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:



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> **Question.** Is it consistent with ZFC that $\mathrm{dv}(\kappa) = \kappa$ for all infinite cardinal numbers $\kappa$?
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**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](https://math.stackexchange.com/questions/732285/are-there-situations-outside-of-set-theory-where-it-would-be-useful-if-mathrm), 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 $2^\kappa$.



Is it consistent with ZFC that $\mathrm{dv}(\kappa) = \kappa$ for all infinite cardinal numbers $\kappa$?
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 $2^{<\kappa} \leq |D|$. 


