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

[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