Hello, can any one help me with this problem?
Verify that, for all subsets A and B of a topological space, we have
____....._.._........_..._......____
A∩B ⊂ A∩B and A \ B ⊂ A \ B ;
check that the inclusions cannot be replaced by equilities.

I just can't figger out how to solve it. The points are just the space between the under line symbols.
Thank you very much!

I will write cl(A) for closure, rather than using the bar.

Checking cl(A∩B) contained in cl(A)∩cl(B) is not too hard. All you need to show is that cl(A)∩cl(B) is a closed set containing A∩B. But A is in cl(A) and B is in cl(B) so A∩B is in cl(A)∩cl(B). Furthermore cl(A)∩cl(B) is closed since it is the intersection of two closed sets.


For the second one, suppose x is in cl(A)\cl(B). To show that x is in cl(A\B), we want to show that x is in every closed set containing A\B. So suppose F is a closed set containing A\B. Then FUcl(B) is a closed set containing A, so x is in FUcl(B) since x is in cl(A). But x is not in cl(B), so it follows that x must be in F.



For counterexamples equality in both you could use A as the set of rationals and B as the set of irrationals, both subsets of the topological space R. Remember that the closure of both sets is R.