Closedness of sets under linear transformation

Let $Y$ be a closed subset of $\mathbb{R}^m$ (in fact $Y$ is convex and compact, but I think the extra assumptions are irrelevant). Let $A \in \mathbb{R}^{n \times n}$ be a non-singular matrix (so $A^{-1}$ exists). Let $C \in \mathbb{R}^{m \times n}$ be any matrix. Is the set $$ Y' = \{ C A x \in \mathbb{R}^m \, : \, x \in \mathbb{R}^n, C x \in Y\} $$ also closed?

Note: just to be clear, the definition of $Y'$ means $Y' = C A X = \{ C A x \in \mathbb{R}^m \, : \, x \in X \}$ where $X = \{ x \in \mathbb{R}^n \, : \, C x \in Y \}$.

Solution 1:

I think we do need $Y$ to be compact. Otherwise take $Y=\{(x,y): x>0, y\ge 1/x\}$, let $C:\mathbb R^3\to\mathbb R^2$ be the projection $(x,y,z)\mapsto (x,y)$, and $A:\mathbb R^3\to\mathbb R^3$ be the reflection $(x,y,z)\mapsto (x,z,y)$. Now $X=\{(x,y,z): x>0, y\ge 1/x\}$, hence $AX=\{(x,y,z): x>0, z\ge 1/x\}$, and $CAX=\{(x,y): x>0\}$.

Assuming $Y$ is compact, we can argue as follows. Note that $X=X_0+\ker C$ where $X_0=\{x\in (\ker C)^\perp: Cx\in Y\}$ is compact. Hence for any linear map $T$ we have $TX=TX_0+T(\ker C)$, where $TX_0$ is compact and $T(\ker C)$ is a linear subspace. Any set of the form (compact set)+(linear subspace) is closed, being the preimage of a closed set under projection.