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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. –  Dave Hartman Dec 12 '12 at 22:28
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2 Answers

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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yes your $2$ is correct there exist such a function say $\frac{1}{x}$ on $(0,1)$. so my original statement is false. –  Bunuelian Trick Dec 12 '12 at 22:31
    
Can't you apply the identity function over X to show that f(X)=X and therefore it is bounded, and then use its inverse to show that X is closed? –  Marra Dec 12 '12 at 22:40
    
@GustavoMarra, ah, that does sound a lot more elegant than my suggestion. I'm not exactly sure I follow how to get that $X$ is closed, though. –  Christopher A. Wong Dec 13 '12 at 15:45
    
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. –  Marra Dec 14 '12 at 17:01
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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.

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