Let [math]\{E_{i}\}[/math] be a sequence of subsets (indexed set of subsets) of [math]X[/math] ([math]E_{i}\subset X[/math] for all [math]i \in \mathbb{N}=\{1,2,...\}[/math]). We define [math]E^{*}[/math] as the set of points [math]x \in X[/math] which belong to [math]X[/math] for infinitely many values of [math]i[/math]. Similarly we say that [math]x \in E_{*}[/math] if [math]x \in E_{i}[/math] except for a finite number of [math]i[/math]s (i.e. we have [math]n \in \mathbb{N}[/math] such that [math]x \in E_{i}[/math] for all [math]i>n[/math]). It follows that [math]E_{*}\subset E^{*}[/math]. The converse is false: if [math]x \in E_{i}[/math] for [math]i=2n,n\in \mathbb{N}[/math] then [math]i=2,4,...[/math] and [math]x \in E^{*}[/math] but [math]x \not \in E_{*}[/math].
Let [math]\{E_{i}\}[/math] such that (s.t.) [math]E_{1}\subset E_{2}\subset E_{3}\subset ...[/math]. Then we prove that [math]\cup_{i=1}^{\infty}E_{i}=E^{*}=E_{*}[/math]. We start from [math]\cup_{i=1}^{\infty}E_{i}=E^{*}[/math]: if [math]x \in \cup_{i=1}^{\infty}E_{i}[/math] then [math]\exists i\in \mathbb{N}[/math] ([math]\cup_{i=1}^{\infty}E_{i}=\cup_{n \in \mathbb{N}}E_{i}[/math]) s.t. [math]x \in E_{i}[/math]. But [math]E_{i}\subset E_{i+n}[/math] for all [math]n \in \mathbb{N}[/math] hence [math]x \in E^{*}[/math]. If [math]x \in E^{*}[/math] then [math]x \in E_{i}[/math] for infinite many values of [math]i[/math] hence for at least one [math]i[/math] we have [math]x \in E_{i}\subset \cup_{i=1}^{\infty}E_{i}[/math]. [math]E^{*}=E_{*}[/math]: we already know that [math]E_{*}\subset E^{*}[/math]. If [math]x \in E^{*}[/math] then [math]x \in E_{i}[/math] for inf. [math]i[/math] hence for at least one [math]i[/math], [math]x \in E_{i}\subset E_{i+n}[/math] for all [math]n \in \mathbb{N}[/math] so we don't care if [math]x \not \in E_{j}[/math] for [math]j=1,..,i-1[/math] and [math]E^{*}\subset E_{*}[/math] follows.
Nathan Allen
Let [math]\{E_{i}\}[/math] s.t. [math]E_{1}\supset E_{2}\supset E_{3}\supset ...[/math]. Then we prove that [math]\cap_{i=1}^{\infty}E_{i}=E^{*}=E_{*}[/math]. We start from [math]\cap_{i=1}^{\infty}E_{i}=E^{*}[/math]: if [math]x \in \cap_{i=1}^{\infty}E_{i}[/math] then [math]x \in E_{i}[/math] for all [math]i \in \mathbb{N}[/math] hence [math]x \in E^{*}[/math]. If [math]x \in E^{*}[/math] then [math]x \in E_{i}[/math] for inf. many [math]i[/math]. Suppose there is an [math]i[/math] s.t. [math]x\not \in E_{i}[/math] (so [math]x \not \in \cap_{i=1}^{\infty}E_{i}[/math]). Then [math]x \not \in E_{i+n}[/math] for every [math]n \in \mathbb{N}[/math] since [math]E_{i+n}\subset E_{i}[/math]; from this [math]x \not \in E^{*}[/math]. Follows that [math]x \in \cap_{i=1}^{\infty}E_{i}[/math] if [math]x \in E^{*}[/math]. Now [math]E^{*}\subset E_{*}[/math]: [math]\cap_{i=1}^{\infty}E_{i}=E^{*}[/math] so showing [math]\cap_{i=1}^{\infty}E_{i}\subset E_{*}[/math] would be proving the same thing. If [math]x \in \cap_{i=1}^{\infty}[/math] then [math]x \in E_{i}[/math] for all [math]i \in \mathbb{N}[/math] hence [math]x \in E_{*}[/math] from the definition of [math]E_{*}[/math].
Charles Rodriguez
If anyone of you guys is interested I can dumb down more of this book and upload the latex code on github or somewhere else.
Caleb Fisher
It would be nice of you...
Caleb Richardson
Practicing your Latex moron?
