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Separation axioms vs subspaces

  1. Jun 8, 2005 #1

    Hurkyl

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    How do separation axioms carry over to subspaces?

    Some are clear -- it's easy to see that if any two points of a space X are separated by neighborhoods, then the same must be true of any subset S of X.

    But what about the nicer ones? Is it true that if S is a subset of a normal space, that S is itself normal?

    This one is less obvious... one example that worries me is this:

    Consider the union S of two open discs in R^2. (that aren't disjoint) Consider the two closed sets formed by restricting the boundaries of the two discs to S. We can't directly appeal to the normality of R^2, because the closure of these sets aren't disjoint in R^2.

    It is still easy to see S is normal, because it's homeomorphic to R^2, but that doesn't help me in the general case of a subset of an aribtrary normal space.
     
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  3. Jun 8, 2005 #2

    NateTG

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    It should be pretty easy to see that subspaces on closed subsets of normal spaces are normal. Or, for that matter that on any normal space [itex]S[/itex], the subspace formed on [tex]S_1[/tex] is normal if, for any two closed sets [tex]U,V[/tex] in [itex]S[/itex] there exists an open set [itex]O[/itex] so that [itex]O\supset U \cap V[/itex] and [itex]O \cap S_1 = \emptyset[/itex]

    It's pretty straightforward to come up with examples of subspaces of normal spaces that aren't normal. For example, consider the space
    [tex]\{0,1,o,c\}[/tex]
    Where the open sets are
    [tex]\emptyset,\{o\},\{0,o\},\{1,o\},\{0,1,o\},\{0,1,o,c\}[/tex]

    Which is normal, since the only closed set that does not contain [itex]c[/itex] is [itex] \emptyset[/itex].

    However, the subspace [itex]\{0,1,o\}[/itex] is not normal since [itex]{0}[/itex] and [itex]{1}[/itex] are both closed, but any open set containing either also contains [itex]o[/itex].
     
  4. Jun 8, 2005 #3

    Hurkyl

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    Bleh. Well, what if the space is nicer? Like a normal T2 space?
     
  5. Jun 8, 2005 #4

    NateTG

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    Hmmm. I have a hard enough time thinking of a non-normal T2 space.
     
  6. Jun 8, 2005 #5

    Hurkyl

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    Well, http://en.wikipedia.org/wiki/Normal_space suggests that if you take the uncountable product of noncompact Hausdorff spaces, it won't be normal.

    Oh, I guess that suggests an example -- take an uncountable product of compact Hausdorff spaces. That should be normal, right? Then, take as a subset the uncountable product of an open subset.


    *sigh* How disappointing. Topology is hard! Maybe I should stick to the world of metric spaces, I understand those better! :smile:
     
  7. Jun 8, 2005 #6

    NateTG

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    I was wondering if the long line was going to figure into this.

    Topology is nice. It's just that negative examples can be really hard to think of.
     
    Last edited: Jun 8, 2005
  8. Jun 8, 2005 #7

    Hurkyl

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    Ack, open sets that are compact? :frown: Well, doesn't the quote from the wikipedia article imply the open subset wouldn't be normal?


    Incidentally, the property I want to be true is this:

    If I have closed sets [itex]A \cup B = C[/itex], then there exists closed sets A' and B' such that [itex]A' \cup B' = X[/itex], [itex] A' \cap C = A[/itex], and [itex]B' \cap C = B[/itex].

    Actually, I don't even need that strong -- I think that it's good enough that if C is a reducible closed set, then there is a way to write X as the union of two closed sets, neither containing C.


    I think that this is provable if the complement of C is a normal space, but I guess I can't guarantee that, even if X is a "nice" space. :frown:
     
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