Proving Connectedness of YUA and YUB with X and Y as Connected Subsets

  • Thread starter MathematicalPhysicist
  • Start date
  • Tags
    Sets
In summary, this problem is trying to show that if two sets are separated, then each set is not contained in the other. However, this proof by contradiction is not correct because the connectedness of X is needed.
  • #1
MathematicalPhysicist
Gold Member
4,699
371
Problem Statement:
Let Y be a subset of X, and X and Y are connected, show that if A and B form a separation 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?

thanks in advance.
 
Physics news on Phys.org
  • #2
Looks to me like "proof by contradiction": assume YUA is NOT connected and show that X is not connected.
 
  • #3
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 separation of Y.
 
  • #4
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:
  • #5
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
loop quantum gravity said:
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 separation of Y.

then what's your problem? You assumed YUA isn't connected and arrived at a contradiction.
 
  • #7
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:
  • #8
HallsofIvy said:
then what's your problem? You assumed YUA isn't connected and arrived at a contradiction.

I just want to see if it's correct, so?
 
  • #9
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
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
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?
 
  • #13
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:

1. How do you define connectedness of sets?

Connectedness of sets refers to the property that a set cannot be divided into two non-empty, disjoint subsets. This means that all points in the set must be "connected" in some way, without any gaps or breaks.

2. What is the significance of proving connectedness between subsets?

Proving connectedness between subsets is important because it helps us understand the relationship between different parts of a set. It also allows us to make conclusions about the behavior of the set as a whole based on the behavior of its subsets.

3. Can you provide an example of proving connectedness of sets?

One example of proving connectedness of sets is showing that the interval [0,1] on the real number line is connected. This can be done by considering two subsets A and B, where A contains all points less than 1/2 and B contains all points greater than or equal to 1/2. It can then be shown that there are no points in A that are adjacent to points in B, proving that the interval is connected.

4. How does the proof of connectedness depend on the choice of X and Y as connected subsets?

The proof of connectedness can depend on the choice of X and Y as connected subsets, as different choices may lead to different conclusions about the connectedness of the sets. However, as long as X and Y are both connected subsets, the proof will follow the same general logic.

5. Are there any practical applications of proving connectedness of sets?

Yes, there are many practical applications of proving connectedness of sets. For example, it is used in topology to classify different types of spaces, and in graph theory to determine the connectivity of networks. It is also important in real-world applications such as understanding the behavior of connected electrical circuits or analyzing the spread of diseases within a connected population.

Similar threads

  • Calculus and Beyond Homework Help
2
Replies
39
Views
4K
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
539
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
961
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
456
Back
Top