My proof for "$\Gamma =\{X\in \mathbf{R}^{n\times n} \mid X \succeq 0, \text{Tr}(X)=1\}$ is compact" This problem comes from: How to prove the compactness of the set of Hermitian positive semidefinite matrices In short, we want to prove > $$\Gamma =\\{X\in \mathbf{R}^{n\times n} \mid X \succeq 0, \text{Tr}(X)=1\\}$$ is compact. Can I prove this in a terse way: > $\Gamma$ is the intersection of PSD cone, which is convex and closed, and the hyperplane $\\{X \mid \text{Tr}(X)=1\\}$, so $\Gamma$ is compact. So it looks like the following graph: !Intersection If not, which part I should say more?

You need some kind of further argument to prove boundedness, in my opinion. I propose this one.

Since the matrices are (symmetric) positive-semidefinite, their eigenvalues are positive. Since the trace is $1$, all the eigenvalues are in $[0,1]$. Therefore, if we consider the operator norm on $\Bbb R^{n\times n}$ $$\lVert A\rVert_2=\sup_{v\
e 0}\frac{\lVert Av\rVert}{\lVert v\rVert}$$ your set is bounded with respect to it since $\lVert A\rVert_2\le\max\limits_{\lambda\in\operatorname{Spec A}}\lvert \lambda\rvert$.

**Addition:** Fact is that you somehow need to prove that your subspace does not cut the cone in a bad way (recall that not only ellypses are conic sections, but parabolas and hyperbolas too!)