2023-11-23 15:38:28 +01:00
|
|
|
\lecture{10}{}{} % Mirko
|
|
|
|
|
|
|
|
Applications of induction and recursion:
|
|
|
|
\begin{fact}
|
|
|
|
For every set $x$ there is a transitive set $t$
|
|
|
|
such that $x \in t$.
|
|
|
|
\end{fact}
|
|
|
|
\begin{proof}
|
2024-02-11 02:43:20 +01:00
|
|
|
\gist{%
|
2023-11-23 15:38:28 +01:00
|
|
|
Take $R = \in $.
|
|
|
|
We want a function $F$ with domain $\omega$
|
|
|
|
such that $F(0) = \{x\}$
|
|
|
|
and $F(n+1) = \bigcup F(n)$.
|
|
|
|
Once we have such a function,
|
|
|
|
$\{x\} \cup \bigcup \ran(F)$
|
|
|
|
is a set as desired.
|
2023-11-24 19:07:12 +01:00
|
|
|
To get this $F$ using the recursion theorem,
|
|
|
|
pick $D$ such that
|
|
|
|
\[
|
|
|
|
(\emptyset, 0, \{x\}) \in D
|
|
|
|
\]
|
|
|
|
and
|
|
|
|
\[
|
|
|
|
(f, n+1, \bigcup\bigcup \ran(f)) \in D.
|
|
|
|
\]
|
|
|
|
The recursion theorem then gives a function
|
|
|
|
such that
|
|
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
|
|
F(0) &=& \{x\},\\
|
|
|
|
F(n+1) &=& \bigcup\bigcup \ran(F\defon{n+1})\\
|
2024-02-11 02:43:20 +01:00
|
|
|
&=& \bigcup \bigcup \{\{x\}, x, \bigcup x, \ldots, \underbrace{\bigcup\nolimits^{n-1} x}_{F(n)}\}
|
2023-11-24 19:07:12 +01:00
|
|
|
= \bigcup F(n),
|
|
|
|
\end{IEEEeqnarray*}
|
|
|
|
i.e.~$F(n+1) = \bigcup F(n)$.
|
2024-02-11 02:43:20 +01:00
|
|
|
}{%
|
|
|
|
\begin{itemize}
|
|
|
|
\item Use recursion ($\in $) to define $F\colon \omega \to V$
|
|
|
|
such that
|
|
|
|
\[
|
|
|
|
F(0) = \{x\}, F(n+1) = \bigcup F(n).
|
|
|
|
\]
|
|
|
|
\item $\{x\} \cup \bigcup \ran(F)$ is as desired.
|
|
|
|
\end{itemize}
|
|
|
|
}
|
2023-11-23 15:38:28 +01:00
|
|
|
\end{proof}
|
2024-02-11 02:43:20 +01:00
|
|
|
\gist{%
|
2023-11-23 15:38:28 +01:00
|
|
|
\begin{notation}
|
|
|
|
Let $\OR$ denote the class of all ordinals
|
|
|
|
and $V$ the class of all sets.
|
|
|
|
\end{notation}
|
2024-02-11 02:43:20 +01:00
|
|
|
}{}
|
2023-11-23 15:38:28 +01:00
|
|
|
\begin{lemma}
|
|
|
|
There is a function $F\colon \OR \to V$
|
|
|
|
such that $F(\alpha) = \bigcup \{\cP(F(\beta)): \beta < \alpha\}$.
|
|
|
|
\end{lemma}
|
|
|
|
\begin{proof}
|
2023-11-24 19:07:12 +01:00
|
|
|
Use the recursion theorem with $R = \in $
|
|
|
|
and $(w,x,y) \in D$ iff
|
|
|
|
\[
|
|
|
|
y = \bigcup \{\cP(\overline{y}) : \overline{y} \in \ran(w)\}.
