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(Created page with "<noinclude> ==Section B== ===Problem B1=== </noinclude>====Part i)==== Suppose that {{M|\mathcal{A}_n}} are algebras of sets satisfying {{M|\mathcal{A}_n\subset \mathcal{...")
 
(Adding final part of question)
 
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===Problem B1===
 
===Problem B1===
 
</noinclude>====Part i)====
 
</noinclude>====Part i)====
−
Suppose that {{M|\mathcal{A}_n}} are [[algebras of sets]] satisfying {{M|\mathcal{A}_n\subset  \mathcal{A}_{n+1} }}. Show that {{M|1=\bigcup_{n\in\mathbb{N} }A_n}} is an algebra.
+
Suppose that {{M|\mathcal{A}_n}} are [[algebras of sets]] satisfying {{M|\mathcal{A}_n\subset  \mathcal{A}_{n+1} }}. Show that {{M|1=\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} is an algebra.
 +
* {{Caution|Is {{M|\subset}} or {{M|\subseteq}} desired?}}
 
=====Solution=====
 
=====Solution=====
 +
# Closed under [[complementation]]: {{M|\forall A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[A^\complement\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n]}}
 +
#* Let {{M|A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} be given.
 +
#** By definition of [[union]]: {{M|\big[A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n\big]\iff\big[\exists i\in\mathbb{N}[A\in\mathcal{A}_i]\big]}}
 +
#*** As {{M|\mathcal{A}_i}} is an algebra of sets itself:
 +
#**** {{M|A^\complement\in\mathcal{A}_i}}
 +
#*** Thus {{M|A^\complement\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}}
 +
#* Since {{M|A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} was arbitrary, we have shown this for all such {{M|A}}, thus {{M|\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} is closed under complementation.
 +
# Closed under [[union]]: {{M|1=\forall A,B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[A\cup B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n]}}
 +
#* Let {{M|A,B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} be given.
 +
#** By definition of [[union]] we see:
 +
#**# {{M|\exists i\in\mathbb{N}[A\in\mathcal{A}_i]}} and
 +
#**# {{M|\exists j\in\mathbb{N}[B\in\mathcal{A}_j]}}
 +
#*** Define {{M|1=k:=\text{Max}(\{i,j\})}}
 +
#**** Now {{M|\mathcal{A}_i\subseteq\mathcal{A}_k}} and {{M|\mathcal{A}_j\subseteq\mathcal{A}_k}} (at least one of these will be strict equality, it matters not which)
 +
#**** Thus {{M|A,B\in\mathcal{A}_k}}
 +
#**** As {{M|\mathcal{A}_k}} is an algebra of sets:
 +
#***** {{M|\forall C,D\in\mathcal{A}_k[C\cup D\in\mathcal{A}_k]}}
 +
#**** Thus {{M|A\cup B\in\mathcal{A}_k}}
 +
#**** So {{M|A\cup B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} (explicitly, {{M|\exists h\in\mathbb{N}[A\cup B\in\mathcal{A}_h]}} - namely choosing {{M|h}} to be {{M|k}} as we have defined it, and we have this {{iff}} {{M|A\cup B}} is in the union, by the definition of [[union]])
 +
#* Since {{M|A,B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} were arbitrary we have shown this for all such {{M|A}} and {{M|B}}. As required.
 
====Part ii)====
 
====Part ii)====
−
Check that if the {{M|A_n}} are all [[sigma-algebras]] that their union need not be an algebra.
+
Check that if the {{M|\mathcal{A}_n}} are all [[sigma-algebras]] that their union need not be a sigma-algebra.
  
 
Is a countable union of sigma-algebras (whether [[monotonic]] or not) an algebra?
 
