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In Euclidean space, the boundary of a subset (Closure - Interior) is always closed, but I would like to know when it's compact as well. Is it when the subset is bounded?

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The boundary of a bounded set is itself bounded. By the Heine-Borel theorem, if a subset of the Euclidean space is bounded then its boundary is compact. –  A. De Luca Jun 25 '12 at 20:24
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Everything that others have said is true. I just want to add that the boundary of the complement of a bounded set is also bounded. After all, everything far out is then in the interior, and thus the boundary is still bounded + closed, and hence by Heine-Borel also compact. –  Jyrki Lahtonen Jun 25 '12 at 20:35
    
hmm, is this necessary? I mean, can we prove that if the boundary of a set is bounded, then either the set itself or its complement are bounded? –  A. De Luca Jun 25 '12 at 20:38
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@A.DeLuca In $\mathbb R^n$ with $n\ge 2$, the above two cases are exhaustive: the boundedness of $\partial A$ implies that either $A$ or $A^c$ is bounded. Not so in $n=1$. The key difference is the connectedness of $\mathbb R^n\setminus B(0,R)$. –  user31373 Jun 25 '12 at 20:38
    
@LeonidKovalev I see, thanks! –  A. De Luca Jun 25 '12 at 20:41
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Community wiki answer-made-of-comments:

The boundary of a bounded set is itself bounded. By the Heine-Borel theorem, if a subset of the Euclidean space is bounded then its boundary is compact. – A. De Luca

Everything that others have said is true. I just want to add that the boundary of the complement of a bounded set is also bounded. After all, everything far out is then in the interior, and thus the boundary is still bounded + closed, and hence by Heine-Borel also compact. – Jyrki Lahtonen

In $\mathbb R^n$ with $n\ge 2$, the above two cases are exhaustive: the boundedness of $\partial A$ implies that either $A$ or $A^c$ is bounded. Not so in $n=1$. The key difference is the connectedness of $\mathbb R^n \setminus B(0,R)$. – Leonid Kovalev

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why is that the boundary of a bounded set is itself bounded? –  palio Jun 26 '12 at 0:37
    
@palio: If the set is contained in some closed ball, so is its closure, and hence its boundary. –  Micah Jun 26 '12 at 4:51
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