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 $2^\kappa$.

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 $2^\kappa$.

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 $\kappa$?

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 $2^\kappa$.