Closure of the PSD matrices of rank k

"$\subseteq$": the right-hand side is closed and contains the set.
- $\{S\ge0\}=\{x^TSx\ge0\ \forall x\}$ is an intersection of closed half-spaces, hence closed.
- $\{\operatorname{rank}S\le k\}$ is the common zero set of all $(k+1)\times(k+1)$ minors (polynomials), hence closed.

So the closure of the set of rank $k$ PSD matrices lies in $\{S\ge0,\operatorname{rank}S\le k\}$.

"$\supseteq$". Let $S\ge0$ with $s=\operatorname{rank}S\le k$. If $s=k$ there is nothing to prove. If $s<k$, the kernel $\ker S$ has dimension $n-s\ge n-k+1\ge k-s$ (we need $k-s\le n-s$, true as $k\le n$). Choose a subspace $W\subseteq\ker S$ of dimension $k-s$ and let $P$ be the orthogonal projection onto $W$ (a symmetric idempotent of rank $k-s$). Put
$$S_\varepsilon=S+\varepsilon P,\qquad\varepsilon>0.$$
- $S_\varepsilon\ge0$ since $S\ge0$ and $P\ge0$.
- $\ker S=(\operatorname{im}S)^\perp$ because $S$ is symmetric, so $W\perp\operatorname{im}S$ and $\operatorname{im}S_\varepsilon=\operatorname{im}S\oplus W$ (on $\operatorname{im}S$, $S_\varepsilon=S$ since $P$ kills $\operatorname{im}S\subseteq W^\perp$; on $W$, $S_\varepsilon=\varepsilon\operatorname{id}$). So $\operatorname{rank}S_\varepsilon=s+(k-s)=k$.
- $S_\varepsilon\to S$ as $\varepsilon\to0$.

Remark. For $k=n$ this says the positive definite matrices are dense in the PSD cone. Lesson. The PSD cone of rank $\le k$ is exactly the closure of the rank-$k$ stratum.