How to prove: Continuous function maps compact set to compact set using real analysis?
i.e. if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, then $f([a,b])$ is closed and bounded.
I have proved the bounded part. So now I need some insight on how to prove $f([a,b])$ is closed, i.e. $f([a,b])=[c,d]$. From Extreme Value Theorem, we know that $c$ and $d$ can be achieved, but how to prove that if $c < x < d$, then $x \in f([a,b])$ ?
Thanks!
Best Answer
Let $\{V_a\}$ be an open cover of $f(X)$. Since $f$ is continuous, we know that each of the sets $f^{-1}(V_a)$ is open. Since $X$ is compact, there are finitely many indices $a_1,...,a_n$ such that $$X\subset f^{-1}(V_{a_1})\cup \cdots\cup f^{-1}(V_{a_n}).$$ Since $f(f^{-1}(E))\subset E$ for every $E\subset Y$, the above implies that $$f(X)\subset V_{a_1}\cup \cdots\cup V_{a_n}.$$ Hence, $f(X)$ is compact.