# Continuous function that maps a compact set

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.

• 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." – Ko Byeongmin Oct 13 '18 at 13:10

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 $$C$$.
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.