If every continuous function $f: A \rightarrow \mathbb{R}$ is bounded, then A is compact.

Question

I'm doing a metric spaces course and got stuck on the proposition:

If $A\subset\mathbb R^n,$ such that every continuous $f:A\to\mathbb R$ is bounded, then $A$ is compact.

I have a feeling that I want to show that if $A\subset\mathbb R^n$ is bounded and closed, then use Heine-Borel Theorem provides compactness. It just seems obvious and I seem to be going around in a circular argument.

Your question is not clear. What do you have to prove? What are $A, C$? — Crostul, Nov 04 '14 at 18:03
http://math.stackexchange.com/questions/114123/if-every-continuous-function-attains-its-maximum-then-the-metric-space-is-comp — Max, Nov 04 '14 at 18:06

Yiorgos S. Smyrlis · Accepted Answer · 2021-11-03T09:34:37.250

5

Since $A\subset \mathbb R^n$, then in order to show that $A$ is compact, it suffices to show that $A$ is bounded and closed.

$A$ is bounded. Clearly the function $f:A\to\mathbb R$, with $f(x)=\|x\|$ is continuous, and as it has to be bounded, then there exists an $M>0$, such that $f(x)=\|x\|\le M$, for all $x\in A$. But this means that $A$ is bounded.

$A$ is closed. If not, then there exists a sequence $\{x_n\}_{n\in\mathbb N}\subset A$, with $x_n\to x_0\not\in A$. Then the function $\,f:A\to\mathbb R$, with $$ f(x)=\frac{1}{\|x-x_0\|}, $$
in continuous and unbounded. A contradiction, and hence $A$ is closed.

Altogether, indeed $A$ is compact.

Note. Another way to obtain contradiction, is by observing that the function $g(x)=\|x-x_0\|$ is also continuous, but it does not attain a minimum. The infimum of $f$ is equal to zero, while $f$ takes only positive values.

edited Nov 03 '21 at 09:34

answered Nov 04 '14 at 20:22

Yiorgos S. Smyrlis

86,581

$f$ should be $f(y) = |x-y|^{-1}$. – daw Nov 04 '14 at 20:29
You are right. Thanks! – Yiorgos S. Smyrlis Nov 04 '14 at 20:42
@YiorgosS.Smyrlis, is $f(x) = \frac{1}{\Vert x- x_{0} \Vert}$ also an example of a function that does not attain its minimum? If not, what would be one? – May 03 '16 at 17:51
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@JessyCat See my edited answer. – Yiorgos S. Smyrlis May 04 '16 at 10:21
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To show $A$ is closed, can we not just consider the constant function $f:x\mapsto c$. As ${c}$ is closed in $\mathbb{R}$, the inverse image $f^{-1}({c})=A$ must be closed owing to the continuity of $f$. – Dimension Entangled Nov 03 '21 at 05:35
@DimensionEntangled Your suggestion shows that $A$ is closed... in $A$. And not as a a subset of $\Bbb R^n$. This does not help here – Didier Jan 03 '23 at 19:06

score 3 · Answer 2 · answered Nov 04 '14 at 18:18

Let $B \subseteq \mathbb{R}$ be any bounded subset. Then "$B$ attains its bounds" means that $B$ is closed.

Call $C(A)= \{ f: \mathbb{R}^n \longrightarrow \mathbb{R} \mbox{ continuous functions} \} $. Then

$$A = \bigcap_{f \in C(A)} f^{-1}(f(A))$$

Since for all $f \in C(A)$ $f(A)$ is closed, $f^{-1}(f(A))$ is closed as well (because $f$ is continuous). So, $A$ is an intersection of closed sets, hence it is closed.

Finally, consider for all $i=1, \dots, n$ the continuous functions $$\pi_i :A \longrightarrow \mathbb{R} \ \ \ \ \ (x_1, \dots, x_n) \mapsto x_i$$

Since all the sets $\{\pi_i(A) \}_{i=1}^n$ are bounded, then so is their product $\prod_{i=1}^n \pi_i(A) \subseteq \mathbb{R}^n$. But $A \subseteq \prod_{i=1}^n \pi_i(A)$, so it is bounded.

Being closed and bounded, $A$ is compact.

$B=[0,1)\cup(1,2]$, attains its bound, but it is not closed! — Yiorgos S. Smyrlis, Nov 04 '14 at 18:38
@YiorgosS.Smyrlis You are right! My proof is in some sense incomplete... =( — Crostul, Nov 04 '14 at 18:48

If every continuous function $f: A \rightarrow \mathbb{R}$ is bounded, then A is compact.

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