195 lines
6.5 KiB
TeX
195 lines
6.5 KiB
TeX
\lecture{07}{2023-11-09}{}
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\begin{lemma}+
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By the axiom of foundation there cannot exist infinite
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descending chains
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$x_1 \ni x_2 \ni x_3 \ni \ldots$
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\end{lemma}
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\begin{proof}
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\todo{TODO!}
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\end{proof}
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\begin{notation}
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From now on, we will write $\alpha, \beta, \ldots$
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for ordinals.
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\end{notation}
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\begin{lemma}
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\begin{enumerate}[(a)]
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\item $0$ is an ordinal, and if $\alpha$ is
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an ordinal, so is $\alpha + 1$.
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\item If $\alpha$ is an ordinal and $x \in a$,
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then $x$ is an ordinal.
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\item If $\alpha, \beta$ are ordinals
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and $\alpha \subseteq \beta$,
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then $\alpha = \beta$ or $\alpha \in \beta$.
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\item If $\alpha$ and $\beta$ are ordinals,
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then $\alpha \in \beta$, $\alpha = \beta$
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or $\alpha \ni \beta$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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We have already proved (a) before.
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(b) Fix $x \in \alpha$. Then $x \subseteq \alpha$.
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So if $y, z \in x$, then $y \in z \lor y = z \lor y \ni z$.
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Let $y \in x$. We need to see $y \subseteq x$.
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Let $z \in y$.
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\begin{claim}
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$z \in x$
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\end{claim}
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\begin{subproof}
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As $\alpha$ is transitive, we have that $z, y, x \in \alpha$.
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Thus $z \in x \lor z = x \lor z \ni x$.
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$z = x$ contradicts Fund:
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Consider $\{x,y\}$. Then $x \cap \{x,y\}$ is non empty,
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as it contains $y$.
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Furthermore $x \in y \cap \{x,y\} $
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$z \ni x$ also contradicts Fund:
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If $x \in z$, then $z \ni x \ni y \ni z \ni x \ni \ldots$.
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$\{x,y,z\}$ yields a contradiction,
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as $y \in x \cap \{x,y,z\}$, $z \in y \cap \{x,y,z\}$,
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$x \in z \cap \{x,y,z\}$.
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So $z \in x$ as desired.
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\end{subproof}
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(c) Say $\alpha \subsetneq \beta$.
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Pick $\xi \in \beta \setminus \alpha$
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such that $\eta \in \alpha$ for every $\eta \in\xi \cap \beta$.
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(This exists by Fund).
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We want to see that $\xi = \alpha$.
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We have $\xi \subseteq \alpha$ by the choice of $\xi$.
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On the other hand $\alpha \subseteq \xi$:
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Let $\eta \in \alpha \subseteq \beta$.
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We have that $\eta \in \xi \lor \eta = \xi \lor \eta \ni \xi$.
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If $\xi \in \eta$, then since $\eta \in \alpha$,
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we get $\xi \in \alpha$ contradicting the choice of $\xi$.
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If $\xi = \eta$, the $\xi = \eta \in \alpha$,
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which also is a contradiction.
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Thus $\eta \in \xi$.
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This yields $\alpha \in \beta$, hence $\alpha$ is an ordinal.
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(d) By (c) if $\alpha$ and $\beta$ are ordinals,
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then $\alpha \subseteq \beta \iff (\alpha = \beta \lor \alpha \in \beta)$.
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We need tho see that if $ \alpha$, $\beta$ are ordinals,
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then $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$.
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Suppose there are ordinals $\alpha$, $\beta$ such that
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this is not the case.
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Pick such an $\alpha$.
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Let $\alpha_0 \in \alpha \cup \{\alpha\}$
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be such that there is some $\beta$
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with $\lnot( \beta \subseteq \alpha_0 \lor \alpha_0 \subseteq \beta)$
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for for all $\gamma \in \alpha_0$,
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$\forall \beta. ( \beta \subseteq \gamma \lor \gamma \subseteq \beta)$.
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Pick $\beta_0$ such that
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\[
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\lnot \left( \beta_0 \subseteq \alpha_0 \lor \alpha_0 \subseteq \beta_0 \right).
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\]
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Consider $\alpha_0 \cup \beta_0$.
