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.

--

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.

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

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:

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

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

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

7. ooh, now I understand.

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