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I am trying to proof the statement below:

Given an $m\times n$ matrix $A$ and a closed set $C \subset R^n$, show that the set $\{Ax, x \in C\}$ is closed.

However, I am not able to proof that statement.

Can someone help me?

Andrés E. Caicedo
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Danilo Perl
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  • Ow, so how can I prove it to be true? Now I'm even more lost lol. I edited the post :) – Danilo Perl Oct 10 '20 at 23:51
  • I apologize for my misleading comment earlier. Here is a counter-example. Let $C={(x,y): xy=1}$. This set is closed. Let $A$ be the matrix corresponding to the projection map $(x,y) \to x$. Then the image is $\mathbb R \setminus {0}$ which is not closed. – Kavi Rama Murthy Oct 11 '20 at 00:06
  • Oh, do not worry. Thanks for the answer :) – Danilo Perl Oct 11 '20 at 01:47

1 Answers1

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We can use the example from this great answer. Choose $$ C = \{(x, y) : y \ge e^x\}\subset \mathbb R^2 $$ so that $C \subseteq \mathbb{R}^2$. [You will want to actually prove that this set is closed, which is time consuming, but not difficult.] Let $A$ be the transformation given by the matrix (with respect to the standard basis) $$ A= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} $$ Notice what $A$ does to a point $(x,y) \in \mathbb{R}^2$: $$ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} 0 \\ y \end{pmatrix} $$ That is, $A$ is the projection onto the $y$-axis. It does not take a bit of thought to see that $AC= \{A x \colon x \in C\}$ is the set $\{(0,y) \colon y \in (0,\infty) \}$. But this set is not closed in $\mathbb{R}^2$. For instance, the point $(0,0) \notin AC$ but every open neighborhood of $(0,0)$ contains points of $AC$. The result is true if we choose the transformation $A$ to be invertible. See the solution linked.

mathematics2x2life
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