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I have no idea how to show whether this statement is false or true:

If every differentiable function on a subset $X\subseteq\mathbb{R}^n$ is bounded then $X$ is compact.

Thank you

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    $\textbf{Hint:}$ Think about how one can characterize compact subsets of $\mathbb{R}^n$. Suppose that $X$ is not compact and try to find a differentiable function on $X$ that isn't bounded.2012-12-12

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Some hints:

  1. By the Heine-Borel property for Euclidean space, $X$ is compact if and only if $X$ is closed and bounded.
  2. My inclination is to prove the contrapositive: If $X$ is not compact, then there exists a differentiable function on $X$ which is unbounded.
  3. If $X$ is not compact, then either it isn't bounded, or it isn't closed. As a first step, perhaps show why the contrapositive statement must be true if $X$ isn't bounded?
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    Well, I'm not sure how to, either. The part where the identity shows that$X$is bounded is OK, but I don't see how the same argument (or the inverse of it) is enought to show compacity. I guess going through "not closed => absurd" can work.2012-12-14
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If $X$ is unbounded, take some variant of $\|x\|$. If $X\subset \mathbb{R}^n$ is bounded but not closed, then there is some $x\in\partial X\cap X'$. To have any smooth functions, $X$ must have interior, so take a small ball $B\subset X$ with $x\in\partial B$.

Now your job boils down to finding a function on the unit ball which goes to $\infty$ as $x\to (1,0,\cdots,0)$ and goes to $0$ as $x\to$ any other point on the boundary.