Enderton. Finish "Peano's Postulates" section.
parent
db6074f1a1
commit
1260c493bc
|
@ -6170,7 +6170,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
|
|||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Theorem 4G}}%
|
||||
\subsection{\unverified{Theorem 4G}}%
|
||||
\label{sub:theorem-4g}
|
||||
|
||||
\begin{theorem}[4G]
|
||||
|
@ -6181,7 +6181,32 @@ Show that $<_L$ is a linear ordering on $A \times B$.
|
|||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Let $T = \{n \in \omega \mid \forall t \in n, t \in \omega\}$.
|
||||
We prove that (i) $T$ is an \nameref{ref:inductive-set} and then (ii) every
|
||||
member of a natural number is itself a natural number.
|
||||
|
||||
\paragraph{(i)}%
|
||||
\label{par:theorem-4g-i}
|
||||
|
||||
First, it vacuously holds that $\emptyset \in T$.
|
||||
Next, let $n \in T$.
|
||||
We must prove that $n^+ \in T$ as well.
|
||||
By definition of the \nameref{ref:successor}, $n^+ = n \cup \{n\}$.
|
||||
That is, either $n^+ = n$ or $n^+ = \{n\}$.
|
||||
If the former, then every member of $n^+$ must be a natural number since
|
||||
this already holds for $n$.
|
||||
If the latter, the only member of $n^+$ is $n$ which is, by definition of
|
||||
$T$, a natural number.
|
||||
Thus $n^+ \in T$.
|
||||
We conclude that $T$ is an inductive set.
|
||||
|
||||
\paragraph{(ii)}%
|
||||
|
||||
Since $T \subseteq \omega$, \nameref{par:theorem-4g-i} and
|
||||
\nameref{sub:theorem-4b} implies $T = \omega$.
|
||||
Thus the member of every natural number is itself a natural number.
|
||||
In other words, $\bigcup \omega \subseteq \omega$.
|
||||
Therefore $\omega$ is indeed a \nameref{ref:transitive-set}.
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
@ -6208,18 +6233,23 @@ Show that $1 \neq 3$ i.e., that $\emptyset^+ \neq \emptyset^{+++}$.
|
|||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 4.2}}%
|
||||
\subsection{\unverified{Exercise 4.2}}%
|
||||
\hyperlabel{sub:exercise-4.2}
|
||||
|
||||
Show that if $a$ is a transitive set, then $a^+$ is also a transitive set.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Suppose $a$ is a \nameref{ref:transitive-set}.
|
||||
By \nameref{sub:theorem-4e}, it follows $\bigcup \left(a^+\right) = a$.
|
||||
By definition of a \nameref{ref:successor}, $a^+ = a \cup \{a\}$.
|
||||
Thus it immediately follows
|
||||
$$\bigcup \left(a^+\right) = a \subseteq a \cup \{a\} = a^+.$$
|
||||
Therefore $a^+$ is indeed a transitive set.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 4.3}}%
|
||||
\subsection{\unverified{Exercise 4.3}}%
|
||||
\hyperlabel{sub:exercise-4.3}
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
|
@ -6231,22 +6261,48 @@ Show that if $a$ is a transitive set, then $a^+$ is also a transitive set.
|
|||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
\paragraph{(a)}%
|
||||
|
||||
Suppose $a$ is a \nameref{ref:transitive-set}.
|
||||
We show that $\bigcup \powerset{a} \subseteq \powerset{a}$.
|
||||
Let $t \in \bigcup \powerset{a}$.
|
||||
By definition of the \nameref{ref:power-set}, there exists some
|
||||
$X \subseteq a$ such that $t \in X$.
|
||||
Thus $t \in a$.
|
||||
Because $a$ is a transitive set, every member of $t$ is a member of $a$.
|
||||
In other words, $t \subseteq a$.
|
||||
Equivalently, $t \in \powerset{a}$.
|
||||
|
||||
\paragraph{(b)}%
|
||||
|
||||
Suppose $\powerset{a}$ is a transitive set.
|
||||
We show that $\bigcup a \subseteq a$.
|
||||
Let $t \in \bigcup a$.
