1. ## Cardinal Numbers

This question has two parts - I've solved the first part, only I can't seem to find a way to solve the second part:

*Let A,B,C be sets. Prove that if B,C ⊆ A , |C|=|B|, and BC= then |A-B|=|A-C|.

*Find, using the first part, the cardinal number of all the real irrational numbers between 0 and 1.

now, I understand that somehow I probably need to use a few sets I know:
Q - the rational numbers, |Q|=Alef 0 (|Q|=|N|)
(0,1) - as required. If I'm not mistaking, it's cardinal number is also Alef 0.

2. Once again without knowing exactly what you have already have done it is impossible to help you.
Usually one of the first theorems done in any treatment of cardinally is Cantor’s theorem:
the open interval (0,1) is uncountable. This is proven with his famous diagonal argument.

Now suppose that $Q^* = \mathbb{Q} \cap (0,1)\;\& \;I^* = (0,1) \cap \mathbb{R}\backslash \mathbb{Q}$.
We know that $Q^*\cup I^*=(0,1)$ because $Q^*$ is countable and $(0,1)$ uncountable this means $I^*$ must be uncountable.

3. use the basic defn of cardinaity. if |C|=|B| then there exists a bijection between them.

now extend this bijection to entire A. for example you may choose to map any element not belonging to B or C to itself. this bijection when restricted to A-B gives you the required answer.

4. oops i am sorry i did not see that u had solved the first part. ok for the second part take A = R, B = Z and C = {rationals between 0 an 1}. now B and C are equicardinal and disjoint. also use the fact that (0,1) and R both have cardinality c. (or just say that they have equal cardinality)

anyway this is a phoney construction just to use the facts provided. use cardinal arithmetic for infinite cardinals instead.