$(k^{<\lambda})^{<\lambda}=k^{<\lambda}$ if $\lambda$ is regular

Question

I'm in need of some help...

Why does $(k^{<\lambda})^{<\lambda}=k^{<\lambda}$ if $\lambda$ is regular? I can't see why... Any hints?

Answer 1

This is clear if $\lambda$ is a successor.

If $\lambda$ is limit, $k^{<\lambda}=\sup_{\tau<\lambda}k^\tau$. There are two cases:

1. The sequence $k^\tau$, $\tau<\lambda$, is eventually constant, that is, there is $\tau_0$ such that $k^\tau=k^\rho$ if $\tau_0\le\tau\le\rho<\lambda$. But then $(k^\tau)^\rho=k^\tau$ for all $\rho<\lambda$ and $\tau\ge\tau_0$.

2. The sequence is not eventually constant, so $k^{<\lambda}$ is the supremum of a strictly increasing $\lambda$ sequence. If $\rho<\lambda$, then any function from $\rho$ to $k^{<\lambda}$ is bounded, as the latter has cofinality larger than $\rho$. But then it is bounded by $k^{\tau}$ for some $\tau$. This shows that $(k^{<\lambda})^\rho$ injects into $k^{<\lambda}$, for any $\rho<\lambda$.

In either case, we have shown that $(k^{<\lambda})^{<\lambda}=k^{<\lambda}$, as needed.