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## Main Question or Discussion Point

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

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