# Cardinals - Exponents

• Aug 24th 2009, 09:15 AM
Cardinals - Exponents
I have a problem with cardinals:

k,j,l are cardinals. 0<k, k=< l

So this, I believe, should be the way to solution:

I need to prove that j^k<=j^l

so, let |a|=j, |b|=k, |c|=l

I need to prove that |a^c| >= |a^b|

( >= means larger or equal)

Now, I understand that I need to create a function from a^b to a^c and show that it is a 1-1 function.

I now that I can say that there is a function $f:b(to)c$ that is a 1-1 function (because i know that b<=c ), but I just can't seem to find a way to use it...

Thank you!~!
• Aug 24th 2009, 10:15 AM
Liwuinan
If $a$ is empty then both $a^b$ and $a^c$ are empty and the assertion holds.
Let $a$ be nonempty and let $s \in a$. Let $g$ be an injective map from $b$ to $c$ and let $c_1=Range(g)$. For every function $f:b \rightarrow a$ we define function $\tilde{f}:c \rightarrow a$ this way:

$\tilde{f}(t) = f(g^{-1}(t))$ if $t \in c_1$
$\tilde{f}(t) = s$ if $t \in c \smallsetminus c_1$

Then mapping $F$ defined as $F(f)=\tilde{f}$ is injective and maps $a^b$ to $a^c$.
• Aug 24th 2009, 10:39 AM
I see, but there's one thing I didn't understand - if $c_1=Range(g)
$
, then what does it mean? That c is this group? :

{ k $| k \in g(b), b \in B$}
• Aug 24th 2009, 10:51 AM
Liwuinan
not quite, Range(g) = {k| (ex. d in b) k = g(d)}
• Aug 24th 2009, 11:25 AM