Proving the Connection of Subsets in Topology

[mathshelp]

I have a question here and I'm not sure what to do as it always confuses me, any help?

Let A,B be closed non-empty subsets of a topological space X with AuB and AnB connected.

(i) Prove that A and B are connected.

(ii) Construct disjoint non-empty disconnected subspaces A,B c R such that AuB is connected.

 

[rasmhop]

Well suppose A is disconnected. Then we can find a separation of A:
A = C \cup D
for disjoint, non-empty, closed sets C,D. Now since
A \cap B = (C \cap B) \cup (D \cap B)
is connected and both C \cap B and D \cap B are closed we can assume without loss of generality that C \cap B = \emptyset (otherwise we would have a separation of A \cap B). Now we have that B and D are both disjoint from C, so B \cup D is disjoint from C which gives us a separation:
A \cup B = (B \cup D) \cup C
which is a contradiction.

ii)
Just consider what it means to be disconnected and connected. For instance,
A = (0,2) \cup (2,3) \qquad B = (0,1)\cup (1,3)
A \cup B = (0,3)

 

[madness]

Shouldn't you be considering sets which are open or closed with respect to each subspace topology? And your sets in (ii) are not disjoint.

 

[rasmhop]

Shouldn't you be considering sets which are open or closed with respect to each subspace topology? And your sets in (ii) are not disjoint.

What set are you referring to. Remember that if Y is a subspace of X, then A is open in Y if and only if A \cap Y is open in X, so if A \subseteq Y then A is open in Y if and only if A is open in X.

About (ii) you're right I must have misread the question as I seem to have striven for open sets. In that case take A=(0,1) U (1,2) and B={1,2}.

 

[madness]

You would be right if Y was closed or if your set didn't contain the boundary points. But the sets here are closed and necessarily contain the boundary points. An obvious counterexample is [-1,1], which has [-1,0) as an open set which is not open in R.

 

[rasmhop]

Yes, you're right I spoke too quickly without thinking however I still don't see where this affects my answer. I only ever use the correct direction: That if A \subseteq Y and A is open in X, then A is open in Y since A=Y\cap A. The same holds true for closed sets. This shows A and B are closed in AUB.

If A is closed in the closed subspace Y of X, then A can be written as A = B\cap Y with B closed in X and therefore A is closed. This shows that closedness is transitive.

C and D are a separation of A so C and D are closed in A, then C \cap A \cap B = C \cap B is closed in A \cap B.

From this we see C is closed in A and therefore also in A U B. In the same way we see D is closed in AUB. Thus B \cap D is closed in AUB so our separation is valid.

 

[madness]

When you say transitive, do you mean reflexive? I've been trying this problem by explicity using the intersections for the open sets of each subspace topology. I came to a problem in showing that the result of plugging my separation of A into AUB satisfied the definition of AUB being disconnected. However I think the first part of your post solved my problem.

 

[rasmhop]

When you say transitive, do you mean reflexive? I've been trying this problem by explicity using the intersections for the open sets of each subspace topology. I came to a problem in showing that the result of plugging my separation of A into AUB satisfied the definition of AUB being disconnected. However I think the first part of your post solved my problem.

I mean transitive (when I said closedness I meant the relation "is closed in"). In the sense that if we write A < B if A is closed in B and A is a subspace of B, then we have:
A < B and B < C imply A < C.

 

[madness]

Yes but what you have written is that
1) if A is contained in Y and contained in X, then A is closed in Y
2) if A is closed in Y and Y is closed in X, then A is closed in X

Ok I just saw what you meant there. Never mind I get it now.