# Finite disjoint unions of intervals?

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 ? –  Hardy 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

$F_2$ 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?
@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
@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