# Homework Help: Connected subspace problem

1. Jun 30, 2017

### PsychonautQQ

1. The problem statement, all variables and given/known data
Let Y be a subspace of X and let both X and Y be connected. If X-Y=AUB where the intersection of A and B is empty, show that YUA is connected.

2. Relevant equations

3. The attempt at a solution
Say YUA = CUD where C and D are disjoint.

Let C_y be the intersection of Y with C and D_y be the intersection of D with Y.
Since A and Y have an empty intersection, Y=D_y U C_y, but since Y is connected this means that either D_y or C_y is empty, or in other words that Y is completely contained in either C or D.

Am I on the right track here? I am quite stuck now

2. Jul 1, 2017

### andrewkirk

Are you sure there isn't some information missing? In particular, I wonder if we were supposed to assume that A and B are open sets.

If not, the statement 'X-Y=AUB where the intersection of A and B is empty' tells us nothing, because for any set A, we get a disjoint union of that form simply by defining $B=X-Y-A$.

3. Jul 1, 2017

### PsychonautQQ

Yes, I'm sorry, A and B are open sets

4. Jul 1, 2017

### PsychonautQQ

So wlog let us assume that C is contained in Y,, if we can show that A must also be contained in C then YUA is connected because D would be empty. I have not yet used the information that X is connected. Perhaps we can a assume that the intersection of A and D is nontrivial and show that this leads to a contradiction by connectedness of X?

5. Jul 1, 2017

### WWGD

You mean $Y \cup A$ is connected as a subspace of X ?

6. Jul 2, 2017

Yeah

7. Jul 4, 2017

### PsychonautQQ

Okay so let me go over where I'm at, if somebody has any advice that'd be swell :D.

Recapping what we know:
Y is a subspace of X, both Y and X are connected.
X-Y=AUB where AiB = empty (i = intersection) and A and B are open sets in X-Y.
WWTS that YUA and YUB are both connected as subspaces of X

Then I supposed en route of a contradiction that YUA = DUC where DiC = empty and D and C are open in YUA.
YiA is empty because A is contained in the complement of Y.
D_y = DiY
C_y = CiY
So Y = D_y U C_y, but since Y is connected that must mean that either D_y or C_y is empty
WLOG let us say that D_y is empty and so Y < C (Y is contained in C).

So if I can somehow show that A is also contained in C, aka t hat DiA = Empty, then D is empty and hence YUA is connected. I haven't yet used the part our assumption that X is connected; So i'm assuming that DiA is not empty and then looking for a reason that this contradicts the connectedness of X.

Is this making sense? I hope so. Here is some more stuff:

D = D' i (YUA)
C = C' i (YUA)
where D' and C' are open in X. We know that such a D' and C' must exist by the definition of D being an open set in YUA where YUA is a subspace of X.

Also,
A = A' i (X-Y)
B = B' i (X-Y)
where A' and B' are open in X. We know that such a A' and B' must exist because X-Y is a subspace of of X and A and B are open sets in X-Y.

8. Jul 4, 2017

### PsychonautQQ

Say that Y is a subspace of X and both X and Y are connected. If X-Y is separable (X-Y = AUB where AiB = empty and A and B are open in X-Y) then show that YUA and YUB are connected.

Okay so let me go over where I'm at, if somebody has any advice that'd be swell :D.

Recapping what we know:
Y is a subspace of X, both Y and X are connected.
X-Y=AUB where AiB = empty (i = intersection) and A and B are open sets in X-Y.
WWTS that YUA and YUB are both connected as subspaces of X

Then I supposed en route of a contradiction that YUA = DUC where DiC = empty and D and C are open in YUA.
YiA is empty because A is contained in the complement of Y.
D_y = DiY
C_y = CiY
So Y = D_y U C_y, but since Y is connected that must mean that either D_y or C_y is empty
WLOG let us say that D_y is empty and so Y < C (Y is contained in C).

So if I can somehow show that A is also contained in C, aka t hat DiA = Empty, then D is empty and hence YUA is connected. I haven't yet used the part our assumption that X is connected; So i'm assuming that DiA is not empty and then looking for a reason that this contradicts the connectedness of X.

Is this making sense? I hope so. Here is some more stuff:

D = D' i (YUA)
C = C' i (YUA)
where D' and C' are open in X. We know that such a D' and C' must exist by the definition of D being an open set in YUA where YUA is a subspace of X.

Also,
A = A' i (X-Y)
B = B' i (X-Y)
where A' and B' are open in X. We know that such a A' and B' must exist because X-Y is a subspace of of X and A and B are open sets in X-Y.

9. Jul 5, 2017

### Staff: Mentor

I have inserted the LaTeX commands to increase readability (and hopefully without errors). It is still hard to read as you deal with a total of $10$ - in words TEN - different sets: $X,Y,A,B,C,D,C',D',C_y,D_y\,$, which disguises your argument quite well. One needs a machete to go through this ...

Last edited: Jul 5, 2017
10. Jul 5, 2017

### Dick

If $Y \subset C$ it must be that $A \cap D$ is nonempty. Try find open sets that separate $A \cap D$ from the rest of $X$ giving you a contradiction to $X$ being connected.

