# Cardinal Inequality Exercise

I'm reading a bit on set theory and there's this elementary question, but I just can't solve it
given the cardinal numbers $$2 \le k \le \lambda$$ and $$\lambda$$ is cardinal of an infinite set
prove that : $$k^\lambda = 2^\lambda$$
I appreciate it if you give me clues rather than directly answering it

• A function from $\lambda$ to $\lambda$ is a member of the power set $P(\lambda \times \lambda)$. Hence $\lambda^{\lambda}=$ $|^{\lambda}\lambda |$ $\le |P(\lambda\times \lambda)|=$ $2^{\lambda\times \lambda}.$ Now if $\lambda\times \lambda=\lambda$ then ...? Commented Jul 13, 2021 at 7:44

Hint: use the inequality $$\kappa^\lambda\leq (2^{\kappa})^{\lambda}$$.
• and after that i have to go with $2^(k\lambda) \le 2^(\lambda\lambda)$ , so in essence proving $k\lambda \le \lambda \lambda$? Commented Jun 29, 2021 at 16:12
• Do you know what is $\kappa\lambda$ equal to?
• yeah it's $\lambda$ , you r right ! thanks!! Commented Jun 29, 2021 at 16:23
• but how do i prove the inequality you used? there is a lemma if A $\le B$ and $C \le D$ then $A^C \le B^D$ but i can't prove it, i know that there is a one to one function like f from A to B and also another one from C to D, can u help? should i open up a new question? Commented Jun 29, 2021 at 16:26
• Note that since $\kappa\subseteq 2^{\kappa}$, every function $\lambda\to\kappa$ is in particular a function $\lambda\to 2^{\kappa}$. So this gives us an injection from $\kappa^{\lambda}$ to $(2^{\kappa})^{\lambda}$.