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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 $r$,

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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.

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  • $\begingroup$ Can you see this ?math.stackexchange.com/questions/171695/… $\endgroup$ – Susan_Math123 Oct 26 '15 at 4:54
  • $\begingroup$ Also, can you prove when both $T$ and $T^{−1}$ are continuous why the linear transformatiom preserves the closedness? $\endgroup$ – Susan_Math123 Oct 26 '15 at 4:57
  • $\begingroup$ @Susan The example given in that answer is more or less the same as what I give here. $\endgroup$ – user99914 Oct 26 '15 at 4:57
  • $\begingroup$ @Susan : Which definition of open/closed are you using? $\endgroup$ – user99914 Oct 26 '15 at 5:02
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    $\begingroup$ A set is closed when it's complement is open. A set is open when for every $x$ in the set you can find a ball centered at $x$ with radius $r>0$ that is contained in the set. $\endgroup$ – Susan_Math123 Oct 26 '15 at 5:06

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