Let $\mathfrak{m,n,p}$ cardinal numbers, show that $\mathfrak{m^{n+p}=m^n\cdot m^p}$.

I believe that the proof is based on showing that there is a bijection between $M^{N_1\cup N_2}$ and $M^{N_1}\times M^{N_2}$, where card $M=\mathfrak{m}$, card $N_1=\mathfrak{n}$ and card $N_2=\mathfrak{p}$, with $N_1\cap N_2=\emptyset$, but I could not find such a bijection. Any idea? Thanks!

  • $\begingroup$ Can you find a map $M^{N_1\cup N_2} \to M^{N_1}$? $\endgroup$ – Daniel Fischer Mar 25 '14 at 19:13

Given a map $f\colon N_1\cup N_2\to M$, this induces two maps $f|_{N_i}\colon N_i\to M$. So you have a map $M^{N_1\cup N_2}\to M^{N_1}\times M^{N_2}$ given by $f\mapsto (f|_{N_1},f|_{N_2})$.

Conversely, suppose you have a pair $(f,g)\in M^{N_1}\times M^{N_2}$. Since $N_1\cap N_2=\emptyset$, this allows you to define a new function $h\colon N_1\cup N_2\to M$, $$ h(x)=\begin{cases} f(x) &&\text{if }x\in N_1,\\ g(x) &&\text{if }x\in N_2. \end{cases} $$

So you have another map $(f,g)\mapsto h$. You can then verify these two maps provide a bijection, so that $M^{N_1\cup N_2}\simeq M^{N_1}\times M^{N_2}$, which gives the equality of the cardinals.

  • 1
    $\begingroup$ Ok, thanks for your help! $\endgroup$ – ZHN Mar 25 '14 at 19:27

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