What is the Cardinality of Sets X, Y, and A, B, C?

[Aeonitis]

Hey guys, this is my first post, (Hi) was just wondering if i could get your help. I'm studying for my repeats and you guys can save me.

If X = {1,2,3,4}, Y = {2,4,6} what is the cardinality of the following sets?

(i) A = {x|x mod 2 = 0 and 0 <=x<=20}
(ii) B = X * X * Y
(iii) C = {(x,y)|x ≠ y and x,y ∈ X}

Please explain your train of thought in solving this. I am trying hard to understand the right way to approach this question quickly, Thank you for your time guys...

 

[Aeonitis]

I've tried them out, with the following answers:-

(i) A = {2,4,6,8,10,12,14,16,18,20}, Therefore A has a cardinality of 10 (elements).
(ii) B = Cartesian Product of 'X times X times Y' or better yet '4 by 4 by 3' elements each to give a total of 48 in cardinality?!
(iii) C = UNSOLVED!

I want to make sure someone agrees with me having the right answers since you're the pros

I really want to know what the '|' symbol stands for or means, as in 'x|x'. Hard to specifically search for in a book.

 

[Aeonitis]

Figured it out. I will post the full answer for future questioneers

(i) A = {2,4,6,8,10,12,14,16,18,20}, Therefore A has a cardinality of 10 (elements).
(ii) B = Cartesian Product of 'X times X times Y' or better yet '4 by 4 by 3' elements each to give a total of 48 in cardinality?!

(iii) C = {(x,y)|x ≠ y and x,y ∈ X}

pairs x,y {such as (1,1),(1,2),etc...} drawn from set X with a cardinality of '4 by 4 = 16' as in the question "x,y ∈ X". Due to the statement 'x ≠ y' pairs can't come in equals, discarding the following four sets (1,1),(2,2),(3,3),(4,4). The end product is 16-4 giving a cardinality of 12 for set 'C'.

 

[JSuarez]

How much is 0 (mod 2)? In the set A, the possible values of x include 0, don't they? Anyway, I'm sure you remember that any number that is 0(mod2) is even and vice-versa.

 

[Aeonitis]

Yes i do, and thanks for pointing that out in any case, it's always the little things that count >_<

 

[ZQrn]

C has the cardinality|X*X| - |x|.

Because for every x in X, there is a pair (x,x), which are exactly the ones not in {(x,y) : x != y /\ x,y in X}