Is a countable union of sigma-algebras (whether [[monotonic]] or not) an algebra?
Line 13: Line 34:
  
 
Check that if {{M|\mathcal{B}_1}} and {{M|\mathcal{B}_2}} are [[sigma-algebras]] that their union need not be an [[algebra of sets]]
 
Check that if {{M|\mathcal{B}_1}} and {{M|\mathcal{B}_2}} are [[sigma-algebras]] that their union need not be an [[algebra of sets]]
 +
=====Solution=====
 +
Suppose all the {{M|\mathcal{A}_i}}s are [[sigma-algebras]] now, and suppose that {{M|\mathcal{A}_{n}\subset\mathcal{A}_{n+1} }} still holds. We wish to show that their [[union]], {{M|\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} is ''not'' a [[sigma-algebra]].
 +
* Our first guess will be that the {{M|\sigma}}-{{M|\cup}}-closed property does not hold. That is:
 +
* {{M|\neg\big[\forall(A_n)_{n\in\mathbb{N} }\subseteq\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[\bigcup_{n\in\mathbb{N} }A_n\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n]\big]}} which is equivalent to (in that {{M|\iff}} or {{iff}}):
 +
** {{M|1=\exists(A_n)_{n\in\mathbb{N} }\subseteq\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n]}} ({{Caution|Things look very similar here! Read with care!}})
 +
* As before: {{M|1=\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n\iff\neg(\exists k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\in\mathcal{A}_k])}} {{M|\iff}} {{M|1=\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k]}}
 +
** So we need to find a {{MSeq|A_n|in=\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} such that {{M|1=\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k]}}
 +
*** As {{M|\mathcal{A}_n\subset\mathcal{A}_{n+1} }} we know {{M|\exists A\in\mathcal{A}_{n+1}[A\notin\mathcal{A}_n]}}, as they're [[proper subsets]] of each other.
 +
**** Thus, define {{M|1=A_n:=X_n}} where {{M|X_n\in\mathcal{A}_{n+1} }} and {{M|X_n\notin\mathcal{A}_n}} (which we have just shown to exist).
 +
***** Now we must prove {{M|1=\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k]}}
 +
****** {{Caution|Not convinced this is all "okay" - I also may have spotted why we're given the hint}}
 +
It is easier to come up with an instance of {{M|\mathcal{A}_n\subset\mathcal{A}_{n+1} }} and prove this for that instance than do it in general.
 +
======Using hint as instance======
 +
Let {{M|1=\mathcal{A}_n:=\mathcal{P}(\{1,2,\ldots,n\}\subset\mathbb{N})}}. Let {{M|1=\mathcal{A}:=\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} for convenience. We claim {{M|\mathcal{A} }} is not a [[sigma-algebra]] and we suggest this based on the {{M|\sigma}}-{{m|\cup}}-closed property. That is we claim:
 +
* {{M|\neg(\overbrace{\forall(A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}[\bigcup_{n\in\mathbb{N} }A_n\in\mathcal{A}]}^{\sigma\text{-}\cup\text{-closed} })}} {{M|1=\iff\exists(A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}]}}
 +
** Notice: {{M|1=\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A} }} {{M|1=\iff\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} {{M|1=\iff\neg(\exists j\in\mathbb{N}[\bigcup_{n\notin\mathbb{N} }A_n\in\mathcal{A}_j])}} {{M|1=\iff\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k]}}
 +
* Combining these we see we want to show: {{M|1=\exists (A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}[\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j]]}} or just: {{M|1=\exists (A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j]}}
 +
Proof: {{M|1=\exists (A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j]}}
 +
