# Math Help - Cardinal exponentiation

1. ## Cardinal exponentiation

I need to prove the following rules on cardinals:

Let κ, μ, ν be cardinal numbers. Prove:

1. κ^(μ + ν) = κ^μ·κ^ν.

2. κ^μ·ν = (κ^μ)^ν.

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How do I start proving such thing? It looks so obvious, but I'm sure there has to be a formal way to prove it.

Please help me!

Thank you

2. Are k, μ, v infinite cardinal numbers? Because otherwise, 1 is obviously not true, unless there is a typo there.

3. I'm sorry, I fixed it now. (a typo)

4. Hi again.
Cardinal power $\kappa^{\lambda}$ is defined as $\kappa^{\lambda}=|^{\lambda}\kappa|$ , that is cardinality of the set of mappings from $\lambda$ to $\kappa$.
So to prove for example (2) it suffices to show that $^{(\nu \times \mu)}\kappa$ has the same cardinality as $^{\nu}(^{\mu}\kappa)$.
For an arbitrary map $f:\nu \times \mu \rightarrow \kappa$ and an element $a \in \nu$ let us define map $f_a : \mu \rightarrow \kappa$ like this: $f_a(t) = f(a,t)$ for $t \in \mu$.
Then, relation $r = \{:a \in \nu \}>:f \in ^{(\nu \times \mu)}\kappa\}$ is an injective mapping from $^{(\nu \times \mu)}\kappa$ to $^{\nu}(^{\mu}\kappa)$. If $g \in ^{\nu}(^{\mu}\kappa)$, then $f=\{:a \in \nu \& \in g(a)\}$ is preimage of $g$ under mapping $r$. So $r$ is injective mapping from $^{(\nu \times \mu)}\kappa$ onto $^{\nu}(^{\mu}\kappa)$.

5. I'm sorry, I lost you in the middle... What do I mainly need to do?

6. Hello,
main idea is to construct a bijection between two sets. Firstly make sure you're ok with these series of equalities:

$\kappa^{\nu \mu}=|^{\nu \mu}\kappa|= \left|^{|\nu \times \mu|}\kappa \right| = |^{(\nu \times \mu)}\kappa|$
and

$(\kappa^{\mu})^{\nu} = |^{\mu}\kappa|^{\nu} = \left|^{\nu}\left|^{\mu}\kappa \right|\right| =\left|^{\nu}(^{\mu}\kappa)\right|$

They justify that all you need to do is to show that $|^{(\nu \times \mu)}\kappa| = \left|^{\nu}(^{\mu}\kappa)\right|$. And that is done by the construction of the relation $r$ in my previous post and contemplating that $r$ is really a bijection.

7. ooh, now I understand.

Thank you, I'll do it now...