connectedness

**cp05** · April 21st 2010, 07:33 AM

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

**Drexel28** · April 21st 2010, 03:33 PM

Originally Posted by cp05

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 $X$ be a topological space and $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of connected subspaces of $X$ such that $\bigcap_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothi ng$ . Then, $\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\Omega$ is connected. To see this, suppose not and that $(E\cap \Omega)\cup (G\cap\Omega)=\Omega$ is a separation with $E,G\subseteq X$ open. Then, since each $E\cap \Omega,G\cap \Omega\ne\varnothing$ we must have that a part of some $U_{\alpha_1}$ and $U_{\alpha_2}$ are in each. But, $U_{\alpha_1}\cap U_{\alpha_2}\ne \varnothing$ this in particular implies that $E\cap U_{\alpha_1},E\cap U_{\alpha_2}\ne\varnothing$ they are open in $U_{\alpha_1}$ and evidently $(E\cap U_{\alpha_1})\cup (G\cap U_{\alpha_1})=U_{\alpha_1}$ but this clearly contradicts that $U_{\alpha_1}$ is connected. The conclusion follows.

**cp05** · April 21st 2010, 05:58 PM

I don't quite understand the proof, sorry

Why is

a separation?
Do we get

since any intersection of connected sets in X aren't empty?

**Drexel28** · April 21st 2010, 06:29 PM

Originally Posted by cp05

I don't quite understand the proof, sorry

Why is

a separation?
I'm making a contradiction and so I'm assuming it's a seperation.
Do we get

since any intersection of connected sets in X aren't empty?

Yes, if it were to be a separation $E\cap \Omega,G\cap \Omega$ they must be non-empy open and disjoint. I forgot to mention disjointness but since $E\cap \Omega,G\cap \Omega$ are disjoint evidently so are $E\cap U_{\alpha_1},G\cap U_{\alpha_1}$

**Drexel28** · April 21st 2010, 06:53 PM

Here, I think I can say it so you'll better understand.

Let $E,G\subseteq X$ be open and be such that $\left(E\cap \Omega \right)\cap\left(G\cap \Omega\right)=\varnothing$ and $\left(E\cup \Omega\right)\cup\left(E\cap\Omega\right)=\Omega$ . We claim that at least one of $E,G$ must be empty. Since the union of their intersections with $\Omega$ is non-empty (assuming $\Omega$ 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 $E$ . Then, we have that $U_{\alpha_1}\cap E\ne\varnothing$

for some $\alpha_1\in\mathcal{A}$ . But, it must be true that $U_{\alpha_1}\subseteq E\cap\Omega$ otherwise we'd have that some point of $U_{\alpha_1}$ is not in $E\cap\Omega$ and since $E\cap\Omega,G\cap\Omega$ cover $\Omega$ it must be that that point is in $G\cap\Omega$ . In other

words, $G\cap U_{\alpha_1}$ and $E\cap U_{\alpha_1}$ are both non-empty. But, they are apparently disjoint, open in $U_{\alpha_1}$ and their union is $U_{\alpha_1}$ . This clearly contradicts that $U_{\alpha_1}$ from where it follows that $U_{\alpha_1}\subseteq E\cap \Omega$ .

Now, the rest is easy. For, suppose that $G\cap \Omega$ is non-empty. Then, by the previous analysis it follows that $U_{\alpha_2}\cap (G\cap \Omega)\ne\varnothing$ and so $U_{\alpha_2}\subseteq G\cap \Omega$ . But, this is a contradiction since $G\cap\Omega$ and

$E\cap\Omega$ are disjoint and $U_{\alpha_1}\cap U_{\alpha_2}$ is non-empty. It follows that $G\cap\Omega$ must be empty and thus no separation of $\Omega$ is possible. The conclusion follows.

**cp05** · April 21st 2010, 06:57 PM

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?

**Drexel28** · April 21st 2010, 06:59 PM

Originally Posted by cp05

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?

Clean it up, but yeah!

**cp05** · April 23rd 2010, 08:30 AM

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?

**Drexel28** · April 23rd 2010, 11:20 AM

Originally Posted by cp05

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?

The empty set is sometimes connected sometimes not depending on the author.

But, we can prove a nice little theorem.

Theorem: Let $E,G\subseteq\mathbb{R}$ be connected and $E\cap G\ne\varnothing$ then $E\cap G$ 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 $\mathbb{R}^2$ , 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.

**cp05** · April 24th 2010, 08:15 PM

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.

**Drexel28** · April 25th 2010, 07:06 PM

Originally Posted by cp05

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.

No, the two do intersect.

I guess an easier example would be two circles in $\mathbb{R}^2$ that intersect at two points. Clearly both are connected but their intersection are two disconnected poitns.

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