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Theorem: Let $K$ be a subset of $\mathbb{R}.$ If $f(K)$ is bounded for any continuous function $f:K\to\mathbb{R}$, then $K$ must be compact.

The same theorem, written in quants: $\forall f\in C(K): $ $f(K)$ is bounded in $\mathbb{R}$ => $K$ is compact.
$K$ is a subset of $\mathbb{R}$, $C(K)$ is the set of all continuous functions defined on $K$

I have seen other solutions, but they use too advanced a level for me. Can you give a proof of this theorem, referring only to results not beyond introductory real analysis?

P.S. Thank you for your time!

Anne Bauval
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Kottoamatsukami
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    I would proceed with contradiction. Assume $f(K)$ is bounded for all continuous $f$, but $K$ is not compact. Then either $K$ is unbounded or $K$ isn't closed, by Heine Borel. Taking $f$ to be the identify shows that $K$ must be bounded, so it must be the case that $K$ isn't closed. Can you use this assumption to find a continuous function $f$ such that $f(K)$ is unbounded, giving you a contradiction? – Matthew H. May 31 '23 at 19:26
  • Are we assuming that $K \subseteq \Bbb R^n$ for some $n \in \Bbb N$? – Robert Shore May 31 '23 at 19:27
  • @RobertShore The question says "... $f$ defined on a subset $K$ of $\Bbb R$" and "$K$ is a subset of $\Bbb R$". – jjagmath May 31 '23 at 19:29
  • @RobertShore, no, I am interested in the case of n=1 – Kottoamatsukami May 31 '23 at 20:09
  • @MatthewH. Thanks! I will try it – Kottoamatsukami May 31 '23 at 20:10
  • "Not beyond introductory analysis"? But what is your (or better, if possible: what are your equivalent) definition(s) of a compact subset of $\Bbb R,$ then? And of a closed subset? Do you have any problem with this answer? (for $n=1$ if you like). – Anne Bauval May 31 '23 at 20:41
  • https://math.stackexchange.com/q/668905/1093844 – Soumik Mukherjee May 31 '23 at 20:44
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    @SoumikMukherjee That link is of far too advanced level for the OP (contrarily to the one proposed just before) – Anne Bauval May 31 '23 at 20:45
  • @AnneBauval
    1. The compact subset was introduced as follows:

    A set $A$ is called compact if a finite set can be extracted from any of its covers by open sets. Then the theorem was introduced: $K$ is a compact if $K$ is bounded and closed

    1. A subset is called closed if its complement is an open set
     – Kottoamatsukami May 31 '23 at 20:49
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    @AnneBauval, yes the OP should check the link that you have shared. I just shared that link in case the OP wants to explore it in future. – Soumik Mukherjee May 31 '23 at 20:49
  • @Kottoamatsukami thanks for this quick answer. All this only for $A\subset\Bbb R,$ right? And don't you have equivalent definitions of a closed subset? Like "$A$ is closed iff for every sequence $(x_n)$ in $A$ which converges in $\Bbb R$, the limit belongs to $A$" (this is the definition used in the answer linked above). – Anne Bauval May 31 '23 at 20:55
  • @AnneBauval Yes, I just asked my professor about that definition. It will be introduced in our next lecture series. Perhaps I should have waited until now, but this theorem really interested me. In other words, your definition can also be used in the proof. – Kottoamatsukami May 31 '23 at 21:03
  • Yes, it is used in the proof previously (twice) linked above. Can we close your post as a duplicate of it, or do you prefer an answer which sticks to your first definition of a closed subset? – Anne Bauval May 31 '23 at 21:05
  • @AnneBauval I prefer the answer for the first definition of a closed subset. – Kottoamatsukami May 31 '23 at 21:09
  • As you like. But you would have certainly been able to prove that these two definitions are equivalent (it is nearly "automatic writing"). – Anne Bauval May 31 '23 at 21:12
  • I just realized that my only reason for not considering that link as a duplicate was imaginary! Its accepted answer actually does not use the sequential characterization of closedness as I claimed. It only relies on your definition, i.e. on the fact that if $F$ is not closed, there exists a $y∉F$ such that $F$ meets every open interval around $y.$ – Anne Bauval Jun 01 '23 at 09:41

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We work by contradiction. We take as our hypothesis that every continuous function on $K$ is bounded. We assume $K$ is not bounded and demonstrate a contradiction. Then we assume that $K$ is not closed and demonstrate another contradiction.

As noted above, if $K$ is unbounded, the identity function on $K$ also is unbounded.

If $K$ is not closed, then $A=\Bbb R \setminus K$ is not open. Choose $x \in A$ such that there is no open ball around $x$ that is contained in $A$. Such an $x$ must exist because $A$ is not open. That means every open ball around $x$, no matter how small, has non-empty intersection with $K$, so you can get as close as you like to $x$ and still stay within $K$.

Then $f: K \to \Bbb R$ defined by $f(y)= \dfrac{1}{y-x}$ is continuous and unbounded on $K$. The function $f$ is continuous on $K$ because it's continuous on $\Bbb R \setminus \{x \}$ and $x \in A \Rightarrow x \notin K$ by the definition of $A$, so $K \subseteq \Bbb R \setminus \{ x \}$. It's unbounded because we just showed that we can make $\vert y - x \vert$ arbitrarily small and still stay within $K$, which means we can make $\vert f(x) \vert =\left \vert \dfrac{1}{y-x} \right \vert$ arbitrarily large within $K$.

In higher dimensions, use the function $\dfrac {1}{d(y, x)}$ to the same effect.

Thus, if all continuous functions on $K$ are bounded, it follows that $K$ must be closed and bounded; in other words, compact.

Robert Shore
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  • One thing I don't understand is how does it follow that the set K is closed? – Kottoamatsukami Jun 03 '23 at 12:15
  • If $K$ is not closed, its complement is not open, so there’s at least one point, which I’m calling $x$, that’s in the complement (which I’m calling $A$) but not in the interior of $A$. Use that point to construct a function that’s both continuous and unbounded on $K$, contradicting our hypothesis that there is no such function. – Robert Shore Jun 03 '23 at 15:17