Limit points of Aintersect B, closer of (AintersectB)

[kathrynag]

Homework Statement

I was given a problem from my book.
a) If y is a limit point of AUB, show that y is either a limit point of A or a limit point of B.
b) Prove that (AUB) closure=A closureUBclosure
c) Does the result about closures in b)extend to infinite unions in sets


I was basically asked to consider this problems with intersections instead of unions.

Homework Equations

The Attempt at a Solution

a)Let y be a limit pt of AintersectB
Then y=limy_n
y_n is an element of A intersect B
So some subsequence (y_n) must lie in A and B
So y is a limit pt of A and B
b)I will consider (AintersectB)closure=Aclosure intersect Bclosure and want to see if this is true.
(AintersectB)closure=AintersectBUL(AintersectB)
Part a says L(AintersectB)=L(A)intersectL(B)
So (AintersectB)closure=A intersect BU L(A) intersect L(B)
=(AUL(A)) intersect (BintersectL(B)
=Aclosure intersect B intersect L(B)
I'm not entirely sure if this is correct
c) I'm not sure how to get started

 

[micromass]

Part a is correct.

For part b you said that a implies
L(A\cap B)=L(A)\cap L(B)

But part a only says that
L(A\cap B)\subseteq L(A)\cap L(B)
The other inclusion is incorrect.
In fact, part b is incorrect: take A=]0,1] and B=[-1,0[.

Part c is also incorrect...

 

[kathrynag]

I don't get why the other inclusion is incorrect:
AintersectB contained in L(A) intersect L(B)
maybe it's where I'm not sure where to go

 

[micromass]

Try A=[-1,0[ and B=]0,1]. Then you will see that L(A\cap B)\neq L(A)\cap L(B). So the other inclusion will be incorrect.

 

[kathrynag]

Ok then I'm a little confused on simplifying this then:
AintersectB contained in L(A) intersect L(B)

 

[micromass]

Wait, I don't follow anymore... Why would you want to do that again?
You know part b is false?? right?

 

[kathrynag]

I understand it's false, but I didn't know if I could go further and find what it would equal
If part b) is false, I can't really make a statement about the result for c) right

 

[micromass]

Ow I see...

But Id really doubt if you could find a nice formula for the closure of A\cap B. At least, I don't know one...

Part c is also false. Take [-1,0[ and ]0,1] together with the collection {[-n,n] | n>1} This will do as counterexample...

 

[kathrynag]

Is there anything I could do with simplifying the closure of A intersect B just to show what it simplifies to? I guess that would make it easier to make a statement.
Could I do something like :
closure A intersect B=A intersect B U L(AintersectB)
Then I would have some sort of formula down to use for c)
Or would it be easier to use a counterexample for c)?

 

[micromass]

Yes, you do have that

\overline{A\cap B}=(A\cap B)\cup L(A\cap B)

But you can't simplify that. And I don't really see how that would be handy for c, but that's just me

I think finding a counterexample for c would be best...

 

[kathrynag]

Ok fair enough. I will try a counterexample

 

[micromass]

Hint: I gave you a counterexample in post 8