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:
- To show it directly using the definition of the closed set.
- 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.
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.