# Proving continuous image of compact sets are compact

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!

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Use the fact that $f$ is continuous plus the Mean Value Theorem (I'm not sure if this is the actual name in english, sorry). –  Leonardo Fontoura Mar 12 '11 at 4:38
@Leonardo: The Mean Value Theorem says that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in(a,b)$ such that $f'(c)(a-b) = f(a)-f(b)$. I think you mean the Intermediate value theorem, which says that if $f$ is continuous on $[a,b]$, and $f(a)\leq k\leq f(b)$, then there exists $c\in[a,b]$ such that $f(c)=k$. –  Arturo Magidin Mar 12 '11 at 4:40
@Lindsay: So you want to show that for every point $k$ between $c$ and $d$, there is a point $c$ in $[a,b]$ such that $f(k)=c$... Sounds like the Intermediate Value Theorem to me... –  Arturo Magidin Mar 12 '11 at 5:00
May I ask why you are not trying to prove this using the very definitions of compactness and continuity? E.g., start with an open covering of the image of $f$, pull it back to an open cover of the domain of $f$, extract a finite subcover, and so forth. Isn't this more straightforward than going through (twice!) the Heine-Borel characterization of compact subsets of $\mathbb{R}^n$ as those that are closed and bounded? –  Pete L. Clark Mar 12 '11 at 5:21
I agree with Peter (since I just was reviewing my notes on this yesterday). Open cover, then finite subcover. Triangle inequality somewhere in there... –  The Chaz 2.0 Mar 12 '11 at 5:35

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 $$X\subset f^{-1}(V_{a_1})\cup ...\cup f^{-1}(V_{a_n}).$$ Since $f(f^{-1}(E))\subset E$ for every $E\subset Y$, the above implies that $$f(X)\subset V_{a_1}\cup...\cup V_{a_n}.$$ Hence, $f(X)$ is compact.