[SOLVED] Cardinal Exponentiation

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April 28th 2010, 04:18 AM
Deadstar

[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 $\kappa$ , $\lambda$ and $\mu$ be three cardinals.

Show that,
$<br /> \kappa^{\lambda + \mu} = \kappa^{\lambda} \kappa^{\mu}$ .
April 28th 2010, 07:55 AM
MoeBlee

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 _______?
April 28th 2010, 08:02 AM
Deadstar

Quote:

Originally Posted by MoeBlee

What is the most natural way you'd think to show equality between cardinals?

bijections.
April 28th 2010, 08:17 AM
MoeBlee

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.
April 28th 2010, 08:59 AM
Deadstar

Quote:

Originally Posted by MoeBlee

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.

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 $A^{B \cup C}$ ~ $A^B A^C$ . A function $f \in A^{B \cup C}$ is a function of 2 variables... $(b \cup c) \to f(b \cup c)$ or something like that.

Then define an element in the RHS as something else and then somehow they'll be equal.
April 28th 2010, 09:16 AM
MoeBlee

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.
April 28th 2010, 09:28 AM
Deadstar

Replace $A^{B \cup C}$ by $A^{B \times \{0\} \cup C \times \{1\}}$ ?
April 28th 2010, 09:51 AM
MoeBlee

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.
April 28th 2010, 09:53 AM
Deadstar

Quote:

Originally Posted by MoeBlee

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}.

You with me so far?

Quote:

Originally Posted by Deadstar

A function $f \in A^{B \cup C}$ is a function of 2 variables... $(b \cup c) \to f(b \cup c)$

Yeah that's what I was trying to show in the quoted post. Probably should have been a bit clearer.
April 28th 2010, 10:07 AM
MoeBlee

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?
April 28th 2010, 10:24 AM
Deadstar

Quote:

Originally Posted by MoeBlee

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?

I don't understand what you mean by kind. Cartesian Product?

Every member has of {g | g is a function from B into A} X {h | h is a function from C into A} has elements from both A and B.
April 28th 2010, 10:32 AM
MoeBlee

Quote:

Originally Posted by Deadstar

Cartesian Product?

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.
April 28th 2010, 10:33 AM
MoeBlee

Quote:

Originally Posted by Deadstar

Every member has of {g | g is a function from B into A} X {h | h is a function from C into A} has elements from both A and B.

No, every element of that set is an ordered pair <g h> where g is some function from B into A and h is some function from C into A.
April 28th 2010, 10:56 AM
Deadstar

Quote:

Originally Posted by MoeBlee

So, for a given f, think about what functions you'd like g and h to be.

I don't get what you're implying here. I can't choose what g and h are.
Since we only define $a \cup b$ as one element I really can't see a way to map it. Surely distinguishing between a and b within $a \cup b$ and then mapping each to g or h respectively is easier.
April 28th 2010, 11:01 AM
MoeBlee

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?

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