Orthogonal projections: compact with empty interior in Sym_n

Idea. The defining equations make $\mathcal P$ closed; boundedness comes from the norm of a projection. For the empty interior one finds symmetric matrices arbitrarily near $P$ whose eigenvalues are not in $\{0,1\}$.

Step 1: closed. $\mathcal P=\{P:P^T-P=0\}\cap\{P:P^2-P=0\}$, an intersection of zero sets of continuous maps.

Step 2: bounded. If $P=P^T=P^2$ and $|x|=1$, then
$$|Px|^2=x^TP^TPx=x^TP^2x=x^TPx=\langle Px,x\rangle\le|Px||x|,$$
so $|Px|\le1$. Thus $\|P\|\le1$ and $\mathcal P$ lies in the closed ball of radius $1$.

Step 3: compact. By the Heine-Borel theorem in $M_n(\mathbb{R})\cong\mathbb{R}^{n^2}$, a closed and bounded set is compact.

Step 4: empty interior in $\mathrm{Sym}_n$. Let $P\in\mathcal P$ and $\varepsilon>0$ small.
- If $P\neq0$: $(1+\varepsilon)P$ is symmetric, and $P$ has eigenvalue $1$, so $(1+\varepsilon)P$ has eigenvalue $1+\varepsilon\notin\{0,1\}$. A projection has all eigenvalues in $\{0,1\}$ (from $P^2=P$), so $(1+\varepsilon)P\notin\mathcal P$.
- If $P=0$: $\varepsilon I$ is symmetric, has eigenvalue $\varepsilon\notin\{0,1\}$, so is not a projection.
In both cases there are symmetric matrices at distance $\varepsilon\|P\|$ (resp. $\varepsilon$) from $P$ outside $\mathcal P$. So $P$ is not an interior point (relative to $\mathrm{Sym}_n$).

Remark.
- $\mathcal P$ is the disjoint union $\bigsqcup_{r=0}^n\mathcal P_r$ where $\mathcal P_r$ is the set of rank-$r$ projections, each compact and connected (a Grassmannian). So $\mathcal P$ has $n+1$ connected components, and the trace $\operatorname{tr}P=\operatorname{rank}P$ is locally constant on it.
- For $n=1$, $\mathcal P=\{0,1\}\subset\mathbb{R}$, which is finite and nowhere dense.