Dense subsets of ${}^{<\kappa} \kappa$, with $\kappa$ regular cardinal

72 Views Asked by At

I have the following problem:

$D\subseteq {}^{<\kappa}\kappa$ is said to be dense if for every $s \in {}^{<\kappa}\kappa$ there is a $t \in D$ s.t. $s \subseteq t$. Prove that if $\kappa$ is regular, then for all collections $\{D_\alpha \ | \ \alpha < \kappa\}$ of dense sets, there exists $x \in {}^\kappa \kappa$ s.t. for every $\alpha < \kappa$ there is an $s \in D$ with $s \subseteq x$.

Now, my problem is that I tried to solve it, but don't see how the hypothesis on the regularity of $\kappa$ is involved.

The sketch of my try is the following

  • Given a collection $\{D_\alpha \ | \ \alpha < \kappa\}$, by transfinite recursion (and AC) I prove the existance of an (increasing) sequence $\{s_\alpha\}_{\alpha < \kappa} \subseteq {}^{<\kappa}\kappa$ s.t.

    • $s_\alpha \in D_\alpha$

    • $s_\alpha \subseteq s_\beta$ for any $\alpha \le \beta$

    • $\text{dom}(s_\alpha) \supseteq \alpha$

  • Then I define $x \in {}^\kappa \kappa$ as $x = \bigcup_{\alpha < \kappa} s_\alpha$, which is the desired $x$

Is the outline right? Notice that I've not mentioned the regularity of the cardinal. Is there maybe a missed requirement of $x$ to be increasing? Thanks

EDIT: I'm not sure whether the notation ${}^\kappa\tau$ is standard, it is the set of functions having $\kappa$ as domain and $\tau$ as codomain, I think it is written in this way to differentiate it from the cardinal exponentiation

1

There are 1 best solutions below

0
On BEST ANSWER

Your outline is fine, but you'll need regularity of $\kappa$ in order to get an appropriate $s_\alpha$ when $\alpha$ is a limit ordinal. If $\kappa$ were singular, say with cofinality $\mu$, then when you get to stage $\mu$ of your inductive construction, you might find that the union of the previously chosen $s_\gamma$'s (for $\gamma<\mu$) might already have length $\kappa$.