Involutions form a closed unbounded set

Idea. The set is defined by $A^2=I$, so it is closed; but a non-orthogonal reflection can be sheared arbitrarily far.

Step 1: closed. $\mathcal I=\{A:A^2-I=0\}$ is the preimage of $\{0\}$ under the continuous map $A\mapsto A^2-I$.

Step 2: unbounded. For $t\in\mathbb{R}$ let
$$S_t=\begin{pmatrix}1&t\\0&-1\end{pmatrix}.$$
Then
$$S_t^2=\begin{pmatrix}1\cdot1+t\cdot0&1\cdot t+t\cdot(-1)\\0&(-1)(-1)\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$
So $S_t\in\mathcal I$ for all $t$, and $\|S_t\|\ge|t|$ (the entry $t$ is at most the norm), which tends to $\infty$. For $n>2$ use $S_t\oplus I_{n-2}$.

Remark.
- The *orthogonal* involutions ($A^2=I$ and $A^T=A$, equivalently $A\in O(n)$ symmetric) form a compact set; the unboundedness comes from non-normal examples. The set $\mathcal I$ is the union of the conjugacy classes of the matrices $\operatorname{diag}(1,\dots,1,-1,\dots,-1)$; the classes containing both signs are unbounded.
- $\mathcal I$ has empty interior too: the eigenvalues of an involution lie in $\{1,-1\}$, while $A+\varepsilon I$ has eigenvalues in $\{1+\varepsilon,-1+\varepsilon\}$, so it is not an involution for small $\varepsilon\ne0$.