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Let $\kappa$ be an infinite cardinal. An ultrafilter ${\cal U}$ on $\kappa$ is said to be uniform if $|R|=\kappa$ for all $R\in{\cal U}$. If ${\cal U}$ is a non-principal ultrafilter on $\kappa$, denote by $b({\cal U})$ the minimal cardinality that a filter base for ${\cal U}$ can have.

If ${\cal U, V}$ are non-principal uniform ultrafilters on $\kappa$, do we necessarily have $b({\cal U}) = b({\cal V})$?

Thanks to Joseph van Name for making me aware of uniform ultrafilters and the fact that this question is only (potentially) interesting when restricted to these.

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    $\begingroup$ You should restrict your question to the uniform ultrafilters to avoid trivial counterexamples. An ultrafilter $\mathcal{U}$ on a cardinal $\kappa$ is said to be uniform if $|R|=\kappa$ whenever $R\in\mathcal{U}$. Here is a related question that answers the case for $\omega$. math.stackexchange.com/questions/1883710/… $\endgroup$
    – Joseph Van Name
    Commented Mar 10, 2019 at 15:01
  • $\begingroup$ Thanks @JosephVanName for your comment! - I will restrict the question to uniform ultrafilters $\endgroup$
    – Dominic van der Zypen
    Commented Mar 10, 2019 at 16:57

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Your number $b(U)$ is usually called the "character" of the ultrafilter $U$.

In general, there may be uniform ultrafilters on the same set with different characters. For example, it is consistent with $2^{\aleph_0}=\aleph_2 $ that some ultrafilters have character $\aleph_1$, others $\aleph_2$.

Also more complicated "character spectra" are possible, according to Sh:915 (Topology and its Applications 158 (2011) 2535-2555)

(Of course, if $2^\kappa=\kappa^+$, then all uniform ultrafilters have character $\kappa^+$.)

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