Continuous function that maps a compact set

Question

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.

Answer 1

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.