Can a low-rank matrix set have nonempty interior? The answer to this question may be super simple, but it is very not obvious to me.
Consider the space $S^n$ of symmetric $n\times n$ matrices. Consider $T\subset S$ the set of rank $n-1$ matrices. Does $T$ have empty interior in $S^n$?
It seems to me that, picking $A\in T$, there is always some perturbation matrix $E$ with $|E|$ arbitrarily small that makes $A+E$ full rank, which leads me to conclude that $T$ must have empty interior in $S^n$, but people smarter than me have told me it shouldn't matter.  
Can anyone correct / affirm my logic? It's possible I just don't have a grasp on the nonempty interior concept.
Thank you!
 A: Yes, your argument is correct; contrapositively, if $T$ has nonempty interior, then there would be some point in $T$ for which there existed no perturbation that takes you out of $T$.
Another way to see this is as follows: The determinant $$\det: S \to \mathbb{F}$$ is a polynomial in the canonical coordinates $(x_{ij})_{i \leq j}$ on $S$ (here I'll assume $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$), and there are symmetric matrices with nonzero determinant, so $\ker \det$ is a proper subvariety of $S$. Thus, its complement is Zariski-open and hence dense in $S$ (in both the Zariski and the standard topology), so $\det \ker$ has empty interior.
Now, any matrix in $T$ has rank $n - 1$, and so has determinant zero. Thus, $T \subset \ker \det$, and so $T$ has empty interior because $\ker \det$.
A: Suppose $A \in T$. Then for sufficiently small $\lambda \neq 0$, we have that 
$A-\lambda I$ is invertible (since there are only a finite number of eigenvalues) and
$A-\lambda I \in S^n$. In particular, the
invertible elements of $S^n$ are dense in $S^n$, hence $T$ cannot contain an
open set.
