Is Y Union A Connected When X Minus Y Splits into Disjoint Subsets A and B?

[Dick]

Let $X$ and $Y$ be connected subsets of some set $U$. Suppose $X-Y=A \cup B$ and there are open sets $N$ and $M$ such that $A \subseteq N$ and $B \subseteq M$ and $N \cap M=\phi$. Now show $Y \cup A$ is connected.

Assume $C$ and $D$ separate $Y \cup A$ such that $Y \subseteq C$. Then clearly $A \cap D$ is nonempty. Then $D \cap N$ (this part contains $A \cap D$) and $C \cup M$ (this contains the rest of $X$) separate $X$. They are disjoint because $C \cap D=\phi$ and $N \cap M=\phi$. This contradicts $X$ being connected hence $Y \cup A$ is connected. QED.

I don't know why this is so elusive...

 

[andrewkirk]

I don't know why this is so elusive...

It's not elusive. It's just a different problem from the one in the OP.

 

[Dick]

It's not elusive. It's just a different problem from the one in the OP.

In what way?? As I've noted before the original problem is the special case $A=N$, $B=M$, $U=X$. The proof of the special case is the same as the proof of the general.

 

[PsychonautQQ]

I am also curious as to why what dick did was a diff problem, it seems legit to me. Still a little worried about how we know A∩D and C∪B are open in X though.

 

[Dick]

I am also curious as to why what dick did was a diff problem, it seems legit to me. Still a little worried about how we know A∩D and C∪B are open in X though.

I'm not sure why. In the context of your formulation aren't $A, B, C, D$ open in $X$?

 

[PsychonautQQ]

I'm not sure why. In the context of your formulation aren't $A, B, C, D$ open in $X$?

Nope, A and B are open in X-Y and C and D are open in YUA.

 

[andrewkirk]

In what way?? As I've noted before the original problem is the special case $A=N$, $B=M$, $U=X$. The proof of the special case is the same as the proof of the general.

The short answer is simply that it has different premises and variable names, therefore it is a different problem.

eg consider the case $U=\mathbb R,\ \ X=[0,2],\ \ Y=\{1\}$, which satisfies the premises of the problem in the OP, but not those of this new problem.

To be a little more helpful, in the new problem, $N,M$ are open in $U$ and disjoint. In the original problem $A,B$ are open in $Y-X$ and disjoint, but we have no reason to expect they are open in $X$, nor do we have any reason to expect that there exist disjoint sets $N,M$ that are open in $X$ such that $A=N\cap (Y-X),\ B=M\cap (Y-X)$. ie it needs to be proven that we can find disjoint $N,M$. Maybe I misunderstood, but I got the impression that it was precisely the task of proving that disjointness that the OP was finding difficult, and I confess that I see no easy way to demonstrate that disjointness either.

Now quite possibly that disjointness can be proven. But that needs to be done, not just assumed, and that makes the proof longer, and hence less straightforward than the proof of the new problem.

 

[Dick]

Now quite possibly that disjointness can be proven. But that needs to be done, not just assumed, and that makes the proof longer, and hence less straightforward than the proof of the new problem.

Yeah, I see what you are saying. Back to the drawing board.

 

[ConfusedMonkey]

I remember this problem being assigned in a topology course I took a few years ago and it was a pain in the ass. I also remember the proof being significantly shorter than andrewkirks. Not saying his proof is wrong - I didn't read it - but there is a more elegant way to go about the problem...

 

[PsychonautQQ]

I remember this problem being assigned in a topology course I took a few years ago and it was a pain in the ass. I also remember the proof being significantly shorter than andrewkirks. Not saying his proof is wrong - I didn't read it - but there is a more elegant way to go about the problem...

Yes Andrew said that he believed his proof could b streamlined a bit. Looks.brilliant to me tho