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Interval => Connected

  1. Apr 15, 2007 #1
    Hi all,

    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.

    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 = { t | [a,t] \subseteq U }
    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,

  2. jcsd
  3. Apr 15, 2007 #2


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    Gold Member

    I don't understand what your N is. When you write "Let N = { t | [a,t] \subseteq U }", it sounds like N is a set but then you go an treat it like a point.

    Here's a suggestion. Suppose the interval is I and that its inf is a and sup is b. Suppose also that I is not connected and that U,V are open subsets of I, neither of which is void and with UuV=I and UnV=void.

    1° Show that the boundary of U must contain a point p other that a and b, otherwise, U is I itself, which would make V void, which is a contradiction.

    2° p must then be in V. How does this imply that V is not open?
  4. Apr 15, 2007 #3
    Oops, N = sup { t | [a,t] \subseteq U }.
  5. Apr 15, 2007 #4


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    Homework Helper

    you need something to work with, like the intermediate value theorem.

    then you can use the easy fact that a set is connected if every continuous function from it to the 2 point set {0,1} is constant.
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