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I am struggling with the question below:

Suppose that $ C \in \Re^m $ is a compact set and $f: C \rightarrow \Re^n$ is a continuous function. Prove that $f(C) := \{y|y=f(x) \text{ for some } x \in C \}$ is a closed set.

How can I prove that $ f(C) $ is closed? I have tried the following:

  1. To show it directly using the definition of the closed set.
  2. To find some contradictions resulting from assuming $f(c)$ is not closed.

... which simply made no avail thanks to my limited knowledge in topology.

I looked up the answers from the "similar questions" tab, but I couldn't understand most of the discussion. I am an Econ undergraduate, and know only the very basics of topology. (i.e. the only metric spaces I am familiar with is the Euclidean metric, etc.)

How can I prove this? Even the slightest help in any form would be appreciated.

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  • $\begingroup$ Mr. Bendit has provided me with a invaluable proof! I leave here that the "sequential compactness" he mentioned could be found in the Mathematics for Economists text (Simon & Blume), which I am pretty sure most of Econ students have. It is stated in page 271 (Theorem 12.14.) Apparently it has its own name -- "Bolzano-Weierstrass Theorem." $\endgroup$ – Ko Byeongmin Oct 13 '18 at 13:10
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You can use sequential compactness, which is typically the compactness used in real analysis and metric space settings. Recall that a set $C$ is (sequentially) compact if, given any sequence $(x_n)_{n=1}^\infty \in C$, there exists a subsequence $(x_{n_k})_{k=1}^\infty$ that converges within $C$.

Also recall that a continuous function is also sequentially continuous, meaning it maps convergent sequences to convergent sequences.

So, consider a sequence $(y_n) \in f(C)$. By definition of $f(C)$, for each $n$, there must exist an $x_n \in C$ such that $f(x_n) = y_n$. But $(x_n)$ is a sequence in $C$, so a subsequence $(x_{n_k})$ must exist that converges to some $x \in C$. The corresponding subsequence of $(y_n)$, i.e. $(y_{n_k})$, is the image of a convergent sequence, and hence is also convergent. In particular, we have $y_n \to f(x) \in f(C)$. Therefore, $f(C)$ is compact.

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  • $\begingroup$ Thank you for your hint! Now let me try this one again by myself... $\endgroup$ – Ko Byeongmin Oct 13 '18 at 1:12
  • $\begingroup$ I actually gave more than a hint, but if it helps you work it out for yourself, then I'm glad. :-) $\endgroup$ – Theo Bendit Oct 13 '18 at 1:14
  • $\begingroup$ Oh, perhaps referring to it as a mere hint was a big understatement, considering you gave me the full proof. I apologize if I annoyed you with my phrasing! I really appreciated your help :) $\endgroup$ – Ko Byeongmin Oct 13 '18 at 1:17
  • $\begingroup$ Oh no, you haven't annoyed me. I tend to much prefer hints. I gave a full solution because it sounded like your own attempts to find similar questions had led you to some less helpful paths, so I thought a full solution would help you back on track (which, by the sounds of it, you are). $\endgroup$ – Theo Bendit Oct 13 '18 at 1:20
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    $\begingroup$ Yes I got it! I can close this thread now :) $\endgroup$ – Ko Byeongmin Oct 13 '18 at 1:29

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