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Let $U \subseteq \mathbb{R}^n$ be open in the usual topology. Is its boundary, $\partial U$, necessarily a topological manifold?

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up vote 4 down vote accepted

No, consider $U = \{ (x,y) \in \mathbb{R}^2 : xy \neq 0 \}$. Then $\partial U = \{ (x,y) : x = 0 \vee y = 0 \}$, and the point $(0,0)$ doesn't have a neighborhood homeomorphic to $\mathbb{R}$.

You can modify this example to get a connected and bounded $U$: consider $$U = \{ (x,y) : x^2 + y^2 < 2 \} \setminus \left( [-1,1]\times0 \cup 0 \times [-1,1]\right)$$

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I see. What if we further constrain $U$ to be connected and/or bounded? – iheap Mar 11 '14 at 9:48
See my edit.${}$ – Najib Idrissi Mar 11 '14 at 9:51
That is, $\mathbb{R}^2 - +$? ;) – Neal Mar 11 '14 at 10:09

I doubt it. Let $Y$ be any nowhere dense closed set in $\mathbb R^n$, and let $U=\mathbb R^n\setminus Y$; then $U$ is an open set and $\partial U=Y$. Now suppose, e.g., that $n=2$ and that the set $Y$ is homeomorphic to the letter Y.

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