If $\mathfrak d$ is the cardinality of the least dominating family in the set of functions from $\omega$ to $\aleph_{\omega}$, what are its possible values with respect to the cardinal $(\aleph_{\omega})^{\omega}$?


Your dominating number is the same as the usual dominating number for families of functions $\omega\to\omega$.

For the one direction, we may associate any function $f:\omega\to\omega$ with a corresponding function $f^*:\omega\to\aleph_\omega$ by defining $f^*(n)=\aleph_{f(n)}$. And if $\cal F$ is a dominating family on $\omega^\omega$, then the set ${\cal G}=\{f^*\mid f\in{\cal F}\}$ is dominating in $(\aleph_\omega)^\omega$, since every function $g:\omega\to\aleph_\omega$ is dominated by $g^+(n)=|g(n)|^+$, which is $f^*$ for some function $f:\omega\to\omega$, which is dominated by a function in ${\cal G}$.

Conversely, if $\cal G$ is a dominating family in $\aleph_\omega^\omega$, then ${\cal F}=\{\bar f\mid f\in {\cal G}\}$ is dominating in $\omega^\omega$, where $\bar f(n)$ is the least $m$ with $f(n)\leq \aleph_m$.

So the two dominating numbers are exactly the same.

  • $\begingroup$ I guess the same idea works on any cardinal of countable cofinality. More generally, this method shows that one should consider dominating numbers on $\kappa$ only for regular $\kappa$, since the corresponding dominating number will be the same as for $\text{cof}(\kappa)$. $\endgroup$ – Joel David Hamkins May 20 '11 at 23:27
  • $\begingroup$ Just to complete Joel's answer by explicitly giving the comparison with $(\aleph_\omega)^\omega$: We trivially have $\mathfrak{d}\leq 2^\omega\leq(\aleph_\omega)^\omega$, but whether $\mathfrak{d}=(\aleph_\omega)^\omega$ is independent of ZFC. $\endgroup$ – Andreas Blass May 21 '11 at 1:15

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