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All intervals are connected

  1. Apr 15, 2007 #1
    Hi all,

    I originally posted this in the analysis/topology forum but I think this might be a more appropriate place for it.

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

    I'm having difficulty proving that all intervals of the real line are
    connected in the sense that they cannot be decomposed as a disjoint
    union of two non-empty open subsets.

    3. The attempt at a solution

    Here is the "proof":

    Suppose X is an interval and

    X = (X intersect U) union (X intersect V)

    where U,V are open and

    X intersect U intersect V = emptyset

    Suppose also we have points a in X intersect U and b in X intersect V with a < b.

    Let N = sup { t | [a,t] \subseteq U }
    Then
    1. a <= N
    2. N < b
    3. N in X (since and X is an interval)

    If N is in U, then since U is open we can find an open interval (N -
    epsilon,N + epsilon) about N which is contained in U. Thus [a, N +
    epsilon/2] is contained in U which is a contradiction. Therefore N
    must be in V. Then [N-eta,N] is contained in V for some eta.

    Now, if N - eta/2 is in U, the we have a contradiction since it is also in V and X.

    How do I show that N - eta/2 is in U?

    Thanks in advance,

    James
     
  2. jcsd
  3. Apr 15, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If N- eta/2 were NOT in U then [a,t] for t> N- eta/2 would not be a subset of U so that N= sup {t | [a,t] is a subset of U} <= N- eta/2 which is impossible.
     
  4. Apr 15, 2007 #3
    Sounds good. Thanks for your help.
     
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