What is the cardinality of the set $F$ of all normal functions $f \colon \omega_1 \to \omega_1$, where $\omega_1$ is the first uncountable ordinal? What is the least cardinality of a subset of $F$ such that every function in $F$ is bounded by some element of the subset?
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asked Nov 25, 2011 at 5:22
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2$\begingroup$ @Vladimir: I just thought I'd mention that it's generally not considered good form to cross post so quickly between here and math.SE (I refer to the similar question math.stackexchange.com/q/85409/11532 ). Usually one should wait more than just a few hours before reposting here an unanswered question from math.SE; I don't think there is a precise amount of time to wait, but I think a few days at least has been suggested in the past (to minimise the chance of people duplicating one another's effort in answering the question(s)). $\endgroup$– Philip BrookerCommented Nov 25, 2011 at 7:48
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$\begingroup$ I think it would also be reasonable to link to similar questions on stackexchange, whenever you are aware of them. $\endgroup$– GoldsternCommented Nov 25, 2011 at 10:39
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$\begingroup$ If we call $d$ the answer to the second question, then it is easy to show (just as in the countable case) that $\aleph_1 < d \leq 2^{\aleph_1}$. I suppose that when $2^{\aleph_1}>\aleph_2$, there are interesting things to say (just as in the countable case) but I´ve never seen those. $\endgroup$– Ramiro de la VegaCommented Nov 25, 2011 at 12:09
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1 Answer
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For both questions, the answer does not change when you remove the word "normal" from the question.
For the first question: There is a 1-1 map $f\mapsto N(f)$ that assigns to each function $f$ a normal function $N(f)$. $N(f)(\alpha)$ just adds up all values of $f$ below $\alpha$. (Or better: of $f+1$, to make it strictly increasing.) So there are $2^{\aleph_1}$ many of them.
For the second question: Let $\mathfrak d(\kappa)$ be the smallest number functions needed to dominate all functions from $\kappa$ to $\kappa$.
- James Cummings and Saharon Shelah, Cardinal invariants above the continuum, Ann. Pure Appl. Logic 75 (1995), no. 3, 251–268, https://doi.org/10.1016/0168-0072(95)00003-Y, arXiv:math/9509228, MR1355135
shows that, just like the continuum functions $\kappa\mapsto 2^\kappa$, also the "dominating" function $\kappa\mapsto \mathfrak d(\lambda)$ can have quite arbitrary behaviour. In particular, both $\mathfrak d(\aleph_1)=2^{\aleph_1}$ and $\mathfrak d(\aleph_1)< 2^{\aleph_1}$ are consistent. (This specific result for $\aleph_1$ may be older, though.)