* Let {{M|1=\mathcal{A}_n:=\mathcal{P}(\{1,\ldots,n\}\subset\mathbb{N})}} - where {{M|\mathcal{P} }} denotes the [[power set]] of the finite set {{M|\{1,\ldots,n\} }} - which is a portion of {{M|\mathbb{N} }} - the naturals.
 +
* We see that we have {{M|1=\mathcal{A}_n\subseteq\mathcal{A}_{n+1} }} or more specifically in fact: {{M|1=\mathcal{A}_n\subset\mathcal{A}_{n+1} }}.
 +
** Define {{MSeq|A_n|in=\mathcal{A} }} as {{M|1=A_n:=\{n\} }}
 +
*** Clearly {{M|1=A_n\in\mathcal{A}_n}} for each {{M|n}}.
 +
*** Let {{M|j\in\mathbb{N} }} be given (so {{M|j}} is arbitrary and we show the following for all {{M|j}})
 +
**** We must show {{M|1=\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j}}
 +
**** '''Lemma: ''' {{M|1=\forall a\forall b\forall c[(a\in b\wedge \forall U\in c[a\notin U])\implies b\notin c]}}
 +
***** Let {{M|a,\ b}} and {{M|c}} be given.
 +
*****# Suppose {{M|(a\in b\wedge \forall U\in c[a\notin U])}} is false, then by the nature of [[logical implication]] we do not care about the truth or falsity of the RHS and we're done
 +
*****# Suppose {{M|(a\in b\wedge \forall U\in c[a\notin U])}} is true. We must show that in this case {{M|b\notin c}}
 +
*****#* Suppose {{M|b\in c}}. Then {{M|1=\exists U\in c[a\in U]}} - namely {{M|1=U:=b}} (as by hypothesis, {{M|a\in b}})
 +
*****#** But this is the [[negation]] of {{M|1=\forall U\in c[a\notin U]}}! Contradicting the hypothesis
 +
*****#* Thus we cannot have {{M|b\in c}} if {{M|(a\in b\wedge \forall U\in c[a\notin U])}} is true.
 +
*****#* So we must have {{M|b\notin c}} - as required.
 +
***** This completes the proof of the lemma.
 +
**** Notice {{M|1=j+1\in\bigcup_{n\in\mathbb{N} }A_n}} (trivial: {{M|1=j+1\in\mathbb{N} }} and {{M|1=A_{j+1}:=\{j+1\} }}, so {{M|j+1\in A_{j+1}\subseteq\bigcup_{n\in\mathbb{N} }A_n}}, thus {{M|1=j+1\in\bigcup_{n\in\mathbb{N} }A_n}})
 +
**** Notice also that {{M|1=\forall U\in \mathcal{A}_j[j+1\notin U]}}. Proof:
 +
***** Let {{M|U\in\mathcal{A}_j }} be given.
 +
****** Then {{M|U\subseteq\{1,\ldots,j\} }}, by the [[implies-subset relation]], {{M|1=x\in U\implies x\in \{1,\ldots,j\} }}.
 +
****** Suppose {{M|j+1\in U}}, then {{M|j+1\in\{1,\ldots,j\} }} which is a contradiction! Thus we cannot have {{M|j+1\in U}}
 +
****** We must have {{M|j+1\notin U}}
 +
**** By the lemma: {{M|1=(a\in b\wedge \forall U\in c[a\notin U])\implies b\notin c}}, in this case:
 +
**** {{M|1=(j+1\in \bigcup_{n\in\mathbb{N} }A_n\wedge \forall U\in \mathcal{A}_j[j+1\notin U])\implies \bigcup_{n\in\mathbb{N} }A_n\notin \mathcal{A}_j}}
 +
*** Thus we see {{M|\bigcup_{n\in\mathbb{N} }A_n\notin \mathcal{A}_j}} is true for all {{M|j\in\mathbb{N} }} (since it was arbitrary)
 +
** For this choice of {{M|(A_n)_{n\in\mathbb{N} } }}
 +
Thus {{M|\bigcup_{n\in\mathbb{N} }\mathcal{A}_n}} ''cannot'' be a [[sigma-algebra]] as it isn't closed under countable union (of pairwise disjoint sets as it happens)
 +
====Part iii)====
 +
There isn't really a part 3 but the last part of part 2 is:
 +
* Is a countable union of [[sigma-algebra|{{sigma|algebras}}]] (whether monotone or not) an [[algebra of sets]]?
 +
Then there's
 +
* Check that if {{M|\mathcal{B}_1}} and {{M|\mathcal{B}_2}} are {{sigma|algebras}} that their union need not be an algebra.
 +
If we do the second one first, we have shown the first, as {{M|1=(\mathcal{B}_1,\mathcal{B}_2,\{\emptyset\},\{\emptyset\},\ldots)}} is a countable collection of sigma-algebras containing {{M|1=\mathcal{B}_1\cup\mathcal{B}_2}} and thus its union cannot be a sigma-algebra.
 +
=====Solution=====
 +
Let {{M|\mathcal{B}_1}} and {{M|\mathcal{B}_2}} be {{sigma|algebras}}. Is {{M|1=\mathcal{B}_1\cup\mathcal{B}_2}} an algebra?
 +
 
