A linear transform of a closed set $E\subset \mathbb{R}^d \to \mathbb{R}^d$ is closed.
I have seen a lot of similar questions here, but none of them exactly addresses the issue. Please if you find it duplicate make sure it is and comment about it.
A set is closed if it's complement is open. A set $E\subset \mathbb{R}^d$ is open if for every $x\in E$ there exists $r>0$ with $B_r(x)\subset E$, where $B_r(x)$ is a ball centered at $x$ with radius $r$,
Best Answer
It is not true. Let $$E = \{(x, y) : y \ge e^x\}\subset \mathbb R^2$$ and $T : \mathbb R^2 \to \mathbb R^2$ be given by $T(x, y) =(0, y)$. Then the image of $E$ is $\{0\} \times (0,\infty)$, which is not closed.
On the other hand, if you assume that $T$ is invertible, then it is true as in this case, both $T$ and $T^{-1}$ are continuous, so in particular, for all sets $A$, $T(A)$ is closed whenever $A$ is closed.
Recall that a mapping $T$ is continuous if $T^{-1} A$ is closed whenever $A$ is closed. So in our case, as $T^{-1}$ is continuous, $$T(E) = \{ Tx \in \mathbb R^n : x\in E\} = \{ x\in \mathbb R^n: T^{-1} x \in E\} = T^{-1}(E)$$ is also closed as $E$ is closed.