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The question

Given a matrix $A_{ m \times n}$ and a closed set $C \in \mathbb{R}^{n}$. the set $B$ defined below is closed?

$$B= \{Ax \mid x \in C\}$$

What I've tried

I believed that $B$ is a closed set, but, I could not prove it so now I'm hunting for any counterexample. I think that there is a matrix $A$ that could blow up the fact that $B$ is a closed set. I tried showing that the statement is false when $A$ is null ( filled with 0 entries) but that resulted in a close set. Any hints or advices are more than welcome .

markvs
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Occhima
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  • here is a simpler counterexample https://math.stackexchange.com/questions/1497944/a-linear-transform-of-a-closed-set-is-closed?noredirect=1&lq=1 – daw Oct 19 '20 at 15:03

1 Answers1

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No, this set is not closed in general. Not even for $C$ being a closed convex cone. Define $$ C=\{ (x,y,z): \ (z-x)^2 + y^2 \le x^2, \ x\ge0 \}, $$ which is some ice-cream cone tangent to the positive $x$-axis. Then project onto the $x=0$ plane, $A=\pmatrix{0&1&0\\0&0&1}$. Then $B=\{(y,z): z>0\} \cup \{(0,0)\}, $ which is not closed.

The set $B$ is closed if - for instance - $C$ is compact, or if $C$ is the set of vectors with non-negative entries. To prove the latter claim is quite non-trivial, see

How do you prove that $\{ Ax \mid x \geq 0 \}$ is closed?

(That question has 10 deleted 'answers'!)

daw
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