# Cardinal Exponentiation

• Apr 2nd 2010, 09:36 AM
Cardinal Exponentiation
Right so, I have this major block with set exponentiation and unfortunately we have to do it in my set theory course with cardinals.

I'm gonna post the question and the answer and post what (after many months of thinking about it) is the best I can do on my own...

Question. If κ, λ and μ are cardinals with 0 < κ ≤ λ. Show that $\mu^{\kappa} \leq \mu^{\lambda}.$

Lecturers Solution. Let κ = P(A), λ = P(B) and μ = P(C). Let f:AB be an injection.
Let k:BA be the surjective map defined in the proof of Prop 1.3.1 which exists because κ > 0.
Then we define a map $C^A \to C^B$ by G(g)(b) = g(k(b)) (+++).

We claim G is injective.

If G(g)=G(g′) then g(k(b))=g′(k(b)).
Since k is surjective k(b) ranges over all possible elements of A and so g(a)=g′(a) for all aA. Hence, g=g

--------------------------------------------------------------------------------------------------------------------

Notice the (+++), this is the part that scrambles my brain, making injections, surjections and bijections.
I'm not quite sure what $g$ is either, I seem to think it's a possible injection that takes an element in $C^A$ to $C^B$.

But, my attempted answer is slightly different and I want to know if it looks mildly right and if not, can you explain what the map in (+++) actually means!

So we have $f: A \to B$ is an injective map.

Define $k: B \to A$ as a surjective map from B to A.

So, for every $a \in A$ there is a $b \in B$ such that $a = k(b)$.

So, now the (Headbang) bit (for me)... We define a map from $C^A \to C^B$ by...
$G: g(a) = g(k(b))$...

I think I'll just stop there though, the rest of the Q is a bit easier, just the defining the map part that gets me.

Having just re-read it I'm pretty sure mine isn't even a map, it's just a renaming of $a$.
• Apr 2nd 2010, 12:21 PM
Drexel28
Quote:

Right so, I have this major block with set exponentiation and unfortunately we have to do it in my set theory course with cardinals.

I'm gonna post the question and the answer and post what (after many months of thinking about it) is the best I can do on my own...

Question. If κ, λ and μ are cardinals with 0 < κ ≤ λ. Show that $\mu^{\kappa} \leq \mu^{\lambda}.$

Solution. Let κ = P(A), λ = P(B) and μ = P(C). Let f:AB be an injection.
Let k:BA be the surjective map defined in the proof of Prop 1.3.1 which exists because κ > 0.
Then we define a map $C^A \to C^B$ by G(g)(b) = g(k(b)) (+++).

We claim G is injective.

If G(g)=G(g′) then g(k(b))=g′(k(b)).
Since k is surjective k(b) ranges over all possible elements of A and so g(a)=g′(a) for all aA. Hence, g=g

Notice the (+++), this is the part that scrambles my brain, making injections, surjections and bijections.
I'm not quite sure what $g$ is either, I seem to think it's a possible injection that takes an element in $C^A$ to $C^B$.

But, my attempted answer is slightly different and I want to know if it looks mildly right and if not, can you explain what the map in (+++) actually means!

So we have $f: A \to B$ is an injective map.

Define $k: B \to A$ as a surjective map from B to A.

So, for every $a \in A$ there is a $b \in B$ such that $a = k(b)$.

So, now the (Headbang) bit (for me)... We define a map from $C^A \to C^B$ by...
$G: g(a) = g(k(b))$...

I think I'll just stop there though, the rest of the Q is a bit easier, just the defining the map part that gets me.

Having just re-read it I'm pretty sure mine isn't even a map, it's just a renaming of $a$.

I will come back later and re-read your proof but you're idea is correct.
• Apr 2nd 2010, 12:46 PM
Quote:

Originally Posted by Drexel28
I will come back later and re-read your proof but you're idea is correct.

Cool thanks, it's not exactly a full proof but I wanted to at least put down my thoughts. Note though, the top proof is NOT MINE, that's the actual answer. My attempt is the stuff down at the bottom.

Kinda ironic me debating set theory in one thread and trying to say it's all wrong then getting stuck on what are probably straightforward questions in this thread...
• Apr 2nd 2010, 01:26 PM
MoeBlee
Deadstar, you lost me when you wrote 'P(A)'. What is 'P'? Power set? I don't see that you need it.

Let '<' and '=< ' stand for the cardinal 'less than' and 'less than or equal to' relations.

Let '^' stand for cardinal exponetiation.

Let '\' stand for binary complement.

Theorem: If u, K, L are cardinals and 0 < K =< L, then u^K =< u^L.

Proof:

Let F = {f | f is a function from K into u}.
Let H = {h | h is a function from L into u}.

It suffices to show an injection from F into H.

Since K is not empty, if u is empty, then F is empty, and we're done.

So suppose u is not empty; so let c be in u.

Let G be a function on F defined by:

G(f) = f union {<x c> | x in L\K}.

It's easy to see that G is an injection.
• Apr 2nd 2010, 01:37 PM
Quote:

Originally Posted by MoeBlee
Deadstar, you lost me when you wrote 'P(A)'. What is 'P'? Power set? I don't see that you need it.

Let '<' and '=< ' stand for the cardinal 'less than' and 'less than or equal to' relations.

Let '^' stand for cardinal exponetiation.

Let '\' stand for binary complement.

Theorem: If u, K, L are cardinals and 0 < K =< L, then u^K =< u^L.

Proof:

Let F = {f | f is a function from K into u}.
Let H = {h | h is a function from L into u}.

It suffices to show an injection from F into H.

Since K is not empty, if u is empty, then F is empty, and we're done.

So suppose u is not empty; so let c be in u.

Let G be a function on F defined by:

G(f) = f union {<x c> | x in L\K}.

It's easy to see that G is an injection.

Thanks for that but to be honest that looks quite different to what we've done and frankly given the amount I'm struggling with this course I'm gonna stay with proofs that look like the one in my post.

I don't know what P stands for, I assume power set. It's not my solution is the one given by the lecturer. My attempt is at the end of the post.

Also, what is <x c>? I have no idea what [f union {<x c> | x in L\K}] means but I have a very slight idea how the injections in my lecturers post were arrived at so I'm gonna have to stick with ones similar to it.
• Apr 2nd 2010, 02:09 PM
MoeBlee
I don't know where 'A', 'B', and 'C' come from or what they are supposed to be in the writeup you mentioned.

My approach is extremely simple and I think the first thing one might think of.

We know that u^K is just the number of functions from K into u.
We know that u^L is just the number of functions from L into u.

So let F be the set of functions from K into u; and let H be the set of functions from L into u.

An injection is obvious:

Let G be the function from F into H defined as follows:

For a function f from K into u, just extend it to a function from L into u by adding to f all the ordered pairs <x c> where x is any member of L but not a member of K and c is some chosen member of u.

The idea is so simple you can even draw it:

For a function f in K, do the following: Draw a circle for K and a separate circle for u. Then put dots inside those circles, then lines from the dots in K to their values under the function f (i.e. each member of K gets matched to some member of u). Then add an extension to the circle for K; this extension has all the members of L that are not in K, and match every dot in that extension to the same member (call it 'c') in u. And you see that you now have a function from L into u. Call that new function 'G(f)'. And it's easy to see that if f and f* are different functions from K into u then G(f) and G(f*) are different.

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