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Is a surface bounding a volume always a closed surface?
A surface bounding a volume is indeed a closed surface when understood in the topological sense, as boundaries are always closed. The discussion highlights that while the boundary of the set ##\{(x,y)\in\mathbb{R}^2\mid x\ge 0\}## is closed, it does not qualify as a closed manifold. The distinction between closed sets and closed manifolds is crucial, as emphasized by references to Michael Spivak's "A Comprehensive Introduction to Differential Geometry" and Bourbaki's terminology. The conversation clarifies that the original question pertained to a metric space rather than a manifold, underscoring the importance of precise definitions in topology.
PREREQUISITESMathematicians, students of topology, and anyone interested in the nuances of differential geometry and manifold theory will benefit from this discussion.
It is a closed manifold. It is not a compact manifold, but as the boundary of a bounding submanifold it is closed. Each neighborhood of ##\{x=0\}## contains points with ##x=0## and ##x > 0## so it's a boundary and boundaries are closed. Considered as manifold in its own right, it is trivially closed.zwierz said:The set ##\{(x,y)\in\mathbb{R}^2\mid x\ge 0\}## has a boundary ##\{x=0\}##. This boundary is a closed subset of the plane but it is not a closed manifold
this does not meet the standard definition:fresh_42 said:It is a closed manifold. It is not a compact manifold
I just have shown that there is a standard meaning of the term "closed manifold". This meaning is broadly used regardless of your disagreement.fresh_42 said:So what?