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Interval ==> Connected

by jdstokes
Tags: connected, interval
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jdstokes
#1
Apr15-07, 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 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 }
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
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quasar987
#2
Apr15-07, 04:42 AM
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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?
jdstokes
#3
Apr15-07, 05:01 AM
P: 527
Oops, N = sup { t | [a,t] \subseteq U }.

mathwonk
#4
Apr15-07, 05:24 AM
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Interval ==> Connected

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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