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?
When is the boundary of a subset of $\mathbb{R^n}$ compact?
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real-analysis
compactness
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3The boundary of a bounded set is itself bounded. By the [Heine-Borel theorem](https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem), if a subset of the Euclidean space is bounded then its boundary is compact. – 2012-06-25
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3Everything 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. – 2012-06-25
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0hmm, 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? – 2012-06-25
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5@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 $\mathbb R^n\setminus B(0,R)$. – 2012-06-25
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0@LeonidKovalev I see, thanks! – 2012-06-25
1 Answers
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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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0why is that the boundary of a bounded set is itself bounded? – 2012-06-26
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0@palio: If the set is contained in some closed ball, so is its closure, and hence its boundary. – 2012-06-26