High School Closed surface bounding a volume V

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SUMMARY

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.

PREREQUISITES
  • Understanding of topological concepts such as boundaries and closed sets
  • Familiarity with the definitions of closed and compact manifolds
  • Basic knowledge of metric spaces and their properties
  • Awareness of differential geometry literature, specifically works by Michael Spivak
NEXT STEPS
  • Study the definitions of closed and compact manifolds in topology
  • Explore Michael Spivak's "A Comprehensive Introduction to Differential Geometry" for foundational concepts
  • Research Bourbaki's terminology regarding quasi-compactness and its implications
  • Examine the differences between closed sets and closed manifolds in mathematical literature
USEFUL FOR

Mathematicians, students of topology, and anyone interested in the nuances of differential geometry and manifold theory will benefit from this discussion.

Pushoam
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Is a surface bounding a volume always a closed surface?
 
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If you understand "bounding" as being a boundary in the topological sense, yes, since boundaries are always closed:
$$
\partial V = \overline{V} \cap \overline{(X-V)}
$$
 
o.k.
Thank you.
 
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
 
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
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.
 
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fresh_42 said:
It is a closed manifold. It is not a compact manifold
this does not meet the standard definition:

6d634c19478c.png

Michael Spivak: A Comprehensive Introduction to Differential Geometry. third edition Vol 1 p.19 Publish of Perish Texas 1999
 
This is only a source of confusion, not really a rigor use of topological terms. After all, a manifold is a topological space in the first place. This determines the terms closed, compact and boundary in a precise way. To use it otherwise is not a good start to learn it. Bourbaki use the term quasi-compact instead of compact, to reserve compactness for Hausdorff spaces. To speak of "non-bounded compact" sets in a context where a metric and a local Euclidean structure is that close, is a crime, not a definition.
 
So what? My book on alternating differential forms contains a proof that boundaries are closed. In topology boundaries are closed.

To call "closed = compact - boundary" is merely a sloppy way of using "compact = closed + bounded" and subtract "bounded", which is wrong by the generality of the first equation and sloppy by the method. Furthermore it only says "compact - boundary" is closed, not that all closed sets are built this way, which makes a huge difference. I assume there is a difference in the meaning between a closed manifold and a closed set and that's where confusion starts and where I want to emphasize, that this distinction has to be made in order to avoid confusion.

Beside all this, there wasn't a manifold in the OP's question at all. Only a metric space, which may be assumed by the term "volume". So to argue about special uses of topological terms in the realm of Riemannian(?!) manifolds doesn't make any sense at all. It simply wasn't part of the question, only part of your answer.
 
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fresh_42 said:
So what?
I just have shown that there is a standard meaning of the term "closed manifold". This meaning is broadly used regardless of your disagreement.
 
  • #11
Thank you all.
Since I haven't studied topology and manifold, I couldn't understand these arguments.
By "bounding", I mean a "boundary "and so I think the answer to my question is yes.
 

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