# How to show that $\mathfrak s \leq \mathfrak d$

I am trying to understand why $\mathfrak s \leq \mathfrak d$. Can anyone state a proof of it? I have a proof , which I don't understand yet. My question regarding that proof is here below:

At the begining of the proof, for every strictly increasing function, $f \in \omega^\omega$, with $f(0) > 0$, a set $\sigma_f$ is defined as follows:

$\sigma_f = \bigcup\{[f^{2n}(0),f^{2n+1}(0));n \in \omega\}$.

My question is: Why is $\sigma_f$ defined this way? Why isn't it possible to define it ths way: $\sigma_f = \bigcup\{[f(2n),f(2n+1));n \in \omega\}$?

Here is the Proof:

Definitions: $\mathfrak s$ is the splitting number which is the smallest cardinality of any splitting family.

A splitting family is a family $\mathcal L \subseteq [\omega]^\omega$ such that each set $y \in [\omega]^\omega$ is split by at least one $x \in \mathcal L$. Also, a set $x \subseteq \omega \space$ splits an infinite set $y \in [\omega]^\omega$ if both $y \cap x$ and $y\setminus x$ are infinite.

A familly $\mathcal D \subseteq \omega^\omega$ is dominating, if for each $f \in \omega^\omega$, there is $g \in \mathcal D$ with $f \leq^*g$. The dominating number $\mathfrak d$, is the smallest cardinality of any dominating family, $\mathfrak d = min \{|\mathcal D|; \mathcal D \space is \space dominating\}$

Thank you, Shir

• We're gonna have the Cichon diagram here soon! – Asaf Karagila Jan 13 '14 at 18:39
• mm.. not sure what cichon means but think I got the point.. :\ – topsi Jan 14 '14 at 3:43
• – bof Jan 14 '14 at 7:42
• Will have a look. Thank you! – topsi Jan 14 '14 at 9:28
• Among many non-Polish sources for Cichon's diagram, there's page 32 of (the unofficial version of) my chapter of the Handbook of Set Theory, available at math.lsa.umich.edu/~ablass/hbk.pdf , but be aware that many people adopt the opposite convention for the direction of the arrows. [My browser or this website rejected my attempts to put the accent on the n in Cichon's name.] – Andreas Blass Jan 15 '14 at 1:52

You could define $\sigma_f$ in the way you suggest, but then you'd have to use a more complicated $f$ later in the proof.
The point of the given definition is that when you apply $f$ to any number in one of the intervals $[f^{2n}(0),f^{2n+1}(0))$ that constitute $\sigma_f$, you get a result in the interval $[f^{2n+1}(0),f^{2n+2}(0))$ that is disjoint from $\sigma_f$. That's the essential ingredient in showing that any infinite set will (for a suitable $f$ as described in the proof), from some point on, meet all of the intervals (the ones in $\sigma_f$ as well as the ones disjoint from $\sigma_f$) and will therefore be split by $\sigma_f$.