To get you started, answer this question:
What is the most natural way you'd think to show equality between cardinals?
Hint: Cardinality is based on the notion of equinumerosity, which is based on the notion of _______?
Right but this is why I posted the question. You clearly use bijections but I never know how to find it. I have never understood inj/sur/bijections, no lecturer/book/website has ever succeeded in making me understand them. It's just something I don't get and don't imagine I ever will nor will need to... But anyway.
It would be like... I want to show that ~ . A function is a function of 2 variables... or something like that.
Then define an element in the RHS as something else and then somehow they'll be equal.
You didn't fully unpack.
Replace the exponentiation A^(B u C) by a certain set (as you already alluded, it's a certain set of functions).
Then replace the multiplication (A^B)*(A^C) by a certain set.
In that result, replace the exponentiation A^B by a certain set (a certain set of functions).
And in that result, replace the exponentiation A^C by a certain set (a certain set of functions).
(Also make a note, for later use, that B and C may be assumed to be disjoint.)
Once you do that, then think of how to define a bijection. I'll help you if you need.
You could do that. But we can make it simpler just by assuming that B and C are disjoint (which we should be able to do, given an appropriate definition of cardinal addition).
So let's assume B and C are disjoint.
I'll do the unpacking:
For A^(BuC) we have:
{f | f is a function from BuC into A}.
For (A^B)*(A^C) we have:
{g | g is a function from B into A} X {h | h is a function from C into A}.
So, we're unpacked now.
So now we want to construct a bijection J from:
{f | f is a function from BuC into A}
onto
{g | g is a function from B into A) X {h | h is a function from C into A}.
So for each f in {f | f is a function from BuC into A} we have to say what J(f) is.
Now I'll let you work out what you think J(f) should be. I'll help you if you need.
f is not really a function of two variables. It's just a function of one variable. For each object x in BuC, we have that f maps x to some member of A. So 'x' is the only variable.
So, for each f in {f | f is a function from BuC into A} we want to map f to a member of
{g | g is a function from B into A} X {h | h is a function from C into A}.
Now, you do have the right idea that both B and C are going to be involved here. All you have to do now is think about f (which is a function from BuC into A) and which member of
{g | g is a function from B into A} X {h | h is a function from C into A}
you want to map f to.
So think about the members of
{g | g is a function from B into A} X {h | h is a function from C into A}.
Now, answering this question will lead you further:
What KIND of set is
{g | g is a function from B into A} X {h | h is a function from C into A} ?
So, what form does every member of
{g | g is a function from B into A} X {h | h is a function from C into A}
have?
Right. So every member is an ordered pair, which is of the form <g h>
where g is a member of {g | g is a function from B into A}
and h is a member of {h | h is a function from C into A}
So, now for each f in {f | f is a function from BuC into A} we need to map f to some ordered pair <g h> as described above. So, for a given f, think about what functions you'd like g and h to be.
For each f in {f | f is a function from BuC into A} you're going to choose an appropriate g and h where <g h> is in
{g | g is a function from B into A} X {h | h is a function from C into A}.
Now, think about f. It's a function whose domain is BuC. But B and C are disjoint. So f is the union of two functions, one function from B into A and another function other from C into A.
So, for a given f, what should g and h be?