<noinclude>
 
<noinclude>
 
==Notes==
 
==Notes==

Latest revision as of 14:09, 23 October 2016

Section B

Problem B1

Part i)

Suppose that [ilmath]\mathcal{A}_n[/ilmath] are algebras of sets satisfying [ilmath]\mathcal{A}_n\subset \mathcal{A}_{n+1} [/ilmath]. Show that [ilmath]\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] is an algebra.

  • Caution:Is [ilmath]\subset[/ilmath] or [ilmath]\subseteq[/ilmath] desired?
Solution
  1. Closed under complementation: [ilmath]\forall A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[A^\complement\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n][/ilmath]
    • Let [ilmath]A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] be given.
      • By definition of union: [ilmath]\big[A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n\big]\iff\big[\exists i\in\mathbb{N}[A\in\mathcal{A}_i]\big][/ilmath]
        • As [ilmath]\mathcal{A}_i[/ilmath] is an algebra of sets itself:
          • [ilmath]A^\complement\in\mathcal{A}_i[/ilmath]
        • Thus [ilmath]A^\complement\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath]
    • Since [ilmath]A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] was arbitrary, we have shown this for all such [ilmath]A[/ilmath], thus [ilmath]\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] is closed under complementation.
  2. Closed under union: [ilmath]\forall A,B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[A\cup B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n][/ilmath]
    • Let [ilmath]A,B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] be given.
      • By definition of union we see:
        1. [ilmath]\exists i\in\mathbb{N}[A\in\mathcal{A}_i][/ilmath] and
        2. [ilmath]\exists j\in\mathbb{N}[B\in\mathcal{A}_j][/ilmath]
        • Define [ilmath]k:=\text{Max}(\{i,j\})[/ilmath]
          • Now [ilmath]\mathcal{A}_i\subseteq\mathcal{A}_k[/ilmath] and [ilmath]\mathcal{A}_j\subseteq\mathcal{A}_k[/ilmath] (at least one of these will be strict equality, it matters not which)
          • Thus [ilmath]A,B\in\mathcal{A}_k[/ilmath]
          • As [ilmath]\mathcal{A}_k[/ilmath] is an algebra of sets:
            • [ilmath]\forall C,D\in\mathcal{A}_k[C\cup D\in\mathcal{A}_k][/ilmath]
          • Thus [ilmath]A\cup B\in\mathcal{A}_k[/ilmath]
          • So [ilmath]A\cup B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] (explicitly, [ilmath]\exists h\in\mathbb{N}[A\cup B\in\mathcal{A}_h][/ilmath] - namely choosing [ilmath]h[/ilmath] to be [ilmath]k[/ilmath] as we have defined it, and we have this if and only if [ilmath]A\cup B[/ilmath] is in the union, by the definition of union)
    • Since [ilmath]A,B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] were arbitrary we have shown this for all such [ilmath]A[/ilmath] and [ilmath]B[/ilmath]. As required.

Part ii)

Check that if the [ilmath]\mathcal{A}_n[/ilmath] are all sigma-algebras that their union need not be a sigma-algebra.

Is a countable union of sigma-algebras (whether monotonic or not) an algebra?

Hint: Try considering the set of all positive integers, [ilmath]\mathbb{Z}_{\ge 1} [/ilmath] with its sigma-algebras [ilmath]\mathcal{A}_n:=\sigma(\mathcal{C}_n)[/ilmath] where [ilmath]\mathcal{C}_n:=\mathcal{P}(\{1,2,\ldots,n\})[/ilmath] where [ilmath]\{1,2,\ldots,n\}\subset\mathbb{N}[/ilmath] and [ilmath]\mathcal{P} [/ilmath] denotes the power set

Check that if [ilmath]\mathcal{B}_1[/ilmath] and [ilmath]\mathcal{B}_2[/ilmath] are sigma-algebras that their union need not be an algebra of sets

Solution

Suppose all the [ilmath]\mathcal{A}_i[/ilmath]s are sigma-algebras now, and suppose that [ilmath]\mathcal{A}_{n}\subset\mathcal{A}_{n+1} [/ilmath] still holds. We wish to show that their union, [ilmath]\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] is not a sigma-algebra.

