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

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