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)

Theorem
Let $\mathcal E$ such that:


 * $\operatorname{card}\left({\N}\right) \le \operatorname{card}\left({\mathcal E}\right) \le \mathfrak c$

Then $\operatorname{card}\left({ \sigma\left({\mathcal E}\right) }\right) = \mathfrak c$.

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

Let $\sigma\left({\mathcal E}\right)$ be the $\sigma$-algebra generated by $\mathcal E$.

Then $\sigma\left({\mathcal E}\right)$ 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 $j \ge 2, j \in \N$:


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

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

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

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

Continue the above process by separately considering whether or not $\beta$ immediately precedes $\alpha$:


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


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

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

Proof of Proposition 1.23
By the definition of union, $\mathcal E_j$ contains $\mathcal E$ for all $j$.

From $\sigma$-Algebra of Countable Sets, $\mathcal E_3$ is a $\sigma$-algebra.

By the construction of $\mathcal E_j$, $\mathcal E_\alpha$ is also a $\sigma$-algebra for all $\alpha \in \Omega$.

So, by induction:


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

is also a $\sigma$-algebra containing $\mathcal E$.

WRONG DIRECTION

Thus $\mathcal E_\Omega \subseteq \sigma\left({\mathcal E}\right)$, by the definition of a generated $\sigma$-algebra.

Let $E$ be an arbitrary $\sigma$-algebra containing $\mathcal E$.

Need: Countable subset of $\Omega$ has an upper bound.

By the definition of $\Omega$, we can define a countable sequence $\left \langle {\alpha_0, \alpha_1, \ldots} \right \rangle$ of initial segments in $\Omega$, where the indexing is surjective.

Then there exists $\mathcal A = \sup \left \langle { \alpha_0, \alpha_1, \ldots } \right \rangle$

Let $\Sigma_i$ be an arbitrary $\sigma$-algebra on $\mathcal E$.

If $\Sigma_i = \mathcal E_{\alpha_i}$ for some $i \in \N$, then $E_i \in \mathcal E_A$ for all $i$, by the definition of supremum.

Then $\displaystyle \bigcup_{i \mathop \in \N} E_i = \mathcal E_B$, where $B$ is the immediate successor of $A$.

Hence $\sigma\left({\mathcal E}\right) \subseteq \mathcal E_\Omega$.

The result follows from equality of sets.

AoC|any sequence in $\Omega$ admits a supremum in $\Omega$

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