I have read that the interior of $S^n_+$ in $S^n$ (real symmetric matrices of size $n\times n$, $n \in \mathbb{N}^*)$ is $S^n_{++}$. I am trying to prove it.
I have that there is no open set bigger than $S^n_{++}$ contained in $S^n_{+}$: if there were such a set, call it $V$, then taking $A \in V \setminus S^n_{++}$ and $\epsilon > 0$ small enough, $A - \epsilon I_n$ would be in $V$ (by openness), hence in $S^n_{+}$ but not in $S^n_{++}$.
But $0$ being an eigenvalue of $A$, $-\epsilon$ would be an eigenvalue of $A - \epsilon I_n$, absurd because this matrix is in $S^n_{+}$ hence has all eigenvalues $\geq 0$.
However I am struggling to prove that $S^n_{++}$ is open. I am tempted to say that $A\mapsto Sp(A)$ is continuous, because $A \mapsto \chi_A$ is continuous (the coefficients of $\chi_A$ are polynomials in coefficents of $A$) and roots of polynomials are continuous wrt the coefficents of the polynomial.
However i have also read that the interior of $S^n_{+}$ in $\mathcal{M}_n(\mathbb{R})$ is empty, and I don't see what goes wrong in my proof in this case, so I guess there is a mistake. Can someone pinpoint it and provide a correct proof, hopefully in the same spirit?
Your argument correctly allows us to deduce that $S^{n}_{++}$ is an open set relative to the set of symmetric matrices. Another argument that works is to note that $A \mapsto \det(A)$ is continuous.
Of course, $S^n_{++}$ is not open relative to the larger space $\mathcal M_n$. In particular, the symmetric matrices form a proper subspace of $\mathcal M_n$, and it follows that symmetric matrices has an empty interior within $\mathcal M_n$, as does any subset of the symmetric matrices.