Suppose $X\subseteq \mathbb{R}^3$ is a real semi-algebraic set. Consider the subspace topology induced by the Euclidean topology on the Zariski closure $\overline{X}.$ Can we guarantee that $X$ has nonempty interior in this topology?
It seems like a semi-algebraic set should have the same dimension, geometrically speaking, as its Zariski closure. Then it seems like the semi-algebraic set would have nonempty interior within its Zariski closure. I don't know enough to confirm/justify my intuition on this.