# Cardinality of cofinal set of normal functions $f \colon \omega_1 \to \omega_1$

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 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)). – 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

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.)

References:

• arxiv

• Math Reviews: MR1355135 (96k:03117)

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