|
|
|
|
\]
|
|
|
|
This function has the following properties:
|
|
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
|
|
F(0) &=& \bigcup \emptyset = \emptyset,\\
|
|
|
|
F(1) &=& \bigcup \{\cP(\emptyset)\} = \bigcup \{\{\emptyset\} \} = \{\emptyset\},\\
|
|
|
|
F(2) &=& \bigcup \{\cP(\emptyset), \cP(\{\emptyset\})\} = \bigcup \{\{\emptyset\}, \{\emptyset, \{\emptyset\} \} \} = \{\emptyset, \{\emptyset\} \},\\
|
|
|
|
\ldots
|
|
|
|
\end{IEEEeqnarray*}
|
|
|
|
|
|
|
|
It is easy to prove by induction:
|
|
|
|
\begin{enumerate}[(a)]
|
|
|
|
\item Every $F(\alpha)$ is transitive.
|
|
|
|
\item $F(\alpha) \subseteq F(\beta)$ for all $\alpha \le \beta$.
|
|
|
|
\item $F(\alpha+1) = \cP(F(\alpha))$ for all $\alpha \in \OR$.
|
|
|
|
\item $F(\lambda) = \bigcup \{F(\beta) :\beta < \lambda\}$
|
|
|
|
for $\lambda \in \OR$ a limit.
|
|
|
|
\end{enumerate}
|
2023-11-23 15:38:28 +01:00
|
|
|
\end{proof}
|
|
|
|
\begin{notation}
|
2024-02-11 02:43:20 +01:00
|
|
|
Usually, one writes $V_\alpha$ for $F(\alpha)$.
|
2023-11-23 15:38:28 +01:00
|
|
|
They are called the \vocab{rank initial segments} of $V$.
|
|
|
|
\end{notation}
|
|
|
|
\begin{lemma}
|
|
|
|
If $x$ is any set, then there is some $\alpha \in \OR$
|
|
|
|
such that $x \in V_\alpha$,
|
|
|
|
i.e.~$V = \bigcup \{V_{\alpha} : \alpha \in \OR\}$.
|
|
|
|
\end{lemma}
|
2023-11-24 19:07:12 +01:00
|
|
|
\begin{proof}
|
|
|
|
We use induction on the well-founded $\in$-relation.
|
|
|
|
Let $A = \bigcup \{V_\alpha : \alpha \in \OR\}$.
|
|
|
|
We need to show that $A = V$.
|
|
|
|
By induction it suffices to prove that for every $x \in V$,
|
|
|
|
if $\{y : y \in x\} \subseteq A$, then $x \in A$.
|
|
|
|
The hypothesis says that for all $y \in x$,
|
|
|
|
there is some $\alpha$ with $y \in V_\alpha$.
|
|
|
|
Write $\alpha_y$ for the least such $\alpha$.
|
|
|
|
By \AxRep, $\{\alpha_y : y \in x\}$
|
|
|
|
is a set and we may let
|
|
|
|
$\alpha = \sup \{\alpha_y : y \in x\} \ge \alpha_y$
|
|
|
|
for all $y \in x$.
|
|
|
|
Then $y \in V_{\alpha_y} \subseteq V_\alpha$
|
|
|
|
for all $y \in x$.
|
|
|
|
|
|
|
|
In other words $x \subseteq V_\alpha$,
|
|
|
|
hence $x \in V_{\alpha+1}$.
|
|
|
|
\end{proof}
|
|
|
|
|
|
|
|
\begin{lemma}[\vocab{Transitive collapse}/\vocab{Mostowski collapse}]
|
|
|
|
\yalabel{Mostowski Collapse}{Mostowski}{lem:mostowski}
|
|
|
|
Let $R$ be a binary set-like relation on a class $A$.
|
|
|
|
Then $R$ is well-founded iff there is a transitive class $B$
|
|
|
|
such that
|
|
|
|
\[
|
|
|
|
(B, \in\defon{B}) \cong (A, R),
|
|
|
|
\]
|
|
|
|
i.e.~there is an isomorphism $F$,
|
|
|
|
that is a function $F\colon B \to A$
|
|
|
|
with $x \in y \iff (F(x),F(y)) \in R$ for $x,y \in B$.
