I have a ton of these that I want to do but I need to see how one is done to get a feel on what to do...

Let , and be three cardinals.

Show that,

.

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- Apr 28th 2010, 05:18 AMDeadstar[SOLVED] Cardinal Exponentiation
I have a ton of these that I want to do but I need to see how one is done to get a feel on what to do...

Let , and be three cardinals.

Show that,

. - Apr 28th 2010, 08:55 AMMoeBlee
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 _______? - Apr 28th 2010, 09:02 AMDeadstar
- Apr 28th 2010, 09:17 AMMoeBlee
Exactly correct.

Now "unpack" the two cardinals in the equation, by using the definitions of cardinal exponentiation, cardinal addition, and cardinal multiplication toward the goal of setting up a bijection. - Apr 28th 2010, 09:59 AMDeadstar
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. - Apr 28th 2010, 10:16 AMMoeBlee
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. - Apr 28th 2010, 10:28 AMDeadstar
Replace by ?

- Apr 28th 2010, 10:51 AMMoeBlee
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. - Apr 28th 2010, 10:53 AMDeadstar
- Apr 28th 2010, 11:07 AMMoeBlee
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? - Apr 28th 2010, 11:24 AMDeadstar
- Apr 28th 2010, 11:32 AMMoeBlee
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. - Apr 28th 2010, 11:33 AMMoeBlee
- Apr 28th 2010, 11:56 AMDeadstar
- Apr 28th 2010, 12:01 PMMoeBlee
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?