Proof of Connectedness of A and B in Topological Space X

[madness]

Homework Statement

Let A, B be closed non-empty subsets of a topological space X with A \cup B and A \cap B connected.

Prove that A and B are connected.

Homework Equations

A set Q is not connected (disconnected) if it is expressible as a disjoint union of open sets, Q = S \cup T

The Attempt at a Solution

I'm trying a proof by contradiction.
By the above definition, a set which is not connected must be open (is this really true?). So start by assuming A is disconnected, ie A = C \cup D for C, D open and disjoint. Then A must be open, but should also be closed. Now consider A \cup B = C \cup D \cup B and A \cap B = C \cup D \cap B. I want to to arrive at a contradiction. The given properties are that A is both open and closed, B is closed, C and D are open and disjoint and A \cup B = C \cup D \cup B and A \cap B = C \cup D \cap B are connected. This all seems very complicated.

 

[rochfor1]

By the above definition, a set which is not connected must be open (is this really true?).

No, thankfully this isn't true. The set [0,1] \cup [2,3] is disconnected and closed. The definition of connectedness requires S and T to be relatively open; that is, open in the subspace topology induced on Q. Keeping this in mind, an equivalent definition of disconnectedness of a subspace Q is that there exist disjoint open sets S and T such that Q \subset S \cup T and Q \cap S \neq \emptyset \neq Q \cap T. The subset relation shere removes your concern about disconnected sets being open.

 

[madness]

Ah right instead of saying "a set is not connected if..." i should have said "a topological space is not connected if...". Hmm does this mean the rest of my reasoning was wrong? I have to replace my C's and D's with intersections over X.

 

[madness]

Ok so thanks to the push in the right direction I seem to have solved it. However I didnt use the fact that A and B or closed or that A \cup B is connected. It was done entirely on the fact that A \cap B is connected. Could this be right?

 

[rochfor1]

No, that's certainly not true. A counterexample is given by A = (1,2) \cup (2,3) and B = [2,3). Here A \cup B = (1, 3) and A \cap B = (2, 3) are connected, but A is not connected. Also, you should try to think of a very simple example that shows that neither A nor B must be connected even it A \cap B is.

 

[madness]

Yeah it took me a while to realize what I missed. I showed that if A is disconnected then A \cap B is expressible as a union of disjoint open sets, but I need to show that they are non-empty sets.

 

[rochfor1]

Yup. Generally homework-style questions don't give you more assumptions than you need, so if you haven't used some/most of your assumptions, you should examine your proof. This heuristic will take you far in math.