Proving Closure of A in Topological Space X

[c16]

Homework Statement

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.

Homework 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

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
Help please!

 

[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

 

[lanedance]

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

 

[lanedance]

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

 

[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)).

 

[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

 

[lanedance]

welcome to PF by the way, hope its helpful

 

[c16]

thanks so much for your help with c! I'd like to try to prove that int is distributive across 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??

 

[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

 

[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
$$c(A^c) = c(X-A) = X - i(A) = i(A)^c$$

 

[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!

 

[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)

 

[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