I'm using Kenneth Kunen's book: The Foundations of Mathematics. I have a doubt about a fact stated in the proof.
Let $A$ be a set, $\mathcal{F}\subseteq \mathcal{P}(A)$ of finite character and $X\in \mathcal{F}$. As every set is well orderable, $A$ is well orderable. Let $\kappa:=|A|$ and $f:\kappa \to A$ a bijection. Let $a_\alpha:=f(\alpha)$, for every $\alpha<\kappa$. Therefore, $A=\{a_\alpha;\alpha<\kappa\}$. Recursively, we define $Y_\beta\subseteq \{a_\xi;\xi<\beta\}$, for every $\beta\leq \kappa$, by:
- $Y_0:=X$.
- $Y_{\alpha+1}:=
\begin{cases}
Y_\alpha\cup\{a_\alpha\}, & \textrm{ if } Y_\alpha\cup\{\alpha\}\in \mathcal{F}\\
Y_\alpha, & \textrm{ if } Y_\alpha\cup\{\alpha\}\not\in \mathcal{F}
\end{cases}$. - $Y_\gamma:=\bigcup \{Y_\alpha;\alpha<\gamma\}$, if $\gamma$ is a limit ordinal.
If we prove that $Y_\beta\in \mathcal{F}$, for every $\beta\leq \kappa$, then $Y:=Y_\kappa\in \mathcal{F}$. Let $Z$ be a set such that $Y\subsetneq Z\subseteq A$. Hence, $Z\setminus Y\neq \varnothing$. Let $\alpha\leq \kappa$ be an ordinal such that $a_\alpha\in Z\setminus Y$. If $Y_\alpha\cup \{a_\alpha\}\in \mathcal{F}$, then $a_\alpha\in Y_{\alpha+1}\subseteq Y$. Hence, $Y_\alpha \cup \{a_\alpha\}\not\in \mathcal{F}$ and then $Z\not\in \mathcal{F}$ (since $Y_\alpha\cup \{a_\alpha\}\subseteq Z$). QED
If we use Transfinite Induction to prove that $Y_\beta\in \mathcal{F}$, for every $\beta\leq \kappa$, we suppose it's false and, therefore, there is a least ordinal $\delta$ such that $Y_\delta\not\in \mathcal{F}$. It was quite easy to prove for the cases where $\delta$ isn't a limit ordinal. For the case where it is, the book says we could use the fact that $\mathcal{F}$ is of finite character, but I still don't realize any contradictions here. How can we prove that every finite subset of $Y_\delta$ is in $\mathcal{F}$, when $\delta$ is a limit ordinal?
Best Answer
If $\delta$ is the minimal such that $Y_\delta\notin\mathcal F$, and $\delta$ is limit, let $B=\{a_0,\ldots,a_{n-1}\}\subseteq Y_\delta$ finite.
Each $a_i$ is in minimal $Y_{\alpha_i}$, and so $B\subseteq \bigcup Y_{\alpha_i}=Y_{\max_{0\le i<n}(\alpha_i)}\subseteq Y_\delta$, so $Y_{\max_{0\le i<n}(\alpha_i)}\in\cal F$(from minimality of $\delta$), so $B\in \cal F$(from finite character), hence $Y_\delta\in \cal F$ (again from finite character), contradiction.