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I'm trying to solve the following question from the real analysis section:

  1. Let $K$ be a nonempty subset of $\mathbb R^n$ where $n > 1$. Which of the following statements must be true?

(I) If $K$ is compact, then every continuous real-valued function defined on $K$ is bounded.

(II) If every continuous real-valued function defined on $K$ is bounded, then $K$ is compact.

(III) If $K$ is compact, then $K$ is connected.

The proof for (I) is standard. I'm trying to see (II) by contradiction.

Is it possible to frame a proof for (II) along these lines:

Suppose $K \subseteq \mathbb R^n$ is not compact. Then there exists an open cover $\mathcal C$ that has no finite subcover. But $f: K \to \mathbb R$ is continuous. (...) Contradiction.

  • @Arthur Yes, I know the Heine-Borel theorem. A subset of $\mathbb R^n$ is compact iff it is closed and bounded. Is it helpful here? –  Aug 15 '20 at 15:04

2 Answers2

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A subset of $\mathbb{R^n}$ is compact if and only if it is closed and bounded, this is a standard result. Now, suppose every continuous real valued function defined on $K$ is bounded. In particular, the function $f(x)=||x||$ is bounded on $K$, hence $K$ is a bounded set.

So we only have to prove $K$ is closed. Well, suppose it isn't. Then there is some point $y\in\overline{K}\setminus K$. Define $f:K\to\mathbb{R}$ by $f(x)=\frac{1}{||x-y||}$. This is a continuous function which isn't bounded, a contradiction.

Mark
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  • Thanks but why are we specifically choosing $y \in \bar K \setminus K$? Wouldn't $f(x)$ be unbounded even if $y \in K$? (Only the domain of $f$ would have to exclude $y$ to make it defined.) –  Aug 15 '20 at 15:16
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    But this wouldn't give us a contradiction. We are assuming that every continuous function defined on $K$ is bounded, not that every continuous function defined on $K\setminus{y}$ is bounded. This is a big difference. For example, in $\mathbb{R}$ the set $[0,1]$ is compact, but $[0,1)$ isn't, and there are continuous functions defined on $[0,1)$ which are unbounded. – Mark Aug 15 '20 at 15:19
  • Actually I'm not even sure how to see that $f(x)$ as you defined is unbounded. –  Aug 15 '20 at 15:19
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    By definition, if $y\in\overline{K}$ then there is a sequence $(y_n)\subseteq K$ such that $y_n\to y$. This means $||y_n-y||\to 0^+$, and thus $f(y_n)=\frac{1}{||y_n-y||}\to\infty$. So $f$ is unbounded. – Mark Aug 15 '20 at 15:21
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    Excellent, thank you! –  Aug 15 '20 at 15:21
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I would just like that to add that if the range was the reals endowed with the bounded metric, $d(x,y)=\frac{|x-y|}{1+|x-y|}$, then the statement is not true for metric spaces even if the $Dom(f)$ satisfied the Heine-Borel property.

Umesh Shankar
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