✳204.7 $\vdash: P \in Ser .\supset. P_1 \in 1 \rightarrow 1$

Which says if $P$ is a series, then $P_1$ is one-one.

✳201.63 $\vdash: P \in trans \cap Rl‘J .\supset. P_1 = P \overset{.}{-}P^2$

Which says if $P$ is transitive and asymetrical, then $P_1=P\overset{.}{-}P^2$


$P_1$ is "immediately precedes" as defined in ✳121.02 :

✳121.02 $P_v = \hat{x} \hat{y}\{ N_0c‘P(x|-|y)=v+_c1 \}$ Df where $v=1$

and $P^2 = P|P$ as defined in ✳34.02

Let R be "less than confined to the real interval [0,1]," then R is transitive, asymmetrical and connected. Therefore, by definition, $R$ is a series. But there is no term in $R$'s field that immediately precedes any other terms. In other words, $R_1$ does not seem to exist. If I'm not mistaken, the premises of 204.7 should include $\overset{.}{\exists}!P_1$ as a hypothesis.

  • $\begingroup$ OK. According to ✳250.21, every term in a well-ordered series(except for the last) has an immediate successor. That means even if ✳204.7 missed the hypothesis, it has no impact on well-ordered series, because $\Omega$ always contains a $P_1$. $\endgroup$ Oct 5 '14 at 3:59
  • $\begingroup$ But I do not understand your : "immediately precedes any other terms". In the set $\mathbb N$ of natural numbers $0$ precedes any other term (different from itself) but not immediately: $0 < 5$ but the "immediate" predeceswsor of $5$ is $4$. Of couse, my "reading" of your question may be wrong... $\endgroup$ Oct 5 '14 at 9:05
  • $\begingroup$ Thanks, @Mauro. $Ser$ is defined at 204.01. A series is a relation that is asymmetrical, transitive and connected. < over the field [0,1] is a series, but 0.5 (for example) has no immediate neighbours. No matter how close its neighbour(say, 0.500001) is to 0.5, there are always infinitely many other terms in between. $\endgroup$ Oct 5 '14 at 9:16
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    $\begingroup$ A series does not have to be discrete. That is why I think $P_1$'s existence is not guaranteed. $\endgroup$ Oct 5 '14 at 9:18
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    $\begingroup$ Ok, but then $[0,1]$ is a series and by Well-ordering Theorem we can order it. Clearly $0$ precedes any other term, but the Th is non-constructive: thus it gives us no clue about the way of finding the immediate predecessor of $1$. $\endgroup$ Oct 5 '14 at 9:32

If I understood well the definition, it is trivial that in $[0,1]$ there is an element which precedes any other element : it is $0$.

But to say that for any element different from $0$ there is an immediate predecessor, we have to use something like the Well-ordering Theorem proved by Zermelo in 1904 (thus known to W&R at time of writing PM) using the Axiom of Choice (the Multiplicative Axiom in PM).

This theorem says that any set can be well-ordered, and so also $[0,1]$.

But the proof is non-constructive: thus it gives us no clue about the way of finding the immediate predecessor of (say) $1$.

  • $\begingroup$ 202.7 has the same problem. $\endgroup$ Oct 5 '14 at 9:52
  • $\begingroup$ I think there is a problem. See 202.7, step (3), the implication is true, but the premises are not necessarily true. In other words, given an $x$, there may not exist a $y$ such that $y(P\overset{.}{-}P^2)x$. This is the kind of mistake W&R have been warning against throughout the whole volumes. $\endgroup$ Oct 5 '14 at 10:03
  • $\begingroup$ But I think this mistake is inconsequential. $\endgroup$ Oct 5 '14 at 10:04
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    $\begingroup$ Surpise! See 72.1 $\vdash. \overset{.}{\Lambda}\in 1\rightarrow 1$. That means even if $P_1$ is $\overset{.}{\Lambda}$, it is still one-one. $\endgroup$ Oct 5 '14 at 21:09
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    $\begingroup$ @GeorgeChen - see page 498 (vol.II) : in a compact (i.e. not discrete) series $P_1 = \overset{.}{\Lambda}$ and, as you have verified : $\overset{.}{\Lambda} \in 1 \to 1$. $\endgroup$ Oct 6 '14 at 7:11

Given a set $\mu$,

Let $Q=\hat{ \alpha}\hat{\beta}\{\alpha,\beta \in Cl_{ex}‘\mu.\beta \subset \alpha.\beta \neq \alpha \}$

Let $S \in \epsilon_\Delta‘Cl_{ex}‘\mu$

Let $R=\hat{\alpha}\hat{\beta}\{\alpha \in Cl_{ex}‘\mu.\beta=\alpha - \iota ‘S‘\alpha \}$

Then $Q_{R\mu}$ is $Q$ confined to the transfinite posterity of $\mu$, and $S^;Q_{R\mu}$ is Well-ordered over the field of $\mu$.

To construct $S^;Q_{R\mu}$, all we need is to construct a selector $S$ which selects one term out of each of $\mu$'s subclasses.

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