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
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.
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.
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?
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.
But if we did not have the condition that , 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.