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If $A$ and $B$ are compact subset of a normed vector space then $A+ B$ is also compact space, right?

Because since $A\times B$ is compact and $f:A\times B \rightarrow A+B$ such that $f(a,b) = a +b $ is continuous by sequentially criteria.

So a continuous image of compact space is compact.

There is a similar question on Mathstack exchange but it is generalize version of normed vector space.

Tina
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Texas
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    Yes, image of a compact space by a continuous map is compact. What you said is correct. – Jakobian Sep 08 '24 at 05:18
  • I would prefer to say that $f \colon X \times X \to X$, $f(u,v)=u+v$, is continuous and $f(A \times B) = A+B$. – azif00 Sep 08 '24 at 06:29
  • @TimonKolt This question has stronger condition. Can you edit the question ? – Texas Sep 08 '24 at 07:30
  • @Texas i dont know how you want me to edit your question, i think it is best if you edit it the way you want. To be honest i think think this question should be closed because it really is a duplicate and to become a non duplicate it would need to become a totally new question. – Tina Sep 08 '24 at 08:54
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    @timonkolt okay thank you – Texas Sep 08 '24 at 10:09

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