How can I count the number of eventually constant functions $\kappa_1\to\kappa_2$?

Question

Given two infinite cardinals $\kappa_1,\kappa_2$, what's the number $\tau$ of functions $f:\kappa_1\to\kappa_2$ that are eventually constant? I think that, if $\tau_0$ is the number of eventually zero functions, then $\tau=\kappa_2\cdot\tau_0=\max\{\kappa_2,\tau_0\}$ but I don't exactly know how to compute $\tau_0$. Well, if $\kappa_1=\aleph_0$, we can compute $\tau_0$ as $$\tau_0=\sum_{n\in\mathbb{N}}\kappa_2^n=\kappa_2$$ But I don't know how to evaluate $\sum_{\alpha<\kappa_1}\kappa_2^{|\alpha|}$ in general. Is it simply $\max\{\kappa_1,\kappa_2\}$? Thanks in advance!

Answer 1

Your reasoning almost correct: $$\tau=\kappa_2\cdot \tau_0 = \kappa_2\cdot \kappa_1^{<\kappa_2} = \kappa_2\cdot\sum_{\alpha<\kappa_1}\kappa_2^{|\alpha|}=\sup_{\alpha<\kappa_1}\kappa_2^{|\alpha|} \text. $$

However, the cardinal on RHS is not simply $\max\{\kappa_1,\kappa_2\}$; its value really depends on the cardinal arithmetic for the specific cardinals below $\kappa_1$ and $\kappa_2$.

For example, it is consistent with ZFC that $2^{\aleph_0}=\aleph_2$. Now $\aleph_0^{<\aleph_1}=\aleph_0^{\aleph_0}=\aleph_2>\max\{\aleph_0,\aleph_1\}$, so there are in fact more than just $\max\{\aleph_0,\aleph_1\}$ eventually constant functions.

Here is another example: $\aleph_\omega^{<\aleph_1}=\aleph_\omega^{\aleph_0}=\aleph_\omega^{\operatorname{cf}(\aleph_\omega)}>\aleph_\omega=\max\{\aleph_\omega,\aleph_1\}$. The inequality is due to König's Lemma.