# Continuous image of compact sets are compact

How to prove: Continuous function $$f : M\to N$$ maps compact set to compact set?

1. The simplest case in real analysis: if $$f: [a,b] \rightarrow \mathbb{R}$$ is continuous, then I need to show that $$f([a,b])$$ is closed and bounded, by the Heine-Borel theorem.

I have proved the boundedness, and I need some insight on how to prove that $$f([a,b])$$ is closed, i.e. $$f([a,b])=[c,d]$$. From the Extreme Value Theorem, we know that $$c$$ and $$d$$ can be achieved, but how to prove that if $$c < x < d$$, then $$x \in f([a,b])$$ ?

1. What if $$M, N$$ are metric spaces?
2. What if $$M$$ and $$N$$ are general topological spaces?
• 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). Mar 12, 2011 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 $f(c)=k$. Mar 12, 2011 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... Mar 12, 2011 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? Mar 12, 2011 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... Mar 12, 2011 at 5:35

This is a proof assuming that $$X$$, $$Y$$ are both general topological spaces.

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 $$X\subset f^{-1}(V_{a_1})\cup \cdots\cup f^{-1}(V_{a_n}),$$ where $$f^{-1}$$ represents the preimage map of $$f$$. 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.

• 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? Dec 5, 2017 at 6:14
• @BIQS No, the argument of cdhanson is correct. Here $f^{-1}$ is the preimage map, not the inverse of $f$. Jan 12, 2018 at 1:09
• Ah, ok.Thanks for clarifying this @Olivier! Jan 12, 2018 at 3:04
• So this is valid for $X$ and $Y$ being any two metric (or even topological) spaces right, where $f: X \rightarrow Y$? Seems so from the proof. Let me know if I've understood something incorrectly. Jul 26, 2021 at 2:38
• Updated to reflect @Olivier’s comment about the preimage of $f$. Feb 6 at 21:12

In this answer we assume that $$M, N$$ are metric spaces.

I feel using the sequence definition is far easier to understand is much more intuitive, and the proof is nice and clean.

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.

• 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? Feb 24, 2018 at 15:43
• @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. Jun 28, 2018 at 14:15
• Technically speaking, you have to choose a sequence from $N$ first and then tie it to the corresponding sequence in $M$. Aug 22, 2018 at 5:13

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

• The downvotes are certainly unjustified. This answer gives exactly what the OP was looking for. Mar 2, 2017 at 20:01
• IVT is connected sets are sent to connected set, right? Dec 28, 2018 at 5:28
• I didn't downvote, but, really, connectivity has nothing to do with the reasons such a result holds... Mar 11, 2023 at 22:32