# Connected Sets.

1. Apr 19, 2008

### MathematicalPhysicist

Problem Statement:
Let Y be a subset of X, and X and Y are connected, show that if A and B form a seperation of X-Y then YUA and YUB are connected.

Attempt at solution:
Well I'm not sure where the fact that X is connected comes to play (perhaps it gurantees us the possibility of X-Y not to be connected).
anyway if we look at: seperations of YUB=CUD and YUA=C'UD'
then Y=(YUB)-B=(C-B)U(D-B), Y=(YUA)-A=(C'-A)U(D'-A)
which are both seperations of Y which is a contradiction to Y being connected, am I correct here or yet again wrong?

any input?

2. Apr 20, 2008

### HallsofIvy

Staff Emeritus
Looks to me like "proof by contradiction": assume YUA is NOT connected and show that X is not connected.

3. Apr 20, 2008

### MathematicalPhysicist

Yes I assumed that YUA isn't connected, but got that then Y isn't connected.
cause if YUA=BUC
then Y=(YUA)-A=(B-A)U(C-A) which is a seperation of Y.

4. Apr 21, 2008

### andytoh

Suppose YUA is not connected, with separation CUD. Since Y is connected, then Y must be a subset of C or D, say Y is a subset of C. Now show that X is not connected.

In the end, you will find that BUC and D are closed subspaces, and thus X = (BUC)UD is a separation of X.

Last edited: Apr 21, 2008
5. Apr 22, 2008

### MathematicalPhysicist

You mean Y should be D-A or C-A, let's say C-A, then because X-Y=AUB
then X=XUX-Y=AU(BUC-A)
which is a separation of X, something like this?

6. Apr 22, 2008

### HallsofIvy

Staff Emeritus
then what's your problem? You assumed YUA isn't connected and arrived at a contradiction.

7. Apr 22, 2008

### andytoh

Are you sure you separated Y correctly?

You will need to use closures in your proof. Make use of the fact that if two sets form a separation, then each set does not meet the closure of the other, i.e. ClA is in X-B, ClB is in X-A, ClC is in X-D, ClD is in X-D.

Suppose YUA is not connected, with separation CUD. Since Y is connected, then Y must be a subset of C or D, say Y is a subset of C. Show that BUC and D are closed subspaces by showing that they equal to their own closure. Then X = (BUC)UD is a separation of X, a contradiction.

Last edited: Apr 22, 2008
8. Apr 22, 2008

### MathematicalPhysicist

I just want to see if it's correct, so?

9. Apr 22, 2008

### andytoh

Your solution is not correct because you didn't separate Y properly (read the precise definition of separation). The correct solution is more complicated than that. The connectedness of X is needed.

10. Apr 24, 2008

### MathematicalPhysicist

So let me see if I get it A and B form a sparation of X-Y=AUB i.e they are open in X-Y and disjoint.
now assume YUA=CUD where C and D are open in YUA and disjoint, now
Y=(YUA)-A=(C-A)U(D-A), but C-A and D-A arent open in Y, then how do i reconcile it.
I understnad that i need to show that BUC is closed in X and D is closed in X, but how?

11. Apr 26, 2008

### MathematicalPhysicist

Ok X=(BUC)UD
and BUC and D are disjoint this I know.
how to show that cl(BUC) equals BUC?
we know that
cl(BUC)=cl(B)Ucl(C)
and B is closed in X-Y thus B=(X-Y)^cl(B)=cl(B)
and C is closed in YUA, then C is closed in A and thus also in X-Y, and again we get that B=cl(B)
for D, we know the same that D is contained in A and thus D=cl(D)
so both these sets are closed.

is this correct?

12. Apr 26, 2008

### andytoh

13. Apr 26, 2008

### MathematicalPhysicist

so the fact that cl(A) is a subset of X-B comes evidently from the fact that:
A and B are disjoint, and (X-Y)-B equals A, then because clA contains A, it must be contained in X-B which contains (X-Y)-B.

Ok I got it, sorry for my thickness, three exams in four days can make me anxious.

btw, great site, wish I knew it a week ago.

Last edited: Apr 26, 2008