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A set $A\in \Bbb{R}^n$ is compact $\iff$ every continuous function on $A$ is bounded. I have a problem proving the direction according to which $A$ is compact. First direction I said: If $A$ is compact, then since it is bounded, so is every continuous function whose domain is $A$, according to Weierstrass. Supposing that every continuous function on $A$ is bounded, assuming by contradiction that there is a set $x_n\in A$, such that $x_n\to x$ but $x\notin A$. For every such $f$, $\lim\limits_{n\to \infty}f(x_n)=f(\lim\limits_{n\to \infty}x_n)=f(x)$ which means I arrive at no contradiction. Maybe it has something to do with the fact that $f$ isn't necessarily defined on $x$ as $x$ is not in $A$? If so, how is it argued? Any help would be much appreciated.(Besides, can you help me with the tags? It is a question from my calculus course, but I am not sure what tag fits the best.)

Brian M. Scott
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Meitar
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    If you would like a hint before you look at the answer, let $x$ be a limit point of $A$ that is not in $A$, and consider the function on $A$ that takes $y\in A$ to $\frac1{|y-x|}$. You also have to show that $A$ is bounded; consider the function giving the distance of a points in $A$ from the origin. – Brian M. Scott Jun 29 '15 at 19:53
  • Thank you very much for your hint. Still, I don't see why it is bounded if that $x$ is in $A$. It still looks like it tends to infinity. How come? – Meitar Jun 29 '15 at 19:57
  • Or maybe if $x$ is in $A$, it isn't continuous in the first place, perhaps? – Meitar Jun 29 '15 at 19:59
  • I see. It isn't continuous in that case. Thank you Brian. – Meitar Jun 29 '15 at 20:00
  • You’ve got it; you’re welcome. – Brian M. Scott Jun 29 '15 at 20:06

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