Set of symmetric positive semidefinite matrices is closed 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.
 A: $S_+^n$ 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.
A: The space $\mathbf{R}^{n \times n}$ is a $(n^2)$-dimensional real vector space, and the space $\mathbf{S}^n$ of symmetric matrices is a linear subspace (this is easy to check). The map $\lambda_{\min} : \mathbf{S}^n \to \mathbf{R}$, given, for example, by $\lambda_{\min}(X) = \min_{\|v\| = 1}v^TXv$ is continuous (with respect to the relative topology on $\mathbf{S}^n$). Now note that 
$$
\mathbf{S}^n_+ = \{X \in \mathbf{S}^n: \lambda_{\min}(X) \geq 0\} = \lambda_{\min}^{-1}([0, \infty)),
$$
which is the continuous preimage of a closed set, thus closed. 
A: First, observe that $S^n_+$ is self-dual (e.g. see Boyd's example 2.24).
Then note that the dual cone, $K^*$ is closed and convex (since, by definition, the dual cone is the intersection of a set of closed halfspaces; and since the intersection of closed sets is closed, and since the intersection of any number of halfspaces is convex).
So, since $S^n_+ = (S^n_+)^*$, and since the latter is closed and convex, we know that $S^n_+$ is closed and convex! (and self-dual!)