  • Our first guess will be that the [ilmath]\sigma[/ilmath]-[ilmath]\cup[/ilmath]-closed property does not hold. That is:
  • [ilmath]\neg\big[\forall(A_n)_{n\in\mathbb{N} }\subseteq\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[\bigcup_{n\in\mathbb{N} }A_n\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n]\big][/ilmath] which is equivalent to (in that [ilmath]\iff[/ilmath] or if and only if):
    • [ilmath]\exists(A_n)_{n\in\mathbb{N} }\subseteq\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n][/ilmath] (Caution:Things look very similar here! Read with care!)
  • As before: [ilmath]\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n\iff\neg(\exists k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\in\mathcal{A}_k])[/ilmath] [ilmath]\iff[/ilmath] [ilmath]\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k][/ilmath]
    • So we need to find a [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \bigcup_{n\in\mathbb{N} }\mathcal{A}_n [/ilmath] such that [ilmath]\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k][/ilmath]
      • As [ilmath]\mathcal{A}_n\subset\mathcal{A}_{n+1} [/ilmath] we know [ilmath]\exists A\in\mathcal{A}_{n+1}[A\notin\mathcal{A}_n][/ilmath], as they're proper subsets of each other.
        • Thus, define [ilmath]A_n:=X_n[/ilmath] where [ilmath]X_n\in\mathcal{A}_{n+1} [/ilmath] and [ilmath]X_n\notin\mathcal{A}_n[/ilmath] (which we have just shown to exist).
          • Now we must prove [ilmath]\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k][/ilmath]
            • Caution:Not convinced this is all "okay" - I also may have spotted why we're given the hint

It is easier to come up with an instance of [ilmath]\mathcal{A}_n\subset\mathcal{A}_{n+1} [/ilmath] and prove this for that instance than do it in general.

Using hint as instance

Let [ilmath]\mathcal{A}_n:=\mathcal{P}(\{1,2,\ldots,n\}\subset\mathbb{N})[/ilmath]. Let [ilmath]\mathcal{A}:=\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] for convenience. We claim [ilmath]\mathcal{A} [/ilmath] is not a sigma-algebra and we suggest this based on the [ilmath]\sigma[/ilmath]-[ilmath]\cup[/ilmath]-closed property. That is we claim:

  • [ilmath]\neg(\overbrace{\forall(A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}[\bigcup_{n\in\mathbb{N} }A_n\in\mathcal{A}]}^{\sigma\text{-}\cup\text{-closed} })[/ilmath] [ilmath]\iff\exists(A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}][/ilmath]
    • Notice: [ilmath]\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}[/ilmath] [ilmath]\iff\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] [ilmath]\iff\neg(\exists j\in\mathbb{N}[\bigcup_{n\notin\mathbb{N} }A_n\in\mathcal{A}_j])[/ilmath] [ilmath]\iff\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k][/ilmath]
  • Combining these we see we want to show: [ilmath]\exists (A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}[\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j]][/ilmath] or just: [ilmath]\exists (A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j][/ilmath]

Proof: [ilmath]\exists (A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j][/ilmath]

