Let $\mathcal{K}\subseteq\mathbb{R}^n$ be a convex cone and let $\mathcal{A}\subseteq\mathbb{R}^n$ be an affine space such that $\mathcal{K}\cap\mathcal{A}\neq\emptyset$. Is it necessarily true that $$ \overline{\mathcal{K}\cap\mathcal{A}} = \overline{\mathcal{K}}\cap\mathcal{A}? $$ (Here $\overline{A}$ denotes the closure of a set $A$.)
Since any affine space $\mathcal{A}$ is closed, the containment $\overline{\mathcal{K}\cap\mathcal{A}} \subseteq \overline{\mathcal{K}}\cap\mathcal{A}$ is trivial.
No. Take $$ K = \{(x,y,z) \in \mathbb R^3 \mid z > 0 \text{ or } (z = 0 \text{ and } x = 0)\}$$ together with $$ A = \{(x,y,z) \in \mathbb R^3 \mid z = 0\}.$$