Positive definite matrices: open cone with closure PSD

Idea. Everything is governed by the continuous function $\lambda_{\min}$. PD is $\{\lambda_{\min}>0\}$ and PSD is $\{\lambda_{\min}\ge0\}$; and shifting by $\pm\varepsilon I$ moves $\lambda_{\min}$ by exactly $\pm\varepsilon$.

Step 1: $\lambda_{\min}$ is continuous. For symmetric $A$ the Rayleigh formula gives $\lambda_{\min}(A)=\min_{|x|=1}x^TAx$. If $A,B$ are symmetric and $|x|=1$,
$$x^TAx=x^TBx+x^T(A-B)x\ge\lambda_{\min}(B)-\|A-B\|,$$
so $\lambda_{\min}(A)\ge\lambda_{\min}(B)-\|A-B\|$; exchanging $A,B$ gives
$$|\lambda_{\min}(A)-\lambda_{\min}(B)|\le\|A-B\|.$$
So $\lambda_{\min}$ is $1$-Lipschitz.

Step 2: open / closed. $\mathrm{PD}=\lambda_{\min}^{-1}((0,\infty))$ is open and $\mathrm{PSD}=\lambda_{\min}^{-1}([0,\infty))$ is closed in $\mathrm{Sym}_n$.

Step 3: $\operatorname{int}\mathrm{PSD}=\mathrm{PD}$. PD is open and contained in PSD, so $\mathrm{PD}\subset\operatorname{int}\mathrm{PSD}$. Conversely let $A\in\mathrm{PSD}$ with $\lambda_{\min}(A)=0$. For every $\varepsilon>0$ the matrix $A-\varepsilon I\in\mathrm{Sym}_n$ has $\lambda_{\min}=-\varepsilon<0$, so it is not PSD, while $\|(A-\varepsilon I)-A\|=\varepsilon$. So no neighbourhood of $A$ lies in PSD: $A\notin\operatorname{int}\mathrm{PSD}$.

Step 4: $\overline{\mathrm{PD}}=\mathrm{PSD}$. PSD is closed and contains PD, so $\overline{\mathrm{PD}}\subset\mathrm{PSD}$. If $A\in\mathrm{PSD}$, then $A+\varepsilon I$ has $\lambda_{\min}=\lambda_{\min}(A)+\varepsilon>0$ for $\varepsilon>0$, so it is in PD, and it tends to $A$ as $\varepsilon\to0$.

Remark.
- Example ($n=2$): $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ is PSD but not PD; $\begin{pmatrix}1&0\\0&-\varepsilon\end{pmatrix}$ is arbitrarily close and not PSD.
- PD is a convex cone, and the identities $\overline{\mathrm{PD}}=\mathrm{PSD}$, $\operatorname{int}\mathrm{PSD}=\mathrm{PD}$ are the standard picture behind semidefinite programming.