Integer matrices are closed and discrete, rational ones dense

Idea. The topology of $M_n(\mathbb{R})\cong\mathbb{R}^{n^2}$ is the product topology, so everything can be checked entry by entry, where it reduces to the facts that $\mathbb{Z}$ is discrete in $\mathbb{R}$ and $\mathbb{Q}$ is dense.

Step 1: discreteness. Use the norm $\|A\|_{\max}=\max_{i,j}|a_{ij}|$ (all norms are equivalent). If $A\neq B$ in $M_n(\mathbb{Z})$, some entries satisfy $a_{ij}\neq b_{ij}$ and both are integers, so $|a_{ij}-b_{ij}|\ge1$. Hence
$$\|A-B\|_{\max}\ge1.$$
So the open ball $B(A,1)$ contains no other integer matrix: every point of $M_n(\mathbb{Z})$ is isolated in $M_n(\mathbb{Z})$.

Step 2: closedness. Let $A_k\in M_n(\mathbb{Z})$ converge to $A\in M_n(\mathbb{R})$. Then $(A_k)$ is Cauchy, so for $k,l$ large $\|A_k-A_l\|_{\max}<1$, which by Step 1 forces $A_k=A_l$. The sequence is eventually constant, so its limit $A$ equals that constant and is an integer matrix. Hence $M_n(\mathbb{Z})$ contains all its limit points.

Step 3: density of $M_n(\mathbb{Q})$. Let $A\in M_n(\mathbb{R})$ and $\varepsilon>0$. For each entry $a_{ij}$ choose $q_{ij}\in\mathbb{Q}$ with $|a_{ij}-q_{ij}|<\varepsilon$ (density of $\mathbb{Q}$ in $\mathbb{R}$). Then $Q=(q_{ij})\in M_n(\mathbb{Q})$ and $\|A-Q\|_{\max}<\varepsilon$.

Remark.
- A closed discrete set has no accumulation points: any bounded part of it is finite. In particular $M_n(\mathbb{Z})\cap\{\|A\|\le R\}$ is finite for each $R$.
- $M_n(\mathbb{Q})$ is dense and countable, so $M_n(\mathbb{R})$ is separable. It also has empty interior (it is countable, and a non-empty open set of $\mathbb{R}^{n^2}$ is uncountable).