g_n \to g_N$. Since $g_n \to g$, w have $g = g_N$. This is a contradiction because $g \in G - \Gamma$ but $g_N \in \Gamma$.

Edit: to answer your second question, the metric is actually used in this argument to get the sequence $g_n$. But you're right that Hausdorff is all that you really need: suppose $G$ is Hausdorff and $\Gamma$ is a non-closed discrete subgroup. Then there is some limit point $g$ of $\Gamma$ such that $g \notin \Gamma$. Take a sequence $g_n \to g$ with each $g_n \in \Gamma$ and use the same argument as in this proof. It's perhaps a little harder to see that $\{g_n g_{n+1}^{-1}\}$ must be convergent, but it's still doable:

Let $U$ be an open set containing $1$. Let $V = \{(x,y) \in G \times G : xy^{-1} \in U\}$. $V$ is also open because $G$ is a topological group. The sequence $\{(g_n, g_{n+1})\}$ converges to $(1,1)$, so there is some $N \in \mathbb{N}$ such that $(g_n, g_{n+1}) \in V$ for all $n \geq N$. This means that $g_n g_{n+1}^{-1} \in U$ for all $n \geq N$, and since $U$ was an arbitrary open neighborhood of $1$, we have that $g_n g_{n+1}^{-1} \to 1$.

Edit 2 The first edit is wrong – the result is still true in general when $G$ is Hausdorff, but my proof assumes that "$\Gamma$ is not closed" implies "there exists a sequence in $G - \Gamma$ which converges to a point in $\Gamma$", which is not true in all topological spaces! We need to use a different argument; thanks to @HennoBrandsma for linking to this relatively simple approach.

diracdeltafunk
  • 14,911
  • 1
    In general you cannot get a sequence but will get a net. I don't see an analogue for $g_n g_{n+1}^{-1}$ as clearly in that general case. So metric seems really needed. – Henno Brandsma May 31 '20 at 07:34
  • Ah you're right, for my argument we actually need $G$ to be a sequential space; equivalently $G$ needs to be a quotient of some metric space. I'm pretty sure the result holds for general Hausdorff groups though – I'll try to come up with a different proof or counterexample. Thanks! – diracdeltafunk May 31 '20 at 07:38
  • 1
    Maybe this helps? It does seem to be a general Hausdorff fact but the proofs are more tricky, and the OP works in a metric setting anyway.. – Henno Brandsma May 31 '20 at 07:44
  • Good find, thank you! – diracdeltafunk May 31 '20 at 07:47
  • 1
    I find the proofwiki proof quite readable and short in this case. (not always :)) – Henno Brandsma May 31 '20 at 07:50