# Closure of Sets Proof

1. Sep 14, 2010

### mynameisfunk

1. The problem statement, all variables and given/known data

See attachment

2. Relevant equations

3. The attempt at a solution

I am not sure how I should approach this first off. I have tried this 3 ways but I always decide they dont work. Click on the other attachment to see my work, It's only the first part of the first one, but I would just like to know if it works.

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Last edited: Sep 14, 2010
2. Sep 14, 2010

### Dick

That looks ok so far, but you might want to elaborate on why if x_i is in A or B for all i, then the limit point x is either a limit point of A or a limit point of B. And you might want to be a little more careful about notation. {x_i} can mean either the whole sequence or it could be a singleton set consisting of the single element x_i. If x_i is an element of A, then {x_i} isn't necessarily an element of A, it's a subset of A. The second half of the proof should be even easier.

If you are trying to prove the closure of the intersection is the intersection of the closures, then stop. It's false. Look for a counterexample.

3. Sep 15, 2010

### mynameisfunk

OK thanks, I am working on the second one now...
How do I know if (A$$\cap$$B) $$\cap$$ (A$$\cap$$B)' is empty?
I am assuming this must be the case in order for the two not to be equal to each other..
It seems like they don't necessarily have to be since a limit of some {x_i} in A can be in B and vice versa? It seems like it is true to me..

4. Sep 15, 2010

### Dick

You might be thinking too hard. Think about two open intervals on the real line. The closures are 'bigger' than the intervals. The closures might intersect even though the intervals don't.

5. Sep 15, 2010

### mynameisfunk

Thanks tons, Dick.