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I CW complex and Hausdorff

  1. Sep 10, 2016 #1
    I am with a query about cw complex. I was thinking if is possible exist a cw complex without being of Hausdorff space. Because i was thinking that when you do a cell decomposition of a space (without being of Hausdorff) you do not obtain a 0-cell. If can exist a cw complex with space without being of Hausdorff, someone can proof?
     
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  3. Sep 10, 2016 #2

    mathman

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  4. Sep 10, 2016 #3

    fresh_42

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    That's what I've found in the books, too. Definitions start with it and is applied to the characteristic function where it is used.
    However, the question what is lost by giving up Hausdorff could be interesting.
     
  5. Sep 14, 2016 #4

    WWGD

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    The Wiki page describes it as being Hausdorff : https://en.wikipedia.org/wiki/CW_complex

    In addition, the restriction of the attaching map to the interior of the cell is a homeomorphism. But the cell, I assume is a Hausdorff space, and being Hausdorff is a topological property, so the image of the interior is Hausdorff. Now you need to deal with the image of the boundary, which " is contained in the union of a finite number of elements of the partition, each having cell dimension less than n."
     
    Last edited: Sep 14, 2016
  6. Sep 14, 2016 #5

    fresh_42

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    In the end it is an abstract concept that arose from geometry, Euclidean geometry. It's about triangulations, chain- and simplicial complexes. And cells which are little cubes. In this sense giving up Hausdorff is somehow artificial and it is not clear - at least to me - what could be gained.
     
  7. Sep 27, 2016 #6
    CW complexes are automatically Hausdorff.

    If X is a CW complex, X is the disjoint union of the interiors of at most countably infintely many open balls Bk (a space homeomorphic to set of points in some Euclidean space whose distance from the origin is ≤ 1) such that each open ball retains its usually topology in the subspace topology.

    Then any two points p, q of X with p ≠ q each lie in the interior of open balls Bi and Bk, respectively (where we cannot exclude the possibility that j = k).

    Whether or not j = k, it follows that there exist open sets U ∋ p and V ∋ q of X, so X is Hausdorff.
     
  8. Oct 4, 2016 #7
    In the last sentence in the above post, in my haste I omitted the word "disjoint" referring to U and V. It should read:

    Whether or not j = k, it follows that there exist disjoint open sets U ∋ p and V ∋ q of X, so X is Hausdorff.

    (Also: infintely → infinitely.)
     
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