I am trying to prove that
$\omega_1 \leq \mathfrak s$
where $\mathfrak s$ is the splitting number which is the smallest cardinality of any splitting family. This statement was left as an exercise to the reader in a book I am reading.
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.
Here is the proof I was thinking about:
Given a splitting family $\mathcal L$, we will show that $\mathcal L$ is uncountable. Suppose that $\mathcal L$ was countable. Then we can write it a $\bigcup_{n \in \omega}U_n$. Take now: $a_1 \in U_1,a_2 \in U_2\setminus U_1,...a_n \in U_n\setminus \{U_1 \cup U_2 \cup ... \cup U_{n-1}\}$. Take $A=\{a_n\}$ We get that for every $n \in \omega$, $U_n \cap A$ is finite. Contradicting our assumption that $\mathcal L$ was a splitting family.
What do you think, is this proof ok? Thank you! Shir
We show that a countable family $U_1,U_2,U_3,\dots$ of subsets of $\omega$ can't be a splitting family.
Choose an infinite set $S_1\in\{U_1,\omega\setminus U_1\}$.
Choose $a_1\in S_1$.
Choose an infinite set $S_2\in\{S_1\cap U_2,S_1\setminus U_2\}$.
Choose $a_2\in S_2\setminus\{a_1\}$
Choose an infinite set $S_3\in\{S_2\cap U_3,S_2\setminus U_3\}$.
Choose $a_3\in S_3\setminus\{a_1,a_2\}$.
וכן הלאה