# Cardinal exponentiation question of infinite cardinals.

I got a little confused with a question about cardinal exponentiation:

Let $\beta$ be an ordinal, and let $K_\alpha$ , $\alpha < \beta$, be infinite cardinals with $K= \sum_{\alpha<\beta}K_\alpha$ and now I have to prove that for every cardinal $\lambda$ $$\lambda^K =\prod_{\alpha<\beta}\lambda^{K_\alpha}$$ Now I know that $k^{\lambda}k^{\mu}=k^{\lambda+\mu}$ for cardinals $k,\lambda,\mu$, so my first attempt was trying induction on $\beta$. However I got stuck when I assumed $\beta$ was a limit ordinal. Am I on the right track with induction or does anyone suggests an alternative way to prove this?

• You're switching between $K$ and $k$. This seems like a typo to me; i think it's supposed to be $K$ all the way. Also, note that there's a huge difference between cardinal and ordinal exponentiation, and it is worthwhile to be very conscious about which one you're using. Once you are, you can start applying definitions to the induction, which I assume is the correct way to go about this. – Arthur Mar 31 '16 at 13:37

You have $\kappa = \sum_{\alpha < \beta} \kappa_{\alpha}$, so let $X$ be a set of cardinality $\kappa$, and $X_{\alpha}$ of cardinality $\kappa_{\alpha}$. Then $$X = \bigsqcup_{\alpha < \beta} X_{\alpha}$$ is a disjoint union.
Let $Y$ be a set of cardinality $\lambda$.
A function $X \to Y$ is specified precisely by its values on each component $X_{\alpha}$ of $X$. That is, given a function $f: X \to Y$, we may produce a unique member of $$\prod_{\alpha < \beta} \{ \text{functions X_{\alpha} \to Y }\}$$ by $$\langle \text{f restricted to X_{\alpha}} \rangle_{\alpha < \beta}$$
This mapping $Y^X \to \prod_{\alpha < \beta} Y^{X_{\alpha}}$ is bijective.