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 $x \in f([a,b])$ ?


  • $\begingroup$ 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). $\endgroup$ – Leonardo Fontoura Mar 12 '11 at 4:38
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    $\begingroup$ @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 $f(c)=k$. $\endgroup$ – Arturo Magidin Mar 12 '11 at 4:40
  • $\begingroup$ @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... $\endgroup$ – Arturo Magidin Mar 12 '11 at 5:00
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    $\begingroup$ 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? $\endgroup$ – Pete L. Clark Mar 12 '11 at 5:21
  • $\begingroup$ I agree with Peter (since I just was reviewing my notes on this yesterday). Open cover, then finite subcover. Triangle inequality somewhere in there... $\endgroup$ – The Chaz 2.0 Mar 12 '11 at 5:35

Lindsay, what you need is the intermediate value theorem, its proof is given in wikipedia.

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    $\begingroup$ The downvotes are certainly unjustified. This answer gives exactly what the OP was looking for. $\endgroup$ – Vishal Gupta Mar 2 '17 at 20:01
  • $\begingroup$ IVT is connected sets are sent to connected set, right? $\endgroup$ – fdzsfhaS Dec 28 '18 at 5:28

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.

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    $\begingroup$ The fact that f is continuous doesn't guarantee that the image of f's inverse is open, much less is even defined. For example, f(x) = 1 is continuous but it's inverse isn't even defined. Maybe the argument here needs to be broken into more cases? $\endgroup$ – bschneidr Dec 5 '17 at 6:14
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    $\begingroup$ @BIQS No, the argument of cdhanson is correct. Here $f^{-1}$ is the preimage map, not the inverse of $f$. $\endgroup$ – Olivier Jan 12 '18 at 1:09
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    $\begingroup$ Ah, ok.Thanks for clarifying this @Olivier! $\endgroup$ – bschneidr Jan 12 '18 at 3:04

I feel using the sequence definition is far easier to understand is much more intuitive, and the proof is nice and clean. But be careful when making the statement that being closed and bounded implies compactness, because this only holds in Euclidean space (see Heine-Borel Theorem for more about this).

Let $f:M\to N$ be a continuous function and $M$ be a compact metric space. Now let $(y_n)$ be any sequence in $f(M)$ (the image of $f$). We need to show that there exists a subsequence $y_{n_{k}}$ that converges to some $y \in f(M)$ as $k\to \infty$.

We choose a sequence $(x_n)\in M$, and since $M$ is compact by definition we have that there exists a subsequence $(x_{n_{k}})$ which converges to some $p\in M$ as $k \to \infty$. Given that a continuous function preserves the convergence of sequences i.e. if $(x_n) \to p$ in $M$, then $f((x_n)) \to f(p)$ in the image, we have that $f((x_{n_{k}})) \to f(p)$. Since $f(p) \in f(M)$, we have that our image is compact and we obtain our desired result.

Hopefully this helps, and feel free to ask any more questions :).

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  • $\begingroup$ how does compactness of image follow at the end? Seems like a leap. Also does the convergent sub/sequence (xk) relate to the bolzano-weierstrass theorem? $\endgroup$ – CogitoErgoCogitoSum Feb 24 '18 at 15:43
  • $\begingroup$ @CogitoErgoCogitoSum One can prove that in a metric space a set $A$ is compact if and only if every sequence in it contains a convergent subsequence. By the way, you cannot apply Bolzano-Weierstrass in infinite-dimensional spaces. $\endgroup$ – ComFreek Jun 28 '18 at 14:15
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    $\begingroup$ Technically speaking, you have to choose a sequence from $N$ first and then tie it to the corresponding sequence in $M$. $\endgroup$ – Cebiş Mellim Aug 22 '18 at 5:13

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