11. Jul 5, 2017

### Staff: Mentor

@fresh_42 Your post looks like your work, not the OP's I find this thread confusing enough without trying to get who wrote what.
Is there some way you could set off your translation like maybe HTML quote tags or something?

@PsychonautQQ -
You have a LOT of objects in your sketch. Fresh is right - using latex would help us poor old fogies read your post. PF has a nice page on latex composing for posts.

12. Jul 5, 2017

### Staff: Mentor

Corrected. I only thought it would be easier to read outside the quotation tags. Sorry.
@PsychonautQQ It is rather simple: Already the usage of $\text{$ math symbol $}$ around your set names would have helped a lot. And whether you type U or $\text{ \cup }$ and i or $\text{ \cap }$ isn't so much trouble. Or just use the symbols offered behind the $\Sigma$ symbol in our post editor, where the most common symbols could be added the same way as smileys can be used.

Last edited: Jul 5, 2017
13. Jul 5, 2017

### PsychonautQQ

Ugh okay guys sorry I've put off learning latex too long I see now it's criticalness

14. Jul 6, 2017

### PsychonautQQ

LaTex coming soon.... So since Y is contained in C it must be that AiD is nonempty other YUA = CUD wouldn't be a separation of YUA since D would be empty; cool i'm following.

Now you say I should find sets that separate AiD from the rest of X, because then I would have found a separation in X which is a contradiction because X is connected by hypothesis. But does the fact that a separation must be formed by disjoint OPEN sets? I suppose I could show that AiD will be open, but how do I know the other sets that form the separation will be as well?

15. Jul 6, 2017

### Dick

$C$ and $D$ are disjoint open sets that contain all of $X$ except for $B$. Replace $D$ with $A \cap D$, that's an open set , right? Does it intersect $B$?

16. Jul 6, 2017

### PsychonautQQ

But C and D are disjoint sets that are open in YUA, not necessarily open in X. That's why I brought in the D' and C' that would be open in X and their intersection with YUA would be D and C, respectively. But then the thing is I don't know if D' and C' are exactly what I need them to be.

Also, B is open in X-Y not necessarily in X.

Right?

17. Jul 6, 2017

### Staff: Mentor

I know separable in a topological context as something very different.
What you might mean is called separated here, so there might be a variation due to translation. However, two sets $A,B$ are separated, if $\overline{A} \cap B = A \cap \overline{B} = \emptyset$, which is different from what you wrote.
As a connected set $X$ to my knowledge is defined as a set that cannot be written as a union of two non-empty separated sets, this tiny differences might play a role.

I just have started to understand what you wrote, so it might be of no harm, but I don't want to do it twice.

18. Jul 6, 2017

### Dick

Right, but you are letting yourself get tangled up in which topology a set is open in, which is obscuring the point. This problem is about connectivity, not openness. Let's state the problem in its most general terms. Let $X$ and $Y$ be connected subsets of some set $U$. So the only open sets we will talk about are open in $U$. Suppose $X-Y=A \cup B$ and there are open sets $N$ and $M$ such that $A \subseteq N$ and $B \subseteq M$ and $N \cap M=\phi$. No need to assume $A$ and $B$ are themselves open, for example. Now show $Y \cup A$ is connected.

19. Jul 6, 2017

### Staff: Mentor

20. Jul 9, 2017

### andrewkirk

I have a solution for this problem (at least I think it's one. Possibly it has flaws that others may spot).
I'm uncertain as to whether this problem is homework though, since it was posted in a non-homework as well as a homework forum. If it is homework, I'd better not post the solution, only hints.

It's easier to post the solution though, because I've already written it (for my own entertainment), and judging an appropriate level for a hint is quite tricky for a complex problem like this.

So before going any further,
@PsychonautQQ Is this homework, or just something you're doing for fun?

21. Jul 10, 2017

### Dick

If the problem is getting 'complex' that's probably a hint you've taken a wrong turn. The OP was almost there.

22. Jul 10, 2017

### PsychonautQQ

Nope, it's not homework, just doing a self-study in topology before my MGRE in..... 9ish months haha. I suspected that this problem was trickier then it appears on the surface, especially when we are using Munkre's definition of connectedness where the two disjoint subsets must be open in whatever topology you are working in.

Last edited: Jul 10, 2017
23. Jul 10, 2017

### PsychonautQQ

The way Munkre's 'First Course in Topology' describes connected sets as what you said but with the added condition that the two disjoint subsets whose union is the whole set must be open sets in whatever topology you're working in. Now if a subspace of the original space is not connected then you will be able to find two subsets contained in the subspace that are open in the subspace topology whose union is the subspace and neither contains a limit point of the other; the fact that neither contains a limit point of the other will be a consequence of the way Munkre's is defining connectedness.

24. Jul 10, 2017

### PsychonautQQ

The way Munkre's 'First course in topology' (first edition) defines connectedness is that you won't be able to find two open disjoint subsets whose union is the whole set. So the definition of connectedness that i'm using involves openness.

25. Jul 10, 2017

### Dick

Right. But $A$ and $B$ themselves don't have to be open. They just have to be separated by open sets, $M$ and $N$ in my restatement.