$\partial (Q A) = Q (\partial A$) for an orthogonal matrix $Q$?

Question

Let $A \subset \mathbb{R}^d$ and let $Q \in \mathbb{R}^{d \times d}$ be an orthogonal matrix. For a set $B \subset \mathbb{R}^d$, denote $Q B:= \{ Qx : x \in B\}$. Does it hold for the boundary of the set $QA$ that $\partial (Q A) = Q (\partial A$)?

Answer 1

Observe that since $Q$ is a linear mapping then it is continuous, and since it is invertible with a continuous inverse $Q^T$, then it is a homeomorphism. Observe too that $\partial A = \overline{A} - Int(A)$. It follows that $Q(Int(A)) = Int(QA)$ and $\overline{QA} = Q(\overline{A})$. Therefore

$$ \partial(QA) \;\; =\;\; \overline{QA} - Int(QA) \;\; =\;\; Q(\overline{A}) - Q(Int(A)) \;\; =\;\; Q(\overline{A} - Int(A)) \;\; =\;\; Q(\partial A). $$

Answer 2

Yes, it holds. The map $Q:\mathbb{R}^n\to\mathbb{R}^n$ is a homeomorphism and hence commutes with taking closures and interiors.