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Does the class of all finite unions of closed-open intervals on $\mathbb{R}$ form a ring on sets?

By a closed-open interval , I mean an interval of the form $[x,y)$

A ring of sets is a non-empty class of sets that is closed under symmetric difference of any pair of sets of the class and under intersection of any pair of sets of the class.

For me, This is untrue. Since $[0,1)$ and $[2,3)$ are in the class but their intersection is not in the class ( since the empty set is not a closed-open interval, is it? )

Am I right? I think that if we added the empty set to the class then the new class forms a boolean algebra of sets.

rschwieb
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FNH
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    $\varnothing={x\in\Bbb R\mid x<x}={x\in\Bbb R\mid 0\leq x<0}=[0,0)$. – Asaf Karagila Sep 19 '14 at 10:48
  • @AsafKaragila, Good trick, but in the book (Intro to topology and modern analysis by Simmons), $[a,b)$ reuqires that $a<b$ so $[0,0)$ is not a closed-open interval according to the definition. – FNH Sep 19 '14 at 11:27
  • Yes, a bit confusing. When it says "all finite unions of" it is supposed to include in particular the union of zero of them. – GEdgar Sep 19 '14 at 14:01

2 Answers2

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$\emptyset=\cup\{[a_i,b_i)|i\in\emptyset\}$ shows that $\emptyset$ is a finite union of closed-open intervals on $\mathbb R$, since the empty set serving as index-set is a finite set.

edit (to make things less fuzzy)

In general if $\mathcal C$ is some set of sets (above the set that contains all half-open intervals in $\mathbb R$) then the class that consists of all (finite) unions of sets in $\mathcal C$ contains the empty set as element. These unions are exactly the sets that can be written as $\cup\mathcal V$ where $\mathcal V$ is a (finite) subset of $\mathcal C$.

Here $a\in\cup\mathcal V\iff a\in V\text{ for some }V\in\mathcal V$.

One of these subsets $\mathcal V$ is the empty set and clearly $\cup\emptyset=\emptyset$.

drhab
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  • But there's nothing in the question indicating that a ring of sets is closed under unions. – Asaf Karagila Sep 19 '14 at 13:01
  • @AsafKaragila The question was: 'does the class of all finite unions of closed-open intervals on $\mathbb R$ form a ring on sets?'. The OP says that he thinks not, because in that case this class must also have $\emptyset$ as element. He thinks that that is not the case. My answer shows him that he is wrong in thinking like that. – drhab Sep 19 '14 at 13:09
  • Oh, d'oh. I missed that part when I re-read it after you answered! :-) – Asaf Karagila Sep 19 '14 at 13:10
  • hmmm , This is good. but I find that ${[a_i,b_i)| i\in \emptyset}$ is fuzzy. If $i\in \emptyset$ then clearly $[a_i,b_i) = \emptyset$. Why not add the empty set to the class as a new element instead of writing things like that? Isn't that better? – FNH Sep 19 '14 at 13:38
  • You are saying '$i\in\emptyset\rightarrow[a_i,b_i)=\emptyset$' or equivalently $i\notin\emptyset\vee[a_i,b_i)=\emptyset$. This is a true statement, but note that $i\notin\emptyset\vee[a_i,b_i)\neq\emptyset$ is also a true statement. This simply because $i\notin\emptyset$ is true. The empty set is - as a finite union - an element of the class mentioned (finite unions of...). So there is no need to add it anymore. You call it good and 'fuzzy'. If the fog has cleared a bit then you can start enjoying the beauty of it :). – drhab Sep 19 '14 at 18:22
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This follows from the argument in the book (Topology and Modern Analysis) on page 12 talking about finite intersections and unions of classes of sets. The class of all finite unions of closed-open intervals is closed under unions. Since the empty class is necessarily a finite subclass its union the empty set belongs to the class.

Ramesh Kadambi
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