Is this surface a compact manifold?

Question

Let $\Omega \subset \mathbb R^3$ denote an open, bounded and connected set with $C^2-$regular boundary $\Gamma=\partial \Omega$.

Is it true that $\Gamma$ is a compact manifold?

Disclaimer: It 's been a long time since I've dealed with manifolds

Any help or hint is appreciated.

Thanks in advance

Answer 1

$\Omega $ is bounded so that $\Gamma$ is bounded.

If $\Gamma$ is $C^2$, then for $x\in \Gamma$, there is open set $U \subset \mathbb{R}^3$ with $x\in U$ s.t. a homemomorphism $f: B\subset \mathbb{R}^3 \rightarrow U$ is $C^2$ where $B$ is unit open ball and $f\{ (x,y,z)\in B| z=0\}=U\cap \Gamma$.

Hence for any two points in $\Gamma \cap U $, it has finite distance wrt intrinsic metric induced from $\mathbb{R}^3$.

Hence $\Gamma$ has a finite diameter. So $\Gamma$ is compact.

Answer 2

If you consider $\Omega \cup \partial \Omega$ then it’s a compact manifold with boundary.

If you consider only $\Omega$ then it’s non compact manifold without boundary.

And $\partial \Omega$ is a compact manifold without boundary (i.e. closed manifold)