Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: All intervals are connected

**Physics Forums | Science Articles, Homework Help, Discussion**