Pag.16. Let [math]\{E_{i}\}[/math] be a sequence of subsets

>>8278063
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.

>>8278064
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].

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.

>>8278080

It would be nice of you...

>>8278063
Practicing your Latex moron?

>>8278063
>>8278064
>>8278072
Fuck off

>>8278080
One of my profs sent me a digital copy of a Measure Theory textbook designed for undergrads. It'd be better than >>8278080 trying to dumb it down for you. Search for Marek Capinski and Ekkehard Kopp Measure, Integral and Probability.

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I have this book, it's bretty good (but kinda advanced, couldn't read it until after i graduated).

>>8278136
>It'd be better than >>8278080 (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.

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>>8278124
>>8278131
(You)

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].

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>>8278246
Don't know what's wrong with that. The preview is alright.

>>8278162
That is one cute kitty.

>>8278159
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.

>>8278251
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

[math]C(n,k)=n!/[k!(n-k)!]
=4\cdot3\cdot2\cdot1/[2\cdot1\cdot(2\cdot1)]
=6[/math]

so

[math]E_{1}\cup E_{2}=E_{1}\cup E_{2}[/math],
[math]E_{1}\cup (E_{1}\cup E_{2})=E_{1}\cup E_{1}\cup E_{2}=E_{1}\cup E_{2}[/math],
[math]E_{1}\cup \emptyset=E_{1}[/math].
[math]E_{2}\cup (E_{1}\cup E_{2})=E_{1}\cup E_{2}[/math],
[math]E_{2}\cup \emptyset=E_{2}[/math].
[math](E_{1}\cup E_{2})\cup \emptyset =E_{1}\cup E_{2}[/math].

All of them belong to [math]R[/math].

>>8278162
I like this блин кoт :DDDDD

>>8278278
[Refer to: https://en.wikipedia.org/wiki/Combination for \(C(n,k)\).]

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]:

[math]E_{1}\backslash E_{2}=E_{1}\cap E_{2}'=E_{1}[/math],
[math]E_{1}\backslash(E_{1}\cup E_{2})=E_{1}\cap (E_{1}\cup E_{2})'=E_{1}\cap(E_{1}'\cap E_{2}')=E_{1}\cap E_{1}\cap E_{2}=\emptyset[/math], ....

All this pedantry is unnecessary but also nice because it teaches you to think in an ordered manner.

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>>8278290
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.

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>>8278865
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].

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>>8278895
[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.

>>8279023
This is p.18(2).
Here is the URL of sharelatex: https://www.sharelatex.com/project/57b10e0cd256288d6ddb97c9.

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>>8279052
p.18(4)

[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.

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>>8279259