4
$\begingroup$

Can someone explain (prove) me why this holds?

To be precise, by club set I mean closed and unbounded set where given definitions are:

Let κ ∈ Card and C ⊆ κ . C is unbounded in κ if ∀α < κ ∃β ∈ C α < β.

C is closed in κ if ∀λ < κ (λ is a limit ordinal ∧ C ∩λ is unbounded in λ→λ ∈ C).

On the other hand, I am supposed to prove that the next thing holds:

Suppose that µ, κ are regular cardinals and f∶µ → κ is a cofinal strictly increasing continuous function. I have to show that f[C] is club in κ for every club C in µ

  • 0
    I've heard of cofinal sets, but what is a cofinal function? Do you mean an unbounded function? – 2017-01-13
  • 0
    Yes, unbounded one – 2017-01-13
  • 2
    @bof A function is cofinal if its range is. If the codomain is a total order is almost the same as unbounded (though it differs if the codomain has a maximum element), but it's completely different if the codomain is not total. – 2017-01-13

2 Answers 2

4

Hint: A cofinal continuous and strictly increasing function $\mu\to \kappa$ extends to a continuous function $\mu+1\to \kappa+1$ (because it implies that $\lim_{\alpha \to \mu} f(\alpha)=\kappa$), which are compact Hausdorff spaces. Being closed in $\kappa$ or $\mu$ is the same as being closed in the order topology. This helps with the "closed" part. The "unbounded" part should be obvious. (In fact, I belive you don't need to know anything about $\mu$ and $\kappa$ here -- they can be any limit ordinals.)

  • 0
    I have also done the unboundedness part, but still I don't get how this hint helps. Sorry for disturbing – 2017-01-13
  • 0
    @Eurydice: Continuous maps between compact Hausdorff spaces are closed. – 2017-01-13
3

HINT: I like tomasz’s argument, but a more elementary one is also possible.

If $f[C]$ is not closed in $\kappa$, let $\alpha=\min\big((\operatorname{cl}f[C])\setminus C\big)$, and let $A=C\cap\alpha$; clearly $\alpha\in\operatorname{cl}A$. Let $B=f^{-1}[A]=\{\xi\in C:f(\xi)<\alpha\}$, and let $\beta=\sup B\in C$. There are two possibilities: either $\beta=\max B\in B$, or $B\subseteq\beta$. Show that each possibility leads to a contradiction.