Asher Collins
Fuck off
Luke Reyes
One of my profs sent me a digital copy of a Measure Theory textbook designed for undergrads. It'd be better than trying to dumb it down for you. Search for Marek Capinski and Ekkehard Kopp Measure, Integral and Probability.
Angel Richardson
I have this book, it's bretty good (but kinda advanced, couldn't read it until after i graduated).
William Adams
>It'd be better than (You) trying to dumb it down for you What I mean is that I'm explaining passages that might be harder to grasp for who is not that much initiated to math.
Plus, Halmos is a wonderful reference book, better than more watered down resources.
Jeremiah Hill
(You)
Austin Long
p.19.
A class (set of sets) [math]R[/math] of subsets of [math]X[/math] is called ring if for every [math]E,F\in R[/math] (could be [math]E=F[/math]) we have
[math]1.[/math] [math]E\cup F \in R\in R[/math], [math]2.[/math] [math]E\backslash F\in R[/math].
If we choose [math]E=F[/math] then [math]\emptyset=E\backslash F \in R[/math] as well. Example: Let [math]E_{1},E_{2}\subset X[/math] s.t. [math]E_{1}\cap E_{2}=\emptyset[/math] (could be [math]X=\mathbb{R}[/math] and [math]E_{1}=(0,1),E_{2}=[2,3)[/math]); define [math]R=\{E_{1},E_{2},E_{1}\cup E_{2},\emptyset\}[/math].
Jace Bennett
Don't know what's wrong with that. The preview is alright.
Lincoln Martin
That is one cute kitty.
Sebastian Hall
Fair enough. If Halmos is that good, I might just get it myself. I spent the summer pushing through Galois theory using Fraleigh's book and Analysis 2 using Munkres' book. Next summer I make the move to Measure Theory and whatever else suits my fancy at the time.
Christian Jenkins
Now we go through all the possible unions and show [math]1.:[/math] we exclude trivial unions such [math]E\cup E[/math]. We have [math]n=4[/math] sets to place in [math]k=2[/math] spaces [math]\cdot \cup \cdot[/math] (the dots). The order ([math]E\cup F[/math] or [math]F\cup E[/math]) is not important since the union is reflexive ([math]E\cup F=F\cup E[/math]). Then
Notice the order: we pick the first from the left then proceed to the right. Then we pick the second and so on. In the same fashion we fill out [math]\cdot \backslash \cdot[/math]:
All this pedantry is unnecessary but also nice because it teaches you to think in an ordered manner.
Xavier Morgan
If [math]R[/math] is a ring then the statement that [math]R[/math] is closed under the formation of [math]\cap[/math] and [math]\backslash[/math] is true. The converse is false: if we have a class of sets [math][/math] closed under the formation [math]\cap[/math] and [math]\backslash[/math], then we do not necessarily have a ring. How can we be sure? By finding such case.
Consider [math]E,F\subset X[/math] s.t. [math]E\cap F=\emptyset[/math] and define [math]R=\{E,F,\emptyset\}[/math]. You can fill out all the [math]\cdot \cap \cdot[/math]s and [math]\cdot \backslash \cdot[/math]s by yourself, ignoring sets like [math]E\cap E[/math] and [math]E \backslash E[/math] since all they do is give the set itself (the first) and the empty set (the second). [math]R[/math] is not closed under [math]\cup[/math] (required when defining a ring) since [math]E\cup F\not \in R[/math] hence [math]R[/math] is not a ring.
Two set operations that both interchangeably define a ring are [math]\cap[/math] and [math]\triangle[/math], that is: if [math]R[/math] is a ring then [math]R[/math] is closed under [math]\cap[/math] and [math]\triangle[/math] (p. 20). Is [math]R[/math] is a class of sets closed under [math]\cap[/math] and [math]\triangle[/math] then we can rewrite [math]\cup[/math] and [math]\backslash[/math] in terms of [math]\cap[/math] and [math]\triangle[/math] (p. 21), showing that [math]R[/math] is a ring.
Levi Peterson
If [math]R[/math] is a class of subsets of [math]X[/math] s.t. for all [math]E,F\in R[/math] follows [math]E\cup F\in R[/math] and for all [math]E\in R[/math] follows [math]E'\in R[/math] then [math]R[/math] is called algebra. We can illustrate and better remember the relation between a ring and an algebra with the following concise proposition: [math]R[/math] is a ring and [math]X\in R[/math] ([math]R=\{X,...\}[/math]) if and only if [math]R[/math] is an algebra. If [math]R[/math] is a ring and [math]X \in R[/math] then for all [math]E\in R[/math], [math]E'=X\backslash E\in R[/math] hence we have an algebra. If [math]R[/math] is an algebra then for [math]E,F\in R[/math] we have [math]E\backslash F=E\cap F' \in R[/math] since [math]F'\in R[/math].
