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Interval ==> Connected 
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#1
Apr1507, 03:46 AM

P: 527

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 nonempty 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 } 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 [Neta,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
Apr1507, 04:42 AM

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P: 4,771

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? 


#3
Apr1507, 05:01 AM

P: 527

Oops, N = sup { t  [a,t] \subseteq U }.



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