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\begin{claim}
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$\alpha_0 \cup \beta_0$
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is an ordinal.
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\end{claim}
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\begin{subproof}
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$\alpha_0 \cup \beta_0$ is clearly transitive.
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Let $\gamma,\delta \in \alpha_0 \cup \beta_0$.
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We claim that $\gamma \in \delta \lor \gamma = \delta \lor \gamma \in \delta$.
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This can only fail if $\gamma \in \alpha_0$ and $\delta \in \beta_0$
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(or the other way around).
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But then $\gamma \in \delta \lor \gamma = \delta \lor \delta \in \gamma$
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by the choice of $\alpha_0$.
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\end{subproof}
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\begin{claim}
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$\alpha_0 = \alpha_0 \cup \beta_0$
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or $\beta_0 = \alpha_0 \cup \beta_0$.
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\end{claim}
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\begin{subproof}
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If that is not the case,
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then $\alpha_0 \in \alpha_0 \cup \beta_0$
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and $\beta_0 \in \alpha_0 \cup \beta_0$.
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$\alpha_0 \in \alpha_0$ violates Fund.
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Hence $\alpha_0 \in \beta_0$.
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By the same argument, $\beta_0 \in \alpha_0$.
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But this violates Fund,
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as $\alpha_0 \in \beta_0 \in \alpha_0$.
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\end{subproof}
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\end{proof}
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\begin{lemma}
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Let $X$ be a set of ordinals,
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$X \neq \emptyset$.
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Then $\bigcap X$ and $\bigcup X$ are ordinals.
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\end{lemma}
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\begin{proof}
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Easy.
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\end{proof}
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It is actually the case that $\bigcap X \in X$:
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Pick $\alpha \in X$ such that $\alpha \subseteq \beta$
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for all $\beta \in X$. This exists
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by Fund and since all ordinals are comparable.
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Then $\alpha = \bigcap X$.
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\begin{notation}
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We write $\min(X)$ for $\bigcap X$
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and $\sup(X)$ for $\bigcup X$.
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\end{notation}
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It need not be the case that $\bigcup X \in X$,
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for example $\bigcup \omega = \omega$.
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% but $\bigcup 2 = 1$.
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\begin{definition}
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An ordinal $\alpha$ is called
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a \vocab[Ordinal!successor]{successor ordinal},
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iff $\alpha = \beta \cup \{\beta\}$
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for some $\beta \in \alpha$.
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Otherwise $\alpha$ is called
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a \vocab[Ordinal!limit]{limit ordinal}.
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\end{definition}
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\begin{observe}
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Note that $\alpha$ is a limit ordinal iff for all
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$\beta \in \alpha$, $\beta + 1 \in \alpha$:
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If there is $\beta \in \alpha$ such that $\beta+1 \not\in \alpha$,
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then either $\alpha = \beta+1$ (i.e.~$\alpha$ is a successor)
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or $\alpha \in \beta+1$,
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in which case $\beta \in \alpha \in \beta \cup \{\beta\} \lightning$.
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Also if $\alpha$ is a successor,
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then by definition there is some $\beta \in \alpha$,
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with $\beta + 1 = \alpha$, so $\beta + 1 \not\in \alpha$.
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\end{observe}
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\begin{notation}
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If $\alpha, \beta$ are ordinals,
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we write $\alpha < \beta$ for $\alpha \in \beta$
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(equivalently $\alpha \subsetneq \beta$).
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We also write $\alpha \le \beta$
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for $\alpha \in \beta \lor \alpha = \beta$
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(i.e.~$\alpha \subseteq \beta$).
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\end{notation}
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\begin{example}
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Limit ordinals:
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\begin{itemize}
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\item $0$,
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\item $\omega$,
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\item $\omega + \omega = \sup(\omega \cup \{\omega, \omega + 1, \ldots\})$,%
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\footnote{To show that this exists, we need the recursion theorem
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and replacement.}
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$\omega + \omega + \omega, \omega + \omega + \omega + \omega, \ldots$
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\end{itemize}
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Successor ordinals:
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\begin{itemize}
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\item $1 = \{0\}, 2 = \{0,1\}, 3, \ldots$
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\item $\omega +1 = \omega \cup \{\omega\} , \omega + 2, \ldots$,
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\end{itemize}
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\end{example}
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