1
$\begingroup$

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?

$\endgroup$
6
  • 2
    $\begingroup$ If $\delta$ is a limit ordinal, then every finite subset of $Y_\delta$ is also a subset of $Y_\alpha$ for some $\alpha<\delta$ (and is therefore in $\mathcal F$). $\endgroup$
    – Andreas Blass
    Jan 30, 2022 at 19:16
  • $\begingroup$ @AndreasBlass How do you prove that "every finite subset of $Y_\delta$ is also a subset of $Y_\alpha$ for some $\alpha<\delta$"? $\endgroup$
    – Gleison Stanlley
    Jan 30, 2022 at 19:21
  • 1
    $\begingroup$ Since the claim is obvious for the empty set, let $s$ be a nonempty finite subset of $Y_\delta=\bigcup_{\alpha<\delta}A_\alpha$. So we have, for each $a\in s$, some $\alpha_a<\delta$ such that $a\in A_{\alpha_a}$. The nonempty finite set of ordinals $\{\alpha_a:a\in s\}$ has a largest element, and that element serves as the required $\alpha$. $\endgroup$
    – Andreas Blass
    Jan 30, 2022 at 20:25
  • $\begingroup$ @AndreasBlass Let $H\subseteq Y_\delta$ be finite and non-empty. Therefore, $\forall x\in H \exists \beta<\delta (x\in Y_\beta)$. Let $\alpha(x):=\min(\{\beta<\delta;x\in Y_\beta\})$, for each $x\in H$. Since $H\neq \varnothing$ is finite, then $\{\alpha(x); x\in H\}$ is a non-empty and a finite set of ordinals. Hence, it has a largest element. Call it $\mu$. You are saying that $H\subseteq Y_\mu\in \mathcal{F}$ and, because $\mathcal{F}$ is of finite character, then $H\in \mathcal{F}$. Hence, $Y_\delta\in \mathcal{F}$, which is a contradiction. But, how do you prove that $H\subseteq Y_\mu$? $\endgroup$
    – Gleison Stanlley
    Feb 1, 2022 at 0:08
  • 1
    $\begingroup$ Each element $x$ of $H$ is in $Y_{\alpha(x)}$, which is a subset of $Y_\mu$ because $\alpha(x)\leq\mu$. (I'm assuming you know that the sequence of $Y$'s is monotone, i.e., $\xi\leq\eta\implies Y_\xi\subseteq Y_\eta$. If you don't know that, prove it by induction on $\eta$.) $\endgroup$
    – Andreas Blass
    Feb 1, 2022 at 1:20

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Let $H\subseteq Y_\delta$ be finite and non-empty. Therefore, $\forall x\in H \exists \beta<\delta (x\in Y_\beta)$. Let $\alpha(x):=\min(\{\beta<\delta;x\in Y_\beta\})$, for each $x\in H$. Since $H\neq \varnothing$ is finite, then $\{\alpha(x); x\in H\}$ is a non-empty and a finite set of ordinals. Hence, it has a largest element. Call it $\mu$. You are saying that $H\subseteq \bigcup\{Y_{\alpha(x);x\in H}\}=Y_\mu\in \mathcal{F}$ and, because $\mathcal{F}$ is of finite character, then $H\in \mathcal{F}$. Hence, $Y_\delta\in \mathcal{F}$, which is a contradiction. $\endgroup$
    – Gleison Stanlley
    Feb 1, 2022 at 0:12
  • $\begingroup$ But, how do you prove that $\bigcup \{Y_{\alpha(x)};x\in H\}=Y_\mu$? $\endgroup$
    – Gleison Stanlley
    Feb 1, 2022 at 0:13
  • 1
    $\begingroup$ @GleisonStanlley By definition, if $α<β$ we have $Y_α⊆Y_β$, so if $(α_i\mid i<Γ≤κ)$ is any sequence of elements $α_i<κ$ we have that $\bigcup Y_{α_i}=Y_{\sup(α_i)}$, if we know that $α_i<δ$ for all $i$, and that $Γ$ is finite, we have that $\sup(α_i)=\max(α_i)<δ$ $\endgroup$
    – ℋolo
    Feb 1, 2022 at 17:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .