8
$\begingroup$

I've been thinking about this question, but to no avail and I've got to ask.

How to show that for $\kappa\geq\aleph_0,$ $\mu=\min\{\lambda: \kappa^{\lambda} > \kappa\}$ is regular?

If I wanted a contradiction, it would suffice to prove $\kappa^{cf(\mu)}>\kappa$, but I don't see how this could be true.

$\endgroup$
  • $\begingroup$ $k^{cf(k)}>k$ for every infinite cardinal $k$. This is Konig's Theorem, sometimes called Konig's Lemma. But $cf(k)$ is not necessarily the least $l$ such that $k^l>k$. $\endgroup$ – DanielWainfleet Apr 24 '16 at 5:47
9
$\begingroup$

Let $\kappa$ be infinite and $\mu=\min\{\lambda\mid \kappa^\lambda>\kappa\}$ (this cardinal exists, since already $\kappa^\kappa>\kappa$). If $\mu$ is singular, write $\mu=\sum_{\alpha<\theta}\mu_\alpha$ where $\theta={\rm cf}(\mu)<\mu$ and the $\mu_\alpha$ form a strictly increasing sequence of regular cardinals. We have $$ \kappa^\mu=\kappa^{\sum_\alpha \mu_\alpha}=\prod_\alpha \kappa^{\mu_\alpha}. $$ Now, by definition of $\mu$, $\kappa^{\mu_\alpha}=\kappa$ for all $\alpha$, and therefore the product above reduces to $\prod_\alpha\kappa=\kappa^\theta$. Again, since $\theta<\mu$, then $\kappa^\theta=\kappa$, and we are done.

The key equality is of course $\kappa^{\sum_\alpha \mu_\alpha}=\prod_\alpha \kappa^{\mu_\alpha}$, which is proved directly: The sum $\sum_\alpha\mu_\alpha$ is the cardinality of the disjoint union $\bigcup_\alpha \mu_\alpha\times\{\alpha\}$, and we can identify any function $\phi$ from this union to $\kappa$ with the sequence of functions $(\phi_\alpha\mid \alpha<\theta)$ given by $\phi_\alpha=\phi\upharpoonright \mu_\alpha\times\{\alpha\}$. This is actually a bijection between ${}^{\bigcup_\alpha \mu_\alpha\times\{\alpha\}}\kappa$ and the Cartesian product $\prod_\alpha {}^{\mu_\alpha\times\{\alpha\}}\kappa$.

$\endgroup$
  • $\begingroup$ Very clear.....................+1 $\endgroup$ – DanielWainfleet Apr 24 '16 at 5:50
  • $\begingroup$ Note that the "key equality" follows from abstract nonsense in any cartesian closed category : $Hom(-,\kappa^{\displaystyle\sum_\alpha \mu_\alpha}) \cong Hom(- \times \displaystyle\sum_\alpha \mu_\alpha, \kappa ) \cong \displaystyle\prod_\alpha Hom(-\times \mu_\alpha, \kappa) \cong \displaystyle\prod_\alpha Hom(-, \kappa^{\mu_\alpha}) \cong Hom(-, \displaystyle\prod_\alpha \kappa^{\mu_\alpha})$, and we can conclude by the Yoneda Lemma (the fact that the product commutes with the direct sum comes from a classical application of Yoneda). These hold whenever the product and coproduct exist. $\endgroup$ – Maxime Ramzi Dec 20 '17 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.