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 $r$,


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.