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Connectedness again

  1. Oct 13, 2010 #1

    radou

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    1. The problem statement, all variables and given/known data

    This one is giving me serious trouble.

    Let Y be a subset of X. Let both X and Y be connected. SHow that if A and B form a separation of X\Y, then Y U A and Y U B are connected.

    3. The attempt at a solution

    I know all the basic definitions and theorems from the chapter about connectedness preceding this exercise section, but any way I try it, I don't seem to get anywhere.

    Any ideas?
     
  2. jcsd
  3. Oct 13, 2010 #2
    When you say "[tex]A[/tex] and [tex]B[/tex] form a separation of [tex]X \setminus Y[/tex]", what does this mean, exactly? My guess would be what I usually call "[tex]\{A, B\}[/tex] disconnect [tex]X \setminus Y[/tex]", that is, [tex]A[/tex] and [tex]B[/tex] are disjoint (relatively) open subsets of [tex]X \setminus Y[/tex], whose union is [tex]X \setminus Y[/tex]. Is this correct?
     
  4. Oct 13, 2010 #3

    radou

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    Well, partially. According to Munkres, there's a subtle difference:

    If X is a topological space, a separation of X is a pair of non empty, disjoint and open subsets U and V of X whose union is X. (This is a definition.)

    If Y is a subspace of X, a separation of Y is a pair of non empty disjoint sets U, V whose union if Y. (This is a Lemma)

    Note that in the Lemma it is not required for the separation sets to be open.

    So, I assume if we're talking about X\Y, which is a subspace of X, the sets in the separation are not required to be open.
     
  5. Oct 13, 2010 #4
    That won't work. You need some kind of relative openness hypothesis, otherwise the result is false. Take [tex]X = \mathbb{R}[/tex], [tex]Y = [0, 1][/tex], [tex]A = (-\infty, -1] \cup (1, 2)[/tex], [tex]B = (-1, 0) \cup [2, \infty)[/tex]. Then [tex]A[/tex] and [tex]B[/tex] form a separation (according to the conditions you gave) of [tex]X \setminus Y = (-\infty, 0) \cup (1, \infty)[/tex], but [tex]Y \cup A = (-\infty, -1] \cup [0, 2)[/tex] and [tex]Y \cup B = (-1, 1] \cup [2, \infty)[/tex] are both disconnected.

    Also, with the definition of "separation" you cite for subspaces, you can give a separation for a connected subspace, such as [tex]X = \mathbb{R}, Y = [0, 1], U = [0, \textstyle\frac12), V = [\textstyle\frac12, 1][/tex]. That doesn't make sense.
     
  6. Oct 14, 2010 #5

    radou

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    Interesting, it's exactly what says in the book. So, in the "separation lemma" for subspaces, the sets should be open too?
     
  7. Oct 14, 2010 #6
    At least relatively open in the subspace, otherwise they are useless for establishing disconnectedness. I don't have a copy of the new edition of Munkres, so I can't guess what he might be thinking there.
     
  8. Oct 14, 2010 #7

    radou

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    I found the solution to this exercise (Ex 23.12)

    http://www.math.ku.dk/~moller/e02/3gt/opg/S23.pdf

    Consider this problem solved, I'll go through it by myself, although I don't like the looks of it, for some reason.

    I posted it if you happen to be interested to look at it regardless.
     
  9. Oct 14, 2010 #8
    OK, I looked it up. The problem is that you missed an important hypothesis in the lemma 23.1, which changes things entirely.

    Lemma 23.1 of Munkres says that a separation of [tex]Y \subset X[/tex] is a disjoint pair of nonempty sets [tex]A, B[/tex] whose union is [tex]Y[/tex], which satisfy the condition that neither of [tex]A, B[/tex] contains a limit point of the other. This latter condition is exactly what you need for [tex]A[/tex] and [tex]B[/tex] to be relatively open (in fact, clopen) in [tex]Y[/tex]. The examples I gave above do not satisfy the latter condition.
     
  10. Oct 15, 2010 #9

    radou

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    You're right, for some reason I forgot to mention this.
     
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