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)

Exercise 1.3
Let $\mathcal M$ be a $\sigma$-algebra on an infinite set $X$.

Then $\mathcal M$ is uncountable.

Proof
Using the Axiom of Choice, choose an indexed countable collection of subsets $A_i \in M$:


 * $\langle A_i \rangle_{i \mathop \in \N} = \left \langle { A_0, A_1, A_2, \ldots }\right \rangle$

such that $\varnothing \in \langle A \rangle$ and $X \in \langle A \rangle$.

Then the hypotheses of Union of Indexed Family of Sets Equal to Union of Disjoint Sets are satisfied:


 * $\displaystyle \bigcup_{i \mathop \in \N} A_i = \bigsqcup_{i \mathop \in \N} F_i $

for an indexed collection of disjoint sets $\langle F_i \rangle_{i \mathop \in \N}$ of subsets of $M$.

By the definition of a $\sigma$-algebra, $\displaystyle \bigsqcup_{i \mathop \in \N} F_i$ is measurable.

By the definition of an indexed family, $\langle F_i \rangle_{i \mathop \in \N}$ corresponds to a mapping $\iota: \N \hookrightarrow \bigsqcup_{i \mathop \in \N} F_i$.

Such a mapping $\iota$ is injective because the $F_i$s are disjoint.

Define the induced map:


 * $\displaystyle \iota^\to : \mathcal P (\N) \hookrightarrow \mathcal P \left({ \bigsqcup_{i \mathop \in \N} F_i}\right)$

This is an injection because Mapping Induced on Power Set by Injection is Injection.

By the definition of a $\sigma$-algebra, the countable union of measurable sets is measurable.

As Countable Union of Countable Sets is Countable, the elements of $\mathcal P \left({ \bigsqcup_{i \mathop \in \N} F_i}\right)$ are sets in $\mathcal M$.

So $\mathcal P \left({ \bigsqcup_{i \mathop \in \N} F_i}\right) \subseteq M$.

Extend $\iota^\to$ by:


 * $\displaystyle \left({\iota^\to}\right)' : \mathcal P (\N) \hookrightarrow \mathcal M$

defined by $\left({\iota^\to}\right)'(x) = \left({\iota^\to}\right)(x)$

From Continuum equals Cardinality of Power Set of Naturals, the cardinality of $\mathcal M$ is at least the Cardinality of Continuum.

So it is uncountable.

What I need to think about:


 * Though Folland states this result in terms of cardinality of the continuum, does that really add anything to the proof? The cardinality of any sigma-algebra is at least continuum. Would it be clearer to the reader without invoking such concepts? KISS principle?

Feel free to leave your thoughts, I'll come back to this later. --GFauxPas (talk) 21:11, 31 May 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