1
$\begingroup$

How it follows for an infinite cardinal $\lambda$ from $\lambda = \lambda^{<\lambda}$ that $\lambda$ is in fact a regular cardinal?

1 Answers 1

5

If $\lambda$ is singular, what can you say about $\lambda^{\operatorname{cf}(\lambda)}$?

  • 0
    $>\lambda$, thank you for a hint... – 2017-02-06
  • 0
    And also $\leq\lambda^{<\lambda}$, don't forget. – 2017-02-06
  • 0
    Sorry, I've deleted my comment:But $\lambda=\lambda^{<\lambda}$ is an assumption.It may be the case for a regular $\lambda$ that $\lambda < \lambda^{<\lambda}$,right? But you have stated the opposite inequality which is always true: $\lambda\leq \lambda^{<\lambda}$ – 2017-02-06
  • 0
    Yes, that is correct. E.g., if CH fails then $\lambda=\aleph_1$ is an example. – 2017-02-06