The groups GL_n and SL_n: open and closed

Idea. Both sets are defined by a condition on the single number $\det A$, and $\det$ is continuous. Topology then does all the work.

Step 1: $\det$ is continuous. By the Leibniz formula
$$\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\,a_{1\sigma(1)}\cdots a_{n\sigma(n)}$$
$\det A$ is a polynomial in the $n^2$ entries of $A$. The coordinate functions $A\mapsto a_{ij}$ are continuous (for any norm on $M_n(\mathbb{R})\cong\mathbb{R}^{n^2}$), and sums and products of continuous functions are continuous. Hence $\det:M_n(\mathbb{R})\to\mathbb{R}$ is continuous.

Step 2: $GL_n$ is open. A matrix is invertible iff $\det A\neq0$, so
$$GL_n(\mathbb{R})=\det{}^{-1}\big(\mathbb{R}\setminus\{0\}\big).$$
The set $\mathbb{R}\setminus\{0\}$ is open in $\mathbb{R}$, so its preimage under the continuous map $\det$ is open.

Step 3: $SL_n$ is closed. By definition
$$SL_n(\mathbb{R})=\det{}^{-1}(\{1\}),$$
and $\{1\}$ is closed in $\mathbb{R}$, so the preimage is closed.

Remark.
- Concretely, the openness of $GL_n$ says: if $A$ is invertible and $B$ is close enough to $A$, then $B$ is invertible. A quantitative version is in the "Inverse stability" problem of Part 1: it suffices that $\|B-A\|<1/\|A^{-1}\|$.
- $SL_n$ is closed but not open (it has empty interior, since $\det(tA)=t^n$ leaves the value $1$ as soon as $t>0$, $t\ne1$, and $tA\to A$ as $t\to1$).
- $SL_n$ is also unbounded for $n\ge2$: $\begin{pmatrix}t&0\\0&1/t\end{pmatrix}\in SL_2$ for all $t>0$. Closed does not mean compact.