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Whenever $\kappa$ is an infinite cardinal number, write $L(\kappa)$ for the powerset of $\kappa$ ordered lexicographically. (Where the "$L$" stands for linear order.) Furthermore, write $B(\kappa)$ for the subchain of $L(\kappa)$ consisting of all $X \subseteq \kappa$ such that $X$ is bounded in $[0,\kappa).$ Finally, by the density value of an infinite cardinal $\kappa$, let us mean the least cardinal $\nu$ such that $B(\kappa)$ has a dense subset of size $\nu$. We'll denote this $\mathrm{dv}(\kappa)$. Then

$$\kappa \leq \mathrm{dv}(\kappa) \leq 2^{<\kappa},$$

since the RHS is the cardinality of $B(\kappa)$ which is clearly dense in itself. Hence ZFC proves that for all strong limit cardinals $\beth_\lambda$, we have that $\mathrm{dv}(\beth_\lambda) = \beth_\lambda.$

However, ZFC cannot prove that $\mathrm{dv}(\kappa) = \kappa$ for all infinite cardinal numbers $\kappa.$ I think I can supply a proof of this if anyone is interested (please comment). Anyway, this motivates my:

Question. Is it consistent with ZFC that $\mathrm{dv}(\kappa) = \kappa$ for all infinite cardinal numbers $\kappa$?

Addendum. Noah requested a proof that $\exists \kappa : \mathrm{dv}(\kappa)>\kappa$ is consistent with ZFC, so here it is.

By Asaf's answer here, we have that ZFC cannot prove "$(\mathcal{P}(\kappa),\subseteq)$ has a subchain of size $2^\kappa$ for all infinite cardinals $\kappa$." Hence to show that ZFC is consistent with the statement $\exists \kappa : \mathrm{dv}(\kappa)>\kappa,$ it suffices to show that if $\kappa$ is an infinite cardinal number with $\mathrm{dv}(\kappa) = \kappa,$ then $\mathcal{P}(\kappa)$ has a subchain of size $2^\kappa$.

So let $\kappa$ denote an infinite cardinal number with $\mathrm{dv}(\kappa) = \kappa$. Furthermore, let $C(\kappa)$ denote the set of all $X \in L(\kappa)$ such that the characteristic function of $X$ is not eventually $1$. Then $B(\kappa) \subseteq C(\kappa) \subseteq L(\kappa).$ Now by hypothesis, $B(\kappa)$ has a dense subset of size $\kappa$, call it $D$. Then $D$ is also dense in $C(\kappa)$. Hence the function $f : C(\kappa) \rightarrow \mathcal{P}(D)$ given by $f(X) = \{Y \in D \mid Y \leq X\}$ is an injection. Thus $\mathcal{P}(D)$ has a subchain of size $|C(\kappa)|$ i.e. of size $2^\kappa$. But since $D$ and $\kappa$ are equipotent, this means that $\mathcal{P}(\kappa)$ has a subchain of size $2^\kappa$.

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Isn't $dv(\kappa)=\kappa$ implied by GCH? –  Noah S May 13 at 7:20
    
@NoahS, that's VERY possible. What's your reasoning? –  goblin May 13 at 7:25
    
It's late, so I might just be wrong, but I've posted it as an answer and it passes the smell test for me. –  Noah S May 13 at 7:31

2 Answers 2

up vote 7 down vote accepted

One can actually show that dv($\kappa$) = $2^{<\kappa}$ outright. Suppose $D$ is any dense subset of $2^{\kappa}$ (one may take $D \subseteq B(\kappa)$, if preferred). I'll argue that $D$ must have size at least $2^{<\kappa}$, which is sufficient for the claim. For any sequence $r \in 2^{<\kappa}$, let $\alpha_r$ denote its length. One may think of $r$ as identifying the interval $I_r = \{x \in 2^{\kappa} \, : \, x \upharpoonright \alpha_r = r\}$. Since $D$ is dense in $2^{\kappa}$, there must be $x \in D$ that hits $I_r$, i.e. some $x$ beginning with the sequence $r$. This shows that if we let $D'$ be the collection of all initial segments of elements of $D$, then $D' = 2^{<\kappa}$. But $|D'| \leq \kappa \cdot |D| = |D|$ which gives $2^{<\kappa} \leq |D|$.

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Your statement "$\forall \kappa[dv(\kappa)=\kappa]$" is a consequence of GCH:

For successor $\kappa$, by GCH at the predecessor of $\kappa$ we have that the set of bounded subsets of $\kappa$ is just $\kappa^{\kappa^-}=({2^{\kappa^-}})^{\kappa^-}=2^{\kappa^-}=\kappa$ (indulging in a bit of awful notation :P).

For limit $\kappa$: Write $\kappa=\sup\{\lambda_0<\lambda_1< . . . <\lambda_{\alpha}< . . .: \alpha\in \eta\}$ with each $\lambda_\eta$ a cardinal. Now, given a bounded subset $B$ of $\kappa$, let $\alpha_B$ be the least $\alpha$ such that $B\subseteq\lambda_{\alpha}$. Let $\mathcal{B}_\alpha=\{B: \alpha_B=\alpha\}$, and note that $2^{<\kappa}=\bigcup_{\alpha\in\eta} \mathcal{B}_\alpha$. By GCH below $\kappa$, each $\mathcal{B}_\alpha$ has size $\lambda_\alpha^+$, so $2^{<\kappa}=\kappa$.

Note that this is all a very coarse argument - all we're doing is counting the set $B(\kappa)$ and showing that your inequality collapses.

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I think you're right. For some reason I kept coming to the conclusion that GCH implies that for successor $\kappa$, we have that $|B(\kappa)| = \kappa^+.$ However, your argument looks impeccable. –  goblin May 13 at 7:39
    
I'd be interested in the proof that ZFC allows $dv(\kappa)>\kappa$, if you don't mind sharing it. –  Noah S May 13 at 7:40
    
No worries, I adjoined it to my question. –  goblin May 13 at 8:15
2  
Isn't it simpler to note that under GCH $2^{<\kappa} = \kappa$? –  Sam Roberts May 13 at 8:15
1  
Fair enough - it just looks to me a little over complicated to merely establish that $2^{<\kappa} = \kappa$. –  Sam Roberts May 13 at 8:21

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