Proving |A x C| ≤ |(A x B) x C|

[jeff1evesque]

Problem:
Let A, V, and C be any nonempty sets. Prove or disprove the following:
$$|A \times C| \leq |(A \times B) \times C|$$


Proof:
Fix b $$\in$$ B.
Set f(a,c) = ((a,b),c) for a $$\in$$ A and c $$\in$$ C.
Now suppose $$f(a_{1}, c_{1}) = f(a_{2}, c_{2})$$.
Thus, $$f(a_{1}, c_{1}) = ((a_{1}, b), c_{1}) = ((a_{2}, b), c_{2}) = f(a_{2}, c_{2})$$
Thus, $$(a_{1}, c_{1}) = (a_{2}, c_{2})$$ since the corresponding ordered triples are equal.
Thus, $$(a_{1}) = (a_{2}),$$ and $$(c_{1}) = (c_{2}),$$ since the ordered pairs above are equal.
Thus, f is one to one.
Finally, $$|A \times C| \leq |(A \times B) \times C|,$$ (by definition of one to one functions).


Questions:
I understand this proof and why b$$\in$$B is fixed, but to prove $$|A \times C| \leq |(A \times B) \times C|,$$ it is not necessary to state to fix b. Since b can be any value in the triple order, it doesn't have to be fixed, in fact it can hold any value from the set B and still fulfill the condition: $$|A \times C| \leq |(A \times B) \times C|$$. If what I am saying is true, is there an easy modification of the proof above? Personally I think by fixing b, we are limiting the scope of the proof.


Thanks,


JL

 

[Office_Shredder]

You need to fix b only so you can define your function properly. You're right, the function could be any

f: (a,c) -> (a,g(a,c),c)

where g is any function that maps (a,c) to B. But why bother saying that mouthful when you can just say "fix b"?