View source for The ring of sets generated by a semi-ring is the set containing the semi-ring and all finite disjoint unions

==Statement==
Given a [[semi-ring of sets]], {{M|\mathcal{F} }}, the ''[[ring of sets|ring]]'' [[ring of sets generated by|generated by]]<ref group="Note">At the time of writing the ring generated by page doesn't exist so I write this note to clarify. 
* The ring of sets generated by an arbitrary collection of sets is the smallest ring of sets containing every element in the arbitrary collection</ref> {{M|\mathcal{F} }}, written {{M|R(\mathcal{F})}} is exactly the set {{M|\mathcal{F} }} and all finite [[union|unions]] of elements of {{M|\mathcal{F} }}{{rMIAMRLS}}.
{{Begin Notebox}}
:Recall the definitions of a [[ring of sets]], a [[semi-ring of sets]] and the [[ring of sets generated by]]:
{{Begin Notebox Content}}
{{:Ring of sets/Definition}}
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{{:Semi-ring of sets/Definition}}
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{{:Ring of sets generated by/Definition}}
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==Proof that {{M|\mathcal{F}_\cup}} is a [[ring of sets|ring]]==
{{Requires proof|msg=Mostly done but some trivial results do not have proof. Some of it was rushed, it would benefit from being read over|review=true}}
A ring is defined to be {{M|\cup}}-closed and {{M|\setminus}}-closed; however showing it is {{M|\setminus}}-closed is easier if we show already it is {{M|\cap}}-closed so we do this instead. {{Caution|It is not yet said on this website that rings are {{M|\cap}}-closed, but they are!}}
# {{M|\mathcal{F}_\cup }} is {{M|\cap}}-closed
#* Let {{M|S,T\in\mathcal{F}_\cup}} be given. We want to show that {{M|S\cap T\in\mathcal{F}_\cup}}
#** First note that this means {{M|1=S=\bigudot_{i=1}^{m_1}S_i}} and {{M|1=T=\bigudot_{j=1}^{m_2}T_j}} for some {{M|m_1,m_2\in\mathbb{N} }} and {{M|S_i,T_j\in\mathcal{F} }}. Now we want to show that {{M|1=\big(\bigudot_{i=1}^{m_1}S_i\big)\cap\big(\bigudot_{j=1}^{m_2}T_j\big)\in\mathcal{F}_\cup}}
#** Secondly, recall the [[distributivity of intersections across unions]], which states:
#*** {{M|1=A\cap(\bigcup_{i=1}^n B_i)=\bigcup_{i=1}^n(A\cap B_i)}}
#** Furthermore, if the {{M|B_i}} are [[pairwise disjoint]], then:
#*** {{M|1=A\cap(\bigudot_{i=1}^n B_i)=\bigudot_{i=1}^n(A\cap B_i)}}<ref group="Note">The proof of this is easy, as [[the intersection of sets is a subset of each set]] we see {{M|1=A\cap B_i\subseteq B_i}} for each {{M|i}}. As the {{M|B_i}} are pairwise disjoint, we see the {{M|A\cap B_i}} must be also</ref>
#** By immediate application of this proposition we see: {{M|1=\big(\bigudot_{i=1}^{m_1}S_i\big)\cap\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{j=1}^{m_2}\Big(\big(\bigudot_{i=1}^{m_1}S_i\big)\cap T_j\Big)}}
#** By [[commutivity of intersection]] we see {{M|1=\big(\bigudot_{i=1}^{m_1}S_i\big)\cap T_j=T_j\cap\big(\bigudot_{i=1}^{m_1}S_i\big)}}
#** By applying [[distributivity of intersections across unions]] again we see:
#*** {{M|1=\big(\bigudot_{i=1}^{m_1}S_i\big)\cap T_j=T_j\cap\big(\bigudot_{i=1}^{m_1}S_i\big)=\bigudot_{i=1}^{m_1}(T_j\cap S_i)}}
#** Then by substitution: {{M|1=\big(\bigudot_{i=1}^{m_1}S_i\big)\cap\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{j=1}^{m_2}\Big(\big(\bigudot_{i=1}^{m_1}S_i\big)\cap T_j\Big)=\bigudot_{j=1}^{m_2}\Big(\bigudot_{i=1}^{m_1}(T_j\cap S_i)\Big)}}
#** By [[comutivity of the union]] and [[commutivity of intersection]] we can write this in a slightly neater way:
#*** {{M|1=\bigudot_{i=1}^{m_1}\Big(\bigudot_{j=1}^{m_2}(S_i\cap T_j)\Big)}}
#** Notice that if {{M|A,B\in\mathcal{F} }} then as {{M|\mathcal{F} }} is a [[semi-ring of sets]], {{M|A\cap B\in\mathcal{F} }} and that we have {{M|m_1}} unions of {{M|m_2}} unions each, so there are {{M|m_1\times m_2}} unions of elements of {{M|\mathcal{F} }} in total. As {{M|m_1,m_2\in\mathbb{N} }} so is their product. 
#* Thus we have expressed {{M|S\cap T}} as an element of {{M|\mathcal{F}_{\cup} }} showing it is {{M|\cap}}-closed.
