Set of real symmetric matrices with signature $(2,1)$ is open Let $S$ be the space of all $3\times 3$ real symmetric matrices, let $B$ be the subset of $S$ with signature $(2,1,0)$. Show that $B$ is open in $S$ in the topology of $\mathbb{R}^6$. My thoughts: since all norms are equivalent in finite dimensional vector spaces, we use the operator topology on $S$ instead. For any matrix $A\in B$, let $\delta$ be the smallest absolute value of the eigenvalues of $A$. Then if $\|A-C\|<\delta$ in the operator norm, I feel that $C$ will not have its eigenvalues deviate from $A$ too much, therefore maintaining the same signature. Is it correct, and how to prove my last statement?

A corrected solution, using the characteristic polynomial:

For any $A\in S$, all zeros of the characteristic polynomial $f_A$ are real. Assume $A\in B$, so $f_A$ has two positive zeros (or one double zero) and one negative zero. If $A'\in S$ is close enough to $A$, then $f_{A'}$ is close to $f_A$, and since all zeros of $f_{A'}$ are real, their signs must be identical to those of the zeros of $f_A$. Thus $A'$ has the same signature.