# Interval ==> Connected

by jdstokes
Tags: connected, interval
 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
 PF Patron HW Helper Sci Advisor P: 4,755 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?
 P: 527 Oops, N = sup { t | [a,t] \subseteq U }.
HW Helper