# {{M|\mathcal{F}_\cup}} is {{M|\setminus}}-closed.
#* Let {{M|S,T\in\mathcal{F}_\cup}} be given, we want to show that {{M|S-T\in\mathcal{F}_\cup}}
#** First note that this means {{M|1=S=\bigudot_{i=1}^{m_1}S_i}} and {{M|1=T=\bigudot_{j=1}^{m_2}T_j}} for some {{M|m_1,m_2\in\mathbb{N} }} and {{M|S_i,T_j\in\mathcal{F} }}; so we really want to show that {{M|1=\big(\bigudot_{i=1}^{m_1}S_i\big)-\big(\bigudot_{j=1}^{m_2}T_j\big)\in\mathcal{F}_\cup}}
#** Note that {{M|1=\big(\bigudot_{i=1}^{m_1}S_i\big)-\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{i=1}^{m_1}\big(S_i-\bigudot_{j=1}^{m_2}T_j\big)}}{{M|1==\bigudot_{i=1}^{m_1}\big(\bigcap_{j=1}^{m_2}(S_i-T_j)\big)}} {{Caution|I didn't just pull this out of thin air, I drew a venn diagram for the case {{M|1=m_1=m_2=1}} and worked from there, adding some notes on why would be good.}}
#*** As {{M|S_i,T_j\in\mathcal{F} }} we know there exists a finite [[sequence]] of elements of {{M|\mathcal{F} }}, say {{MSeq|W_{ijk}|k|1|n_{ij}|in=\mathcal{F} }} such that {{M|1=S_i-T_j=\bigudot_{k=1}^{n_{ij} }W_{ijk} }}.
#**** Define {{M|1=W_{ij}:=\bigudot_{k=1}^{n_{ij} }W_{ijk} = S_i-T_j}}, then {{M|W_{ij}\in\mathcal{F}_\cup}}
#** Thus: {{M|1=\big(\bigudot_{i=1}^{m_1}S_i\big)-\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{i=1}^{m_1}\big(\bigcap_{j=1}^{m_2}(S_i-T_j)\big)}}{{M|1==\bigudot_{i=1}^{m_1}\big(\bigcap_{j=1}^{m_2}W_{ij}\big)}} (really we should say {{M|1=\bigudot_{i=1}^{m_1}\big(\bigcap_{j=1}^{m_2}\big(\bigudot_{k=1}^{n_{ij} }W_{ijk}\big)\big)}} but it nicer with just {{M|W_{ij}\in\mathcal{F}_\cup}})
#** We have already shown that {{M|\mathcal{F}_\cup}} is {{M|\cap}}-closed, thus {{M|1=\bigcap_{j=1}^{m_2}W_{ij}\in\mathcal{F}_\cup}}
#*** There exists {{MSeq|V_{il}|l|1|n_i|in=\mathcal{F} }} (notice the {{M|V_{il}\in\mathcal{F} }} not {{M|\mathcal{F}_\cup}}) such that {{M|1=\bigudot_{l=1}^{n_i}V_{il}=\bigcap_{j=1}^{m_2}W_{ij} }}
#** So: {{M|1=\big(\bigudot_{i=1}^{m_1}S_i\big)-\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{i=1}^{m_1}\big(\bigcap_{j=1}^{m_2}W_{ij}\big)}}{{M|1==\bigudot_{i=1}^{m_1}\big(\bigudot_{l=1}^{n_i}V_{il}\big)}}
#** This is a finite disjoint union of elements of {{M|\mathcal{F} }} with {{M|1=\sum_{i=1}^{m_1}n_i}} terms, thus {{M|1=S-T=\big(\bigudot_{i=1}^{m_1}S_i\big)-\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{i=1}^{m_1}\big(\bigudot_{l=1}^{n_i}V_{il}\big)\in\mathcal{F}_\cup}}
#* Since {{M|S}} and {{M|T}} were arbitrary we have completed the proof that {{M|\mathcal{F}_\cup}} is {{M|\setminus}}-closed
# {{M|\mathcal{F}_\cup}} is {{M|\cup}}-closed
#* Let {{M|S,T\in\mathcal{F}_\cup}} be given
#** Note {{M|1=S\cup T=(S-T)\udot(S\cap T)\udot(T-S)}} and each term of this union is in {{M|\mathcal{F}_\cup}} 
#* This completes the proof.
==Proof that {{M|\mathcal{F}_\cup}} is the ring generated by {{M|\mathcal{F} }}==
{{Requires proof|grade=A|msg=It's quite easy, just assume that {{M|\mathcal{F}_\cup}} isn't the ring generated by {{M|\mathcal{F} }}, that means the generated ring is a proper subset of {{M|\mathcal{F}_\cup}}, so there is an element in {{M|\mathcal{F}_\cup}} not in{{M|R(\mathcal{F})}}. That element must be the disjoint union of a finite number of elements from {{M|\mathcal{F} }}, that violates that {{M|R(\mathcal{F})}} is closed under unions! As the union of these elements is not in the generated ring!}}
==Purpose==
The main use of this theorem is to aid with extending a [[pre-measure on a semi-ring]] to a [[pre-measure]]. Defining a pre-measure on a semi-ring is often very easy, where as defining one on a ring is very tedious.
==See also==
* [[Types of set algebras]]
==Notes==
<references group="Note"/>
==References==
<references/>
{{Measure theory navbox|plain}}
{{Theorem Of|Measure Theory}}