# Connected sets

1. Oct 8, 2008

### math8

If A is a connected subset of a disconnected set X s.t. X=MUN , M,N nonempty closed disjoint sets, how do we show, A is either contained in M or in N?

I can start a proof, but then, I am kind of stuck.
I would go by contradiction and say A intersection M is non empty and A intersection N is non empty. Hence A would be the union of 2 non empty disjoint sets. But since A is connected, A intersection M and A intersection N cannot be both open. So without loss of generality, say A intersection M is not open. Hence X\(A intersection M) is not closed.
Then I get stuck. Any help?

2. Oct 9, 2008

### Tac-Tics

X is disconnected. That means, there are at least two nonempty disjoint open-and-closed sets. You are conveniently given these two sets. They are M and N.

If A is connected, that means that, relative to the topology on X, there is only one non-empty disjoint open-and-closed set. Namely, A itself.

Now, suppose that A intersects both M and N. What can you say about AnM and AnN? For both, is the set empty? Is it closed?

3. Oct 9, 2008

### math8

I guess AnM and AnN are both clopen. But since A is connected, one of the AnM or AnN has to be empty. Say, AnM is empty. So ACN.

Is that right?

4. Oct 10, 2008

### ice109

assume that AnM and AnN are non empty and A is connected. Since M and N are a disconnection of X they are open in X and by the subspace topolgoy A'=AnM and A''=AnN are open and disjoint in A and form a disconnection of A. contradiction QED