show if A and B are connected in (M,d) with AnB!=empty set, AUB is also connected.

that just seems obvious.....how would I go about proving this one?

thanks

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- April 21st 2010, 07:33 AMcp05connectedness
show if A and B are connected in (M,d) with AnB!=empty set, AUB is also connected.

that just seems obvious.....how would I go about proving this one?

thanks - April 21st 2010, 03:33 PMDrexel28
Let's prove it in general. Let be a topological space and be a collection of connected subspaces of such that . Then, is connected. To see this, suppose not and that is a separation with open. Then, since each we must have that a part of some and are in each. But, this in particular implies that they are open in and evidently but this clearly contradicts that is connected. The conclusion follows.

- April 21st 2010, 05:58 PMcp05
I don't quite understand the proof, sorry

Why is http://www.mathhelpforum.com/math-he...156f55fd-1.gif a separation?

Do we get http://www.mathhelpforum.com/math-he...fd3b22c8-1.gif since any intersection of connected sets in X aren't empty? - April 21st 2010, 06:29 PMDrexel28
- April 21st 2010, 06:53 PMDrexel28
Here, I think I can say it so you'll better understand.

Let be open and be such that and . We claim that at least one of must be empty. Since the union of their intersections with is non-empty (assuming is

non-empty, but then the conclusion is immediate) we must have that at least one of them is non-empty, assume WLOG that it's . Then, we have that

for some . But, it must be true that otherwise we'd have that some point of is not in and since cover it must be that that point is in . In other

words, and are both non-empty. But, they are apparently disjoint, open in and their union is . This clearly contradicts that from where it follows that .

Now, the rest is easy. For, suppose that is non-empty. Then, by the previous analysis it follows that and so . But, this is a contradiction since and

are disjoint and is non-empty. It follows that must be empty and thus no separation of is possible. The conclusion follows. - April 21st 2010, 06:57 PMcp05
I think I kind of get it....

But if I just had 2 sets, like A and B, I would say suppose AUB is disconnected.

So there exist sets X and Y such that

XnAUB != YnAUB != empty

XnAUB U YnAUB = AUB

XnAUB n YnAUB = empty

so some part of A, B is in X, and also some other part of A, B is in Y since XnAUB, YnAUB is not empty and their intersection is empty (so the same part of A, B can't be in both Y and X)

so XnA, YnA is not empty

so XnA U YnA = A??

So A is disconnected because XnA and YnA are disjoint?

And same goes for B

Is this correct? :) - April 21st 2010, 06:59 PMDrexel28
- April 23rd 2010, 08:30 AMcp05
would the intersection or A and B be connected?

I'm kind of stuck on this because I don't know if the empty set is connected or not. For example, if A=(0,1) and B=(2,3) then their intersection would be empty....but is that connected? - April 23rd 2010, 11:20 AMDrexel28
The empty set is sometimes connected sometimes not depending on the author.

But, we can prove a nice little theorem.

Theorem: Let be connected and then is connected.

Proof:This follows since the intersection of two intervals is an interval.

In fact this is true in any linear continuum. But, this is a very special case.

Permit me if you will to describe a picture for you.

Imagine two crescent rolls (look here) it is easy to think of their general shape projected into , right? Now, think about taking two of them and having the concave sides face each other. Now, move them until just their tips are touching. Clearly each crescent roll is connected but their intersection will be the area in the overlapping tips, but this is clearly disconnected. - April 24th 2010, 08:15 PMcp05
But if we did not have the condition that http://www.mathhelpforum.com/math-he...2c2570a0-1.gif, then the intersection of E and G would not be connected, correct?

I'm looking at your crescent roll example for that: the intersection of the crescent rolls is empty....kind of like the intersection of two open sets, so the empty set in this case would be disconnected. - April 25th 2010, 07:06 PMDrexel28