User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

Well-Ordered Transitive Subset is Equal or Equal to Initial Segment

Woset is Isomorphic to Unique Ordinal

Historical note
Ernst Zermelo exposited the axiom of choice in order to prove the Well-Ordering Theorem.

He himself did not invent the idea of choice; he claimed it was implicitly used by many great mathematicians in many contexts.

Formulation 1 and formulation 2 are written in a letter by Zermelo to David Hilbert, postmarked September 24, 1904:


 * ''Der vorliegende Beweis beruht auf der Voraussetzung [...] daß es auch für eine unendliche Gesamtheit von Mengen immer Zuordnungen gibt, bei denen jeder Menge eines ihrer Elemente entspricht, oder formal ausgedrückt, daß das Produkt einer unendlichen Gesamtheit von Mengen, deren jede mindestens ein Element enthält, selbst von Null verschieden ist. Dieses logische Prinzip läßt sich zwar nicht auf ein noch einfacheres zurückführen, wird aber in der mathematischen Deduktion überall unbedenklich angewendet.

''


 * The present proof is based on the presumption [...] that even for an infinite assembly of sets, there are always assignments in which each set corresponds to one of its elements, or formally expressed, that the product of an infinite assembly of sets, each of which holds at least one element, is itself different from zero. This logical principle can’t, however, be reduced to something even simpler, but is used without second thought everywhere in the mathematical deduction.

Zermelo wrote formulation 3 in


 * Eine Menge $S$, welche in eine Menge getrennter Teile $A, B, C, \ldots$ zerfällt, deren jeder mindestens ein Element enthält, besitzt mindestens eine Untermenge $S_1$, welche mit jedem der betrachteten nile $A, B, C, \ldots$ genau ein Element gemein hat.


 * A set $S$, which seperates into a set of divided parts $A, B, C, \ldots$, each of which contains at least one element, has at least one subset $S_1$, which in turn has in common with each part $A,B,C,\ldots$ exactly one element.

Proposition P.17
Let $\left({S, \preceq_S}\right)$ and $\left({T, \preceq_T}\right)$ be well-ordered sets.

Then:


 * $\left({S, \preceq_S}\right)$ is order isomorphic to $\left({T, \preceq_T}\right)$

or:


 * $\left({S, \preceq_S}\right)$ is order isomorphic to an initial segment in $\left({T, \preceq_T}\right)$

or:


 * $\left({T, \preceq_T}\right)$ is order isomorphic to an initial segment in $\left({S, \preceq_S}\right)$

Proof of Proposition P.17
Let $\inf$ denote the smallest element of a woset.

Define:


 * $S' = S \cup \text{ initial segments in } S$


 * $T' = T \cup \text{ initial segments in } T$


 * $\mathcal F = \left\{ { f:S' \to T' \ \vert \ f \text{ is an order isomorphism} } \right\}$

The following three lines need surgical precision:

Every chain in $S'$ has an upper bound, because it defines an initial segment.

Likewise for $T'$.

Furthermore, the ordered product defined on $S' \times T'$ has the property that every chain in $S' \times T'$ has an upper bound.

Then $\mathcal F$ is not empty, because it at least contains the following mapping between singletons:


 * $f_0: \left\{ { \inf S} \right\} \to \left\{ { \inf T} \right\}$

Thus the hypotheses of Zorn's Lemma are satisfied.

Let $f_1$ be a maximal element of $\mathcal F$. Call its domain $A$ and its codomain $B$.

Suppose $A$ is an initial segment $I_a$ in $S$ and $B$ is an initial segment $I_b$ in $T$.

Then $f_1$ can be extended by defining $f_1\left({a}\right) = b$.

This would contradict $f_1$ being maximal, so it cannot be the case that both $A$ and $B$ are initial segments.

Either:


 * $A = S$, with $S$ order isomorphic to an initial segment in $T$

or:


 * $B = T$ with $T$ order isomorphic to an initial segment in $T$

or:


 * both $A = S$ and $B = T$ with $S$ order isomorphic to $T$.

Proposition P.17

Proposition 1.23
Let $\mathcal E \subseteq \mathcal P(E)$ be a set of subsets of $E$.

Then the $\sigma$-algebra generated by $\mathcal E$ can be constructed inductively.

The construction is as follows:


 * $\mathcal E_0 = \mathcal E$


 * $\mathcal E_1 = \mathcal E_0 \cup \left({\mathcal P(E) \setminus \mathcal E_0}\right)$

For $j \ge 2, j \in \N$:


 * $\mathcal E_j = \left \{ { \mathcal S \in \mathcal P\left({\mathcal E_{j-1} }\right) : \mathcal S \text { or } \mathcal S^\complement \text{ is countable} }\right\}$

that is, the $\sigma$-algebra of countable and co-countable subsets of sets in $\mathcal E_{j-1}$.


 * $\mathcal E_{\omega} = \displaystyle \bigcup_{j \mathop \in \N} \mathcal E_j$

Let $\Omega$ denote the set of countable ordinals.

Let $\alpha, \beta$ be initial segments in $\Omega$.

Continue the above process by:


 * $\mathcal E_\alpha = \begin{cases} \left \{ { \mathcal S \in \mathcal P\left({\mathcal E_\beta}\right) : \mathcal S \text { or } \mathcal S^\complement \text{ is countable} }\right\} & \alpha \text{ has an immediate predecessor } \beta \\ \bigcup_{\beta \mathop \prec \alpha } \mathcal E_\beta & \text{ otherwise } \end{cases}$


 * $\mathcal E_{\Omega} = \displaystyle \bigcup_{\alpha \mathop \in \Omega} \mathcal E_{\alpha}$

Then $\sigma\left({\mathcal E}\right) = \mathcal E_{\Omega}$.

Question: The structure for finite indices is not strictly necessary because it is subsumed in the construction over cardinals, because the induction over $\Omega$ subsumes that case. Folland brings it as a motivation before expositing the theorem and proving it. What should I do with that introduction on PW? --GFauxPas (talk) 12:10, 4 June 2018 (EDT)

Eventually
User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

User:GFauxPas/Sandbox/NumberTheory