# If $A$ is compact, is then $f(A)$ compact?

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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Yes, that assumption is fine; that's exactly what it means for $y_n$ to lie in $f(A)$. – Qiaochu Yuan Feb 21 '12 at 19:04
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? – David Mitra Feb 21 '12 at 19:05
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. – Ilya Feb 21 '12 at 19:07
Producing the $(x_n)$ may require the Axiom of Choice, but that is harmless enough. – André Nicolas Feb 21 '12 at 19:08
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! – checkmath Feb 21 '12 at 19:12

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.
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. – Patrick Da Silva Feb 21 '12 at 19:41