Are All Intervals on the Real Line Connected?

[jdstokes]

Hi all,

I originally posted this in the analysis/topology forum but I think this might be a more appropriate place for it.

Homework Statement

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.

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

 

[HallsofIvy]

Hi all,

I originally posted this in the analysis/topology forum but I think this might be a more appropriate place for it.

Homework Statement

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.

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

If N- eta/2 were NOT in U then [a,t] for t> N- eta/2 would not be a subset of U so that N= sup {t | [a,t] is a subset of U} <= N- eta/2 which is impossible.

 

[jdstokes]

Sounds good. Thanks for your help.