I just got my exam back, and I still cannot understand this question:

Given a continuous function $f:A\subseteq\mathbb{R}\to\mathbb{R}$, show that if $A$ is a compact set, then its image, $f(A)$, is also compact.

I know that a set $A\subseteq\mathbb{R}$ is compact if every sequence in $A$ has a subsequence that converges to a limit that is also in $A$, and I know that a function $f$ is continuous on $A$ if for every $(x_n)\subseteq A$ such that $x_n\to c\in A$, it follows that $f(x_n)\to f(c)$. Therefore, all that I need to do is show that for every $(y_n)\subseteq f(A)$, there is a subsequence $(y_{n_k})$ such that $y_{n_k}\to y\in f(A)$.

Can I then make the assumption that for any sequence $(y_n)\subseteq f(A)$, there is a sequence $(x_n)\subseteq A$ such that $y_n=f(x_n)$? If so, I could then continue by stating that since $A$ is compact, there is a subsequence $(x_{n_k})$ such that $x_{n_k}\to x\in A$, and since $f$ is continuous, $f(x_{n_k})\to f(x)$. I believe that this yields the required subsequence $(y_{n_k})$ of $(y_n)$ such that $y_{n_k}=f(x_{n_k})\to f(x)=y\in f(A)$.

What do you guys think? Is this a sound approach? Thanks in advance.

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    $\begingroup$ Yes, that assumption is fine; that's exactly what it means for $y_n$ to lie in $f(A)$. $\endgroup$ Feb 21, 2012 at 19:04
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    $\begingroup$ That's correct. What were you concerned with in regards to the exam, though? Did you offer that argument on the exam? What feedback did you get? $\endgroup$ Feb 21, 2012 at 19:05
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    $\begingroup$ Of course, you can make such assumption since $f(A) = \{y\in \mathbb R: \exists x \in A\text{ s.t. }y = f(x)\}$ by definition. All the rest follows from the continuity of $f$. Your idea seems to be right - please write formally your proof. $\endgroup$
    – Ilya
    Feb 21, 2012 at 19:07
  • $\begingroup$ Producing the $(x_n)$ may require the Axiom of Choice, but that is harmless enough. $\endgroup$ Feb 21, 2012 at 19:08
  • $\begingroup$ You idea is correct, write it down in exactly the way you put in the exam, so we can see if it was correct when you put there! $\endgroup$
    – checkmath
    Feb 21, 2012 at 19:12

2 Answers 2


That's perfectly correct. The only qualm I can imagine any reasonable grader would have with your argument, is that you are using the fact that compact metric spaces are sequentially compact; whereas, the result can by proven directly from the open cover definition of compactness: if $\cal A$ is an open cover of $f(A)$, then ${\cal B}=\{f^{-1}(A) |A\in {\cal A}\}$ is an open cover of $A$. Extract a finite subcover $\{f^{-1}(A_1),\ldots,f^{-1}(A_k)\}$ of $A$ from $\cal B$ to obtain the finite subcover $\{A_1\ldots A_k\}$ from $\cal A$ of $f(A)$.


There is a slightly different approach that doesn't depend on sequential compactness: Let $({{U_{\alpha}}})_{\alpha \in J}$ be an countable open cover of $f(A)$. Since $f^{-1}f(A) \supseteq A$, $A$ is contained in the inverse image of the open cover; in other words, $A \subseteq f^{-1}(\bigcup_{\alpha \in J}{{U_{\alpha}}}) = \bigcup_{\alpha \in J}f^{-1}({U_\alpha})$. $f$ is continuous, so these sets are still open in $A$.

As $A$ is compact and we have an open cover of $A$, there exists a finite subcover of the $U_{\alpha}$ that covers $A$; call it $(f^{-1}(A_1), ..., f^{-1}(A_n))$. Now $f(A) \subseteq f(\bigcup_{i = 1}^{n} f^{-1}(A_i)) \subseteq \bigcup_{i=1}^{n}A_i$, and we have a finite subcover for our open cover of $f(A)$. As this holds for a general open cover of $f(A)$, the space is compact.

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    $\begingroup$ You said "countable open cover of $f(A)$". This is not enough to show that $f(A)$ is compact ; you need an arbitrary open cover of $f(A)$. But your proof works just fine if you remove the word "countable", because you never assume it. Understand why it doesn't have to go there, and then remove this word, please. $\endgroup$ Feb 21, 2012 at 19:41
  • $\begingroup$ haha yes, it is of course a blatant typo. I was trying to finish the answer before anyone commented, but I failed at that as well. With all due respect, I'm leaving it up this way, because: a) if I remove the errant word, your comment won't make any sense, b) this is essentially David's answer but submitted slightly later so it's innocuous, and c) the proof by showing that R is second-countable, and as f(A) is the subspace of a second-countable space, it is second-countable, hence Lindelof, with compactness following from the above, is hilarious to me. Point-set topology, a la Linderholm. $\endgroup$
    – JakeR
    Feb 22, 2012 at 7:03

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