Brody Watson
[math]E_{*}= \cup_{n=1}^{\infty}\cap_{m=n}^{\infty}E_{m}[/math]: if [math]x \in E_{*}[/math] then [math]\exists k \in \mathbb{N}[/math] s.t. [math]x \in E_{k+i}[/math] for all [math]i \in \mathbb{N}[/math]. Then [math]x \in \cap_{m=k}^{\infty}E_{m}\subset \cup_{n=1}^{\infty}\cap_{m=n}^{\infty}E_{m} [/math]. If [math]x \in \cup_{n=1}^{\infty}\cap_{m=n}^{\infty}E_{m}[/math] then [math]\exists n \in \mathbb{N}[/math] s.t. [math]x \in \cap_{m=n}^{\infty}E_{m}[/math] hence [math]x \in E_{n+i}[/math] for all [math]i \in \mathbb{N}[/math], so [math]x \in E_{*}[/math]. [math]\cap_{n=1}^{\infty}\cup_{m=n}^{\infty}E_{m}=E^{*}[/math]: if [math]x \in \cap_{n=1}^{\infty}\cup_{m=n}^{\infty}E_{m}[/math] then [math]x \in \cup_{m=n}^{\infty}E_{m}[/math] for all [math]n \in \mathbb{N}[/math]. If [math]x \not \in E^{*}[/math] then [math]\exists k \in \mathbb{N}[/math] s.t. [math]x \not \in E_{k+i}[/math] for all [math]i \in \mathbb{N}[/math], but put [math]n=k+1[/math] and [math]x \in \cup_{m=k+1}^{\infty}E_{m}[/math] so [math]x \in E_{\tilde{m}}[/math] for [math]\tilde{m}>k[/math] which is a contradiction. If [math]x \in E^{*}[/math] it is true for all [math]n[/math] that [math]x \in \cup_{m=n}^{\infty}E_{m}[/math]: if [math]x \not \in\cup_{m=n}^{\infty}E_{m}[/math] then [math]\not \exists E_{\tilde{m}}[/math], [math]\tilde{m}\geq m[/math] s.t. [math]x \in E_{\tilde{m}}[/math]. [math]x \not \in E^{*}[/math] follows.
[math]\{E\}_{i=1}^{\infty}=\{E_{1},E_{2},E_{3},...\}=\{B,A,B,...\}[/math]. If [math]x \in E_{*}[/math] then [math]\exists i \in \mathbb{N}[/math] s.t. [math]x \in E_{i+n}[/math] for all [math]n \in \mathbb{N}[/math]. If [math]i+1[/math] even then [math]x \in E_{i+1}=A[/math], if not [math]x \in B[/math]. The reasoning is complete considering [math]E_{i+2}[/math]. If [math]x\in A\cap B[/math] then [math]x \in E_{i}[/math] for all [math]i \in \mathbb{N}[/math]. [math]x \in E^{*}[/math]: if [math]x \not \in A\cup B[/math] then [math]x \not \in E_{i}[/math] for all [math]i \in \mathbb{N}[/math] so [math]x \not \in E^{*}[/math]. If [math]x \in A\cup B[/math], whether [math]x \in A[/math] or [math]x \in B[/math], [math]x \in E^{*}[/math] follows.
p.18(6) [math](E_{*})'=\mbox{lim sup}_{n}E'_{n}[/math]: if [math]x \in (E^{*})'[/math] then [math]x \not \in E_{*}\subset E^{*}[/math] then [math]\exists k \in \mathbb{N}[/math] s.t. [math]x \not \in E_{k+i}[/math] for all [math]i \in \mathbb{N}[/math]. This is how [math]\mbox{lim sup}_{n}E'_{n}[/math] is defined. The proof of [math](E^{*})'=\mbox{lim inf}_{n}E'_{n}[/math] should be trivial at this point. [math]F\backslash E_{*}=\mbox{lim sup}_{n}F\backslash E_{n}[/math]: if [math]x \in F\backslash E_{*}[/math] then [math]x\in F[/math] and [math]x\not \in E_{*}\subset E^{*}[/math] so [math]\exists k \in \mathbb{N}[/math] s.t. [math]x \not \in E_{k+i}[/math] for all [math]i \in \mathbb{N}[/math] so [math]x \in \mbox{lim sup}_{n}E'_{n}[/math]. The proof of the second statement follows the same reasoning.
