The special linear group is totally disconnected

Idea. $SL_n(\mathbb{Z})$ is a *discrete* subspace, so every subset is open in it, and a connected subset of a discrete space has at most one point.

Step 1: the subspace is discrete. Take the maximum norm $\|A\|_\infty=\max_{i,j}|a_{ij}|$ (all norms on $M_n(\mathbb{R})$ are equivalent, so the topology does not depend on this choice). If $A\neq B$ are integer matrices, some entry differs by an integer different from $0$, so
$$\|A-B\|_\infty\ge1.$$
Hence the open ball $B(A,1)$ contains no other integer matrix, and $\{A\}=B(A,1)\cap SL_n(\mathbb{Z})$ is open in $SL_n(\mathbb{Z})$.

Step 2: connected subsets are points. Let $S\subset SL_n(\mathbb{Z})$ be connected with two distinct points $A,B$. The set $\{A\}$ is open in $S$ (by step 1) and also closed in $S$ (singletons are closed), and it is neither empty nor equal to $S$. That disconnects $S$, a contradiction. So $S$ has at most one point.

Remark. The same proof shows that any subset of $M_n(\mathbb{Z})$ is totally disconnected. This is a first example of a *discrete subgroup* of a Lie group. The quotient $SL_n(\mathbb{R})/SL_n(\mathbb{Z})$ is the space of unimodular lattices, a central object of number theory.