|
||||
Then there exists some $b \in a$ such that $t \in b$.
|
||||
Since $\{b\}$ is a member of $\powerset{a}$ and $\powerset{a}$ is a
|
||||
transitive set, $b \in \powerset{a}$.
|
||||
That is, $b \subseteq a$.
|
||||
Thus $t \in a$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 4.4}}%
|
||||
\subsection{\unverified{Exercise 4.4}}%
|
||||
\hyperlabel{sub:exercise-4.4}
|
||||
|
||||
Show that if $a$ is a transitive set, then $\bigcup a$ is also a transitive set.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Suppose $a$ is a transitive set.
|
||||
We show that $\bigcup\bigcup{a} \subseteq \bigcup a$.
|
||||
Let $t \in \bigcup\bigcup{a}$.
|
||||
Then there exists some $b \in \bigcup{a}$ such that $t \in b$.
|
||||
Since $a$ is transitive, $\bigcup{a} \subseteq a$.
|
||||
Thus $b \in a$ and $t \in \bigcup a$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 4.5}}%
|
||||
\subsection{\unverified{Exercise 4.5}}%
|
||||
\hyperlabel{sub:exercise-4.5}
|
||||
|
||||
Assume that every member of $\mathscr{A}$ is a transitive set.
|
||||
|
@ -6259,19 +6315,47 @@ Assume that every member of $\mathscr{A}$ is a transitive set.
|
|||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
\paragraph{(a)}%
|
||||
|
||||
Suppose every member of $\mathscr{A}$ is a transitive set.
|
||||
We show that $\bigcup\bigcup{\mathscr{A}} \subseteq \bigcup{\mathscr{A}}$.
|
||||
Let $t \in \bigcup\bigcup{\mathscr{A}}$.
|
||||
Then there exists some $b_1 \in \bigcup{\mathscr{A}}$ such that $t \in b_1$.
|
||||
Likewise there exists some $b_2 \in \mathscr{A}$ such that $b_1 \in b_2$.
|
||||
By hypothesis, $b_2$ is transitive meaning $t \in b_2$.
|
||||
Thus $t \in \bigcup{\mathscr{A}}$.
|
||||
|
||||
\paragraph{(b)}%
|
||||
|
||||
Suppose every member of nonempty set $\mathscr{A}$ is a transitive set.
|
||||
We show that $\bigcup\bigcap{\mathscr{A}} \subseteq \bigcap{\mathscr{A}}$.
|
||||
Let $t \in \bigcup\bigcap{\mathscr{A}}$.
|
||||
Then there exists some $b \in \bigcap{\mathscr{A}}$ such that $t \in b$.
|
||||
Thus $b$ is a member of every member of $\mathscr{A}$.
|
||||
By hypothesis, every member of $\mathscr{A}$ is a transitive set meaning
|
||||
$t$ must be a member of every member of $\mathscr{A}$.
|
||||
In other words, $t \in \bigcap{\mathscr{A}}$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 4.6}}%
|
||||
\subsection{\unverified{Exercise 4.6}}%
|
||||
\hyperlabel{sub:exercise-4.6}
|
||||
|
||||
Prove the converse to Theorem 4E: If $\bigcup \left(a^+\right) = a$, then $a$ is
|
||||
a transitive set.
|
||||
Prove the converse to \nameref{sub:theorem-4e}: If
|
||||
$\bigcup \left(a^+\right) = a$, then $a$ is a transitive set.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Let $a$ be a set such that $\bigcup \left(a^+\right) = a$.
|
||||
Then
|
||||
\begin{align*}
|
||||
\bigcup \left(a^+\right)
|
||||
& = \bigcup (a \cup \{a\}) & \textref{ref:successor} \\
|
||||
& = \bigcup a \cup \bigcup \{a\} & \textref{sub:exercise-2.21} \\
|
||||
& = a. & \text{by hypothesis}
|
||||
\end{align*}
|
||||
But $$\bigcup{a} \subseteq \bigcup a \cup \bigcup \{a\} = a.$$
|
||||
Thus $a$ is indeed a \nameref{ref:transitive-set}.
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
|
Loading…
Reference in New Issue