|
|
|
|
\end{lemma}
|
|
|
|
\begin{proof}
|
|
|
|
``$\impliedby$'' Suppose that $R$ is ill-founded
|
|
|
|
(i.e.~not well-founded).
|
|
|
|
Then there is some $(y_n : n < \omega)$ such that $y_n \in A$
|
|
|
|
and $(y_{n+1}, y_n) \in R$ for all $n < \omega$.
|
|
|
|
But then if $F$ is an isomorphism as above,
|
|
|
|
\[
|
|
|
|
F^{-1}(Y_{n+1}) \in F^{-1}(Y_n)
|
|
|
|
\]
|
|
|
|
for all $n < \omega$ $\lightning$
|
|
|
|
|
|
|
|
``$\implies$ '' Suppose that $R$ is well-founded.
|
|
|
|
We want a transitive class $B$ and a function $F\colon B \leftrightarrow A$
|
|
|
|
such that
|
|
|
|
\[
|
|
|
|
x \in y \iff (F(x), F(y)) \in R.
|
|
|
|
\]
|
|
|
|
Equivalently $G\colon A \leftrightarrow B$
|
|
|
|
with $(x,y) \in R$ iff $G(x) \in G(y)$ for all $x,y \in A$.
|
|
|
|
|
|
|
|
In other words, $G(y) = \{G(x) : (x,y) \in R\}$.
|
|
|
|
Such a function $G$ and class $B$ exist by the recursion theorem.
|
|
|
|
\end{proof}
|
|
|
|
|
|
|
|
\begin{lemma}[\vocab{Rank function}]
|
|
|
|
Let $R$ be a well-founded and set-like binary relation
|
|
|
|
on a class $A$.
|
|
|
|
Then there is a function $F\colon A \to \OR$,
|
|
|
|
such that for all $x,y \in A$
|
2024-01-09 05:05:25 +01:00
|
|
|
\[(x,y) \in R \implies F(x) < F(y).\]
|
2023-11-24 19:07:12 +01:00
|
|
|
\end{lemma}
|
|
|
|
\begin{proof}
|
|
|
|
By the recursion theorem,
|
|
|
|
there is $F$ such that
|
|
|
|
\[
|
|
|
|
F(y) = \sup \{F(x) + 1 : (x,y) \in R\}.
|
|
|
|
\]
|
|
|
|
This function is as desired.
|
|
|
|
\end{proof}
|
|
|
|
|
|
|
|
This does not skip any ordinals,
|
|
|
|
as $F(y)$ is the least ordinal $> F(x)$
|
|
|
|
for all $(x,y) \in R$.
|
|
|
|
Thus $\ran(F)$ is transitive.
|
|
|
|
So either $\ran(F) = \OR$
|
|
|
|
or $\ran(F) \in \OR$.
|
|
|
|
This $F$ is called the \vocab{rank function} for $(A, R)$.
|
|
|
|
\begin{notation}
|
|
|
|
\[
|
|
|
|
\rk_R(x) = \|x\|_R \coloneqq F(x),
|
|
|
|
\]
|
|
|
|
and
|
|
|
|
\[
|
2024-01-10 22:22:28 +01:00
|
|
|
\rank(R) \coloneqq \ran(F).
|
2023-11-24 19:07:12 +01:00
|
|
|
\]
|
|
|
|
|
|
|
|
In the special case that $R$ is a linear order on $A$,
|
|
|
|
hence a well-order,
|
|
|
|
$\rank(R)$ is called the \vocab{order type} of $R$
|
|
|
|
(or of $(A,R)$), written $\otp(R)$.
|
|
|
|
\end{notation}
|
2023-11-23 15:38:28 +01:00
|
|
|
|