I am self-studying Boyd & Vandenberghe's Convex Optimization.
Example 2.15 (page 43) states that the symmetric positive semi-definite cone $S^n_+$ is a proper cone. This necessitates, amongst other things, that it is closed.
I am not sure how to show that $S^n_+$ is closed, particularly because this set consists of matrices, which I am less comfortable working with.
The most relevant question I have found that may have some relation to this one is here; I am not sure how to act on the answer of this question for I am not sure of whether the functions $f_1$ and $f_2$ as defined in the answer are relevant to my task.
Best Answer
$S_+^n$ closed follows quite elementarily from definition, rather than by using topological properties of the eigenvalues of a matrix.
Let $x\in\Bbb R^n$ and consider the following linear maps:
$G:\Bbb R^{n\times n}\to \Bbb R^{n\times n}$, $G(A)=A-A^t$.
$q_x:\Bbb R^{n\times n}\to \Bbb R$, $q_x(A)=x^tAx$.
Since all these maps are continuous, $$S^n_+:=\ker G\cap \bigcap_{x\in\Bbb R^n} q_x^{-1}[0,\infty)$$ must be closed, because it's intersection of closed sets. As an additional observation, this is also an intersection of preimages of convex cones by linear maps, and thus a convex cone.