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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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@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 ). 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)). – Philip Brooker Nov 25 '11 at 7:48
I think it would also be reasonable to link to similar questions on stackexchange, whenever you are aware of them. – Goldstern Nov 25 '11 at 10:39
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. – Ramiro de la Vega Nov 25 '11 at 12:09
up vote 9 down vote accepted

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) showed 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.)


  • arxiv

  • Math Reviews: MR1355135 (96k:03117)

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