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B Closed surface bounding a volume V

  1. May 19, 2017 #1
    Is a surface bounding a volume always a closed surface?
     
  2. jcsd
  3. May 19, 2017 #2

    fresh_42

    Staff: Mentor

    If you understand "bounding" as being a boundary in the topological sense, yes, since boundaries are always closed:
    $$
    \partial V = \overline{V} \cap \overline{(X-V)}
    $$
     
  4. May 19, 2017 #3
    o.k.
    Thank you.
     
  5. May 21, 2017 #4
    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
     
  6. May 21, 2017 #5

    fresh_42

    Staff: Mentor

    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.
     
    Last edited: May 21, 2017
  7. May 21, 2017 #6
    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
     
  8. May 21, 2017 #7

    fresh_42

    Staff: Mentor

    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.
     
  9. May 21, 2017 #8
  10. May 21, 2017 #9

    fresh_42

    Staff: Mentor

    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.
     
  11. May 21, 2017 #10
    I just have shown that there is a standard meaning of the term "closed manifold". This meaning is broadly used regardless of your disagreement.
     
  12. May 22, 2017 #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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