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Let $O = [0, \infty)$ and $F_1$ the class of all intervals of the type $[a, b)$ or $[a, \infty)$, where $0 \le a < b < \infty$. Let $F_2$ be the class of all finite disjoint unions of intervals of $F_1$. Show that $F_1$ is not a field and $F_2$ is a field but not a sigma field.

What does "finite disjoint unions of intervals" mean in this context ? Does that mean $F_2$ is empty should the word disjoint be in there ?

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Finite disjoint unions of intervals can be translated unions of finitely many pairwise disjoint intervals. – Brian M. Scott Feb 11 '12 at 7:15
@BrianM.Scott thanks so much for that i also assumed the same but as per the questions i was under the impression there are no pairwise disjoint intervals in F1 as the all had a ? I guess the question only makes sense as per Yuval Filmus's take on it otherwise it is ill-formed ? – Comic Book Guy Feb 11 '12 at 7:20
Yuval’s reading is really the only possible one; $[a,b)$ and $[a,\infty)$ in the definition of $F_1$ are used to describe the types of intervals that included in that collection; there is no implication that $a$ or $b$ is a fixed constant $-$ quite the opposite, actually. – Brian M. Scott Feb 11 '12 at 7:53
up vote 2 down vote accepted

$F_2$ is the set of all finite disjoint unions of intervals, that is $$[a_1,b_1) \cup \cdots \cup [a_n,b_n),$$ where $b_i \leq a_{i+1}$ and $b_n$ could be $\infty$. The restriction $b_i \leq a_{i+1}$ ensures that the intervals are disjoint. Some examples: $$ [1,2) \cup [3,4), \quad [5,6), \quad \emptyset, \quad [7,8) \cup [8,9) \cup [9,\infty). $$

But in fact, $F_2$ is also the set of all finite (unrestricted) unions of intervals. They don't have to be disjoint. Do you see why?

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i get the finiteness part and the union , not so sure what u meant to imply by unrestricted. Also as per the question i typed are n't all the intervals sharing the minimal value of a and the upper values b is the one changing such that the condition 0 <= a < b < inf held. – Comic Book Guy Feb 11 '12 at 7:18
@Hardy: No, the $a$ is not a fixed constant. $[a,b)$ and $[a,\infty)$ are just prototypes illustrating the two types of intervals included in $F_1$. – Brian M. Scott Feb 11 '12 at 7:54
Thanks guys that was very help full. If u do n't mind could u please possibly validate my answer too based on this reading F1 is not a field since complements and Finite intersections are not closed in F1. As per F2 it has finite unions but what about null set and O and finite intersections ? – Comic Book Guy Feb 11 '12 at 8:07
@Hardy indeed $F_1$ fails to be closed under complements. But $F_2$ has complements and finite intersections (try to show it!), but not infinite unions (or intersections, equivalently). – Henno Brandsma Feb 11 '12 at 10:09
@HennoBrandsma Thanks mate, a little thought along with re-reading Yuval Filmus does show that F2 has complements and finite intersections. – Comic Book Guy Feb 11 '12 at 20:56

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