# Topology proof

1. Jul 28, 2010

### c16

1. The problem statement, all variables and given/known data
Let X be a topological space. If A is a subset of X, the the boundary of A is closure(A) intersect closure(X-A).
a. prove that interior(A) and boundary(A) are disjoint and that closure(A)=interior(A) union boundary(A)
b. prove that U is open iff Boundary(U)=closure(U)-U
c. If U is a nonempty open set in X, is it true that U=Interior(closure(U))? show it or give a counter example.

2. Relevant equations
-the complement of a closed set is open
- the closure of a set is defined as the intersection of all closed set containing A
- the interior of a set is the union of all open sets contained in A

3. The attempt at a solution
a. To prove that Int(A) and Boundary(A) are disjoint, assume the opposite. Let x be in both Int(A) and Bound(A). Then, if x is in Int(A), since Int(A) is open, there exists an open ngbd U of x such that U is contained in Int(A). Now, x in Bound(A) means that x is both in closure(A) and in closure(X-A). If x is in the closure of X-A, then every open ngbd of x intersects X-A. But, U is contained in A, so this is a contradiction. Thus Int(A) and Bound(A) are disjoint. (Am I missing anything here??)
Now, I'm not sure how to prove this equality, although I understand it intuitively. I thought about doing it like this: Int(A) union Bound(A)=Int(A) union (clos(A) intersect clos(X-A))= (Int(A) union clos(A)) intersect (Int(A) unions clos(X-A))=clos(A) intersect (Int(A) union (clos(X-A)). Now, I just need to show that Int(A) union clos(X-A) =X, but I'm not sure how to do that.
b. Assume Bound(U)=clos(U)-U, and U is not open. Then, U(complement) is not closed. Then, Bound(U)=clos(U) intersect U(complement) is not necessarily closed... it seems like this is getting a bit hazy..
Assume U is open. Then, clos(U)-U=clos(U)-Int(U)= clos(U) union (Int(U) complement)... again I need to show that Int(U) complement = clos(X-U)
c. pretty sure this is true, don't quite know how to prove it

2. Jul 28, 2010

### lanedance

a) equality - how about trying to show any point in closure(A) is either in the interior or the boundary and then vice versa, helped by the fact int & bound are disjoint

3. Jul 29, 2010

### lanedance

for b) this is equivalent to showing U open iff U = interior(U)

4. Jul 29, 2010

### lanedance

for C) use closure(U)=interior(U) union boundary(U)

5. Jul 29, 2010

### c16

I'm not sure how C) helps me... so if closure(U)=Interior(U) union Bound(U), then Int(closure(U))=Int(Int(U) union Bound(U))=Int(U union Bound(U)) because U is open. Not sure where to go from here... what I really need to show is that Int(U)=Int(Closure(U)).

6. Jul 29, 2010

### lanedance

So as you've said
Int(closure(U))=Int(Int(U) union Bound(U))_____________from a)

Int(Int(U) union Bound(U))=Int(U union Bound(U)) _______as U is open

let S = U union Bound(U))
now U & Bound(U) are disjoint as U is open
first consider a point in U, you should be able to show this is in interior(S) as U is open
now consider a point in Bound(U), can you show it is not in interior(S)...

if so you have shown int(S) = U, which should do it

Alternatively, though equivalenty, I think you could probably show both:
- the interior is distributive across unions (think about unions of open sets..)
- the interior of a boundary is the empty set
which would get you there as well

7. Jul 29, 2010

### lanedance

welcome to PF by the way, hope its helpful

8. Jul 29, 2010

### c16

thanks so much for your help with c!! I'd like to try to prove that int is distributive accross unions eventually, but would you help me with part (a) first? I have that Int(A) and Bdy(A) are disjoint, and now I need to prove that Closure(A)=Int(A) union Bdy(A). I know that if x is in Int(A), x is in closure(A) since Int(A)\subset Closure(A). Also, I know that if x is in Bdy(A), x is in closure(A) by the definition of boundary. Thus, I know that Int(A) union Bdy(A) \subset closure(A). I'm having trouble proving it the other way too. If x is in closure(A), maybe I can assume it is neither in Int(A) or Bdy(A) and try to get a contradiction??

9. Jul 29, 2010

### lanedance

so just to change notation a little, as its a bit tough going, so to streamline things, let talk about the set A instead of U, and let:
boundary = b(A)
interior = i(A)
closure = c(A)
interestcion = $\cap$
interestcion = $\cup$

then for a), first to show they are disjoint
$$b(A) = c(A) \cap c(A-X)$$

then consider
$$i(A) \cap b(A) = i(A) \cap c(A) \cap c(X-A)$$

now clearly
$$i(A) \cap c(A) = i(A)$$

that gives
$$i(A) \cap b(A) = i(A) \cap c(X-A) = i(A) \cap c(A^c)$$
where A^c is the complement of A
so it only remains to show that is the empty set, to show they are disjoint

Last edited: Jul 30, 2010
10. Jul 30, 2010

### lanedance

for the next part how about directly
$$i(A) \cup b(A)$$
$$i(A) \cup ( c(A) \cap c(A^c))$$
$$(i(A) \cup c(A)) \cap (i(A) \cupc(A^c))$$
$$c(A) \cap (i(A) \cup c(A^c))$$

now say you have a point x in X, not in the right hand side
$$x \notin i(A) \cup c(A^c)$$

can you show that is a contradiction in itself...

alternatively i think you could try and show
$$c(A^c) = c(X-A) = X - i(A) = i(A)^c$$

11. Jul 30, 2010

### c16

I think I've got it all, except for the last part of part c. I'm trying to show that if S=U union bdy(U), then Int(S)=U. Remember that U is open. Obviously, I have no trouble showing that if x is in U, x is in Int(S). However, how do I show that if x is in Bdy(U), x is not in Int(S)? I've shown that if x is in Bdy(U), x is not in U. But, then saying that x is not in Int(S) right off the bat would be assuming what I'm trying to prove!

12. Jul 31, 2010

### lanedance

so consider an any open ball B contained in c(U)
if it has an element in the boundary b(U)
then B must intersect X-U as youve said
but as the ball is open then its a contradiction as then the ball is not fully contained in c(U)
so every open set must be contined in U = i(U)

13. Jul 31, 2010

### lanedance

to further explain, the boundary is a closed set and coincides for U and X-U by def'n

consider open set B containing a point from the boundary
the intersection of B with the boundary is not open
and in general B must conatin points from the interior of both U & X-U

Last edited: Aug 1, 2010