3
$\begingroup$

Let X be a compact metric space. Show that a continuous function $f:X\rightarrow\mathbb{R}$ is bounded.

Attempt: So compact $\implies$ closed and bounded, but it is not always the case that if $A$ is bounded that $f(A)$ is bounded. So I think I need to show that $X$ compact $\implies$ $\exists M\in\mathbb{R}$ : $|f(x)|\leq M$ (definition of a bounded function). Maybe letting $M=\sup(X)$ would work? I'm just not sure that I'm allowed to take a $\sup$ of an arbitrary metric space.

$\endgroup$
1
  • $\begingroup$ typo -- meant to say "a" continuous function. $\endgroup$
    – Emir
    Feb 12, 2012 at 21:45

4 Answers 4

8
$\begingroup$

The image of a compact set under a continuous map is compact. All compact sets in $\mathbb{R}$ are bounded. QED.

$\endgroup$
1
  • $\begingroup$ and I had just proven that the image of a compact set under a continuous map is compact. d'oh! thanks. $\endgroup$
    – Emir
    Feb 12, 2012 at 21:46
7
$\begingroup$

You can do this without using the fact that $f(X)$ is compact (though it does need the result that compactness implies sequential compactness in metric spaces):

If $f$ is not bounded, then there is a sequence $(x_n)_n$ in $X$ with $$\tag{1}|f(x_n)|>n,\quad\text{ for each }n=1,2,\ldots.$$
Since $X$ is compact, there is a subsequence $(x_{n_k})_k$ that converges to $x\in X$. Now, by $(1)$, it follows that $f(x_{n_k})$ does not converge to $f(x)$; which implies that $f$ is not continuous on $X$.


Or, you could argue as follows:

Consider the open balls $O_n=\{ y\in \Bbb R: |y|<n\}$ in $\Bbb R$. Then, since $f$ is continuous, $\{ f^{-1}(O_n): n=1,2,\ldots\}$ is an open cover of $X$. Since $X$ is compact, and since $f^{-1}(O_m)\subset f^{-1}(O_n)$ whenever $n>m$, there is an $N$ so that $f^{-1}(O_N)$ covers $X$. This implies that $f$ is bounded.

I prefer the latter argument, as a modification could be made to prove that the continuous image of a compact set is compact.

$\endgroup$
3
$\begingroup$

Your definition of compactness (closed and bounded) works for $\Bbb{R}$ and $\Bbb{R}^n$ (and other finite dimensional spaces), but it is not the general definition.

In a topological space $X$, a set $K$ is said to be compact if from any open cover ${O_\alpha}$ of $X$ (i.e. $O_\alpha$ are open set, and their union covers $K$) we can extract a finite subcover (i.e. we can choose a finite number of open sets from the initial cover, which still covers $K$).

Using this definition it is rather easy to prove that the image of a compact set through a continuous function remains compact.

So you have $f: X \to \Bbb{R}$, and $X$ is compact. Pick an open cover $(O_\alpha)$ for $f(X)$ (the image of $X$ under $f$). Then $(f^{-1}(O_\alpha))$ is an open cover for $X$.

$X$ is compact, so we can extract a finite subcover $(f^{-1}(O_k))_{k=1}^n$. From here it follows that $(O_k)_{k=1}^n$ covers $f(X)$, so we have found a finite subcover.

$\endgroup$
0
$\begingroup$

David Mitra has already said this, but it is worth saying it more explicitly, because it is a prototypical use of compactness. For each $n \in \mathbb{N}$ let $A_n$ be $f^{-1}(-n,n)$, the inverse image of the interval $(-n,n)$. This is an open cover of $X$: open because $f$ is continuous and a cover because $\mathbb{R}$ is Archimedian. Thus the cover has a finite subcover $F$. Let $m$ be maximal such that $A_m \in F$. Then $f$ is bounded by $m$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.