Let $\Omega \subset \mathbb{R}^N$ be an open set.

I know bounded boundary doesn't imply bounded set, but what if we consider the boundary of an open connected set of class $C^1$ (i.e. the boundary $\partial \Omega$ is locally the graph of a $C^1$ function)-

If the boundary of a connected open set $\Omega$ of class $C^1$ is bounded then $\Omega$ is bounded?

  • $\begingroup$ Have you thought about this? For instance, have you thought about an example where the boundary is bounded but $\Omega$ is not, and considered whether that example can be modified to have smooth boundary? (Or whether it already has a smooth boundary?) $\endgroup$ – David C. Ullrich Jul 9 '16 at 14:45
  • $\begingroup$ @DavidC.Ullrich Yes, for example $\mathbb{R}^2 - R$, where $R$ is a closed segment. But the definition of $C^1$ boundary are not met for the two points at the end of the segment, does it? $\endgroup$ – D1X Jul 9 '16 at 14:47
  • $\begingroup$ True, that example does not answer the question. But there are very simple analogous examples. I mean can you think of any bounded smooth curves? $\endgroup$ – David C. Ullrich Jul 9 '16 at 14:52

Consider $\Omega = \{ x \in \Bbb R^n | \| x \| > 1 \}$

It is open connected, and $\partial \Omega$ is $C^1$ but $\Omega$ is not bounded


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