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I am aware of the Heine-Borel theorem, which says that closed and bounded is the same as compact in $\mathbb{R}^n$. My question is: are there (connected) surfaces without boundary embedded in $\mathbb{R}^3$ that are contained in a ball of finite size and yet are not compact? In other words, to check if a bounded surface is closed is it enough to check that it has no boundary?

One possible counterexample might be Alexander's horned sphere, but I'm having trouble determining whether or not it's compact.

Thanks!

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If I understand your question correctly, there are plenty of counterexamples - the simplest ones are probably the open disc $$\{(x,y,0)\in\mathbb{R}^3\mid x^2+y^2<1\}$$ and the open cylinder $$\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2=1,-1<z<1\}.$$ Both are connected, non-closed (hence non-compact), bounded surfaces in $\mathbb{R}^3$ without boundary.

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  • $\begingroup$ I see -- I'm just confusing "without boundary" with "without 'holes'." Thanks. $\endgroup$ Oct 1 '11 at 6:27

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