  • Let [ilmath]\mathcal{A}_n:=\mathcal{P}(\{1,\ldots,n\}\subset\mathbb{N})[/ilmath] - where [ilmath]\mathcal{P} [/ilmath] denotes the power set of the finite set [ilmath]\{1,\ldots,n\} [/ilmath] - which is a portion of [ilmath]\mathbb{N} [/ilmath] - the naturals.
  • We see that we have [ilmath]\mathcal{A}_n\subseteq\mathcal{A}_{n+1}[/ilmath] or more specifically in fact: [ilmath]\mathcal{A}_n\subset\mathcal{A}_{n+1}[/ilmath].
    • Define [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{A} [/ilmath] as [ilmath]A_n:=\{n\}[/ilmath]
      • Clearly [ilmath]A_n\in\mathcal{A}_n[/ilmath] for each [ilmath]n[/ilmath].
      • Let [ilmath]j\in\mathbb{N} [/ilmath] be given (so [ilmath]j[/ilmath] is arbitrary and we show the following for all [ilmath]j[/ilmath])
        • We must show [ilmath]\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j[/ilmath]
        • Lemma: [ilmath]\forall a\forall b\forall c[(a\in b\wedge \forall U\in c[a\notin U])\implies b\notin c][/ilmath]
          • Let [ilmath]a,\ b[/ilmath] and [ilmath]c[/ilmath] be given.
            1. Suppose [ilmath](a\in b\wedge \forall U\in c[a\notin U])[/ilmath] is false, then by the nature of logical implication we do not care about the truth or falsity of the RHS and we're done
            2. Suppose [ilmath](a\in b\wedge \forall U\in c[a\notin U])[/ilmath] is true. We must show that in this case [ilmath]b\notin c[/ilmath]
              • Suppose [ilmath]b\in c[/ilmath]. Then [ilmath]\exists U\in c[a\in U][/ilmath] - namely [ilmath]U:=b[/ilmath] (as by hypothesis, [ilmath]a\in b[/ilmath])
                • But this is the negation of [ilmath]\forall U\in c[a\notin U][/ilmath]! Contradicting the hypothesis
              • Thus we cannot have [ilmath]b\in c[/ilmath] if [ilmath](a\in b\wedge \forall U\in c[a\notin U])[/ilmath] is true.
              • So we must have [ilmath]b\notin c[/ilmath] - as required.
          • This completes the proof of the lemma.
        • Notice [ilmath]j+1\in\bigcup_{n\in\mathbb{N} }A_n[/ilmath] (trivial: [ilmath]j+1\in\mathbb{N}[/ilmath] and [ilmath]A_{j+1}:=\{j+1\}[/ilmath], so [ilmath]j+1\in A_{j+1}\subseteq\bigcup_{n\in\mathbb{N} }A_n[/ilmath], thus [ilmath]j+1\in\bigcup_{n\in\mathbb{N} }A_n[/ilmath])
        • Notice also that [ilmath]\forall U\in \mathcal{A}_j[j+1\notin U][/ilmath]. Proof:
          • Let [ilmath]U\in\mathcal{A}_j [/ilmath] be given.
            • Then [ilmath]U\subseteq\{1,\ldots,j\} [/ilmath], by the implies-subset relation, [ilmath]x\in U\implies x\in \{1,\ldots,j\}[/ilmath].
            • Suppose [ilmath]j+1\in U[/ilmath], then [ilmath]j+1\in\{1,\ldots,j\} [/ilmath] which is a contradiction! Thus we cannot have [ilmath]j+1\in U[/ilmath]
            • We must have [ilmath]j+1\notin U[/ilmath]
        • By the lemma: [ilmath](a\in b\wedge \forall U\in c[a\notin U])\implies b\notin c[/ilmath], in this case:
        • [ilmath](j+1\in \bigcup_{n\in\mathbb{N} }A_n\wedge \forall U\in \mathcal{A}_j[j+1\notin U])\implies \bigcup_{n\in\mathbb{N} }A_n\notin \mathcal{A}_j[/ilmath]
      • Thus we see [ilmath]\bigcup_{n\in\mathbb{N} }A_n\notin \mathcal{A}_j[/ilmath] is true for all [ilmath]j\in\mathbb{N} [/ilmath] (since it was arbitrary)
    • For this choice of [ilmath](A_n)_{n\in\mathbb{N} } [/ilmath]

Thus [ilmath]\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] cannot be a sigma-algebra as it isn't closed under countable union (of pairwise disjoint sets as it happens)

Part iii)

There isn't really a part 3 but the last part of part 2 is:

Then there's

  • Check that if [ilmath]\mathcal{B}_1[/ilmath] and [ilmath]\mathcal{B}_2[/ilmath] are [ilmath]\sigma[/ilmath]-algebras that their union need not be an algebra.

If we do the second one first, we have shown the first, as [ilmath](\mathcal{B}_1,\mathcal{B}_2,\{\emptyset\},\{\emptyset\},\ldots)[/ilmath] is a countable collection of sigma-algebras containing [ilmath]\mathcal{B}_1\cup\mathcal{B}_2[/ilmath] and thus its union cannot be a sigma-algebra.

Solution

Let [ilmath]\mathcal{B}_1[/ilmath] and [ilmath]\mathcal{B}_2[/ilmath] be [ilmath]\sigma[/ilmath]-algebras. Is [ilmath]\mathcal{B}_1\cup\mathcal{B}_2[/ilmath] an algebra?


Notes

References


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