Though every open set in R is a disjoint union of countably many open intervals, it is not true that every closed set is a disjoint union of closed intervals. For suppose that C is any in nite collection of disjoint open intervals. Part (b) Give an example of an uncountable collection of disjoint open intervals, or argue that no such collection exists.

Write $a\sim b$ if the closed interval $[a, b]$ or $[b, a]$ if $b

Give an example of a countable collection of disjoint open intervals.

Suppose there is some set of disjoint intervals of strictly positive length, whose union is the open unit interval. This allows you to form an infinite binary tree whose vertices correspond to (some of) the closed intervals, such that in-order (see link below) ordering of tree vertices corresponds to the natural ordering on intervals.

The collection f(n;n+ 1) : n2Ngsu ces.

Between any two closed intervals, there would have to be more closed intervals. This must be a countable set, so we can enumerate it: let. 2) Let Ix be the smallest neighborhood of x. To …

Therefore any set of disjoint intervals of strictly positive length must be countable, since we can inject any such set into. $G$ is therefore the union of disjoint equivalence classes. No such collection exists. A set B is called a G set if it can be written as the countable intersection of open sets. The following link shows that opne set can is countable disjoint union of open intervals: Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. Open Set is Union of Disjoint Open Intervals 1) Let x be any rational number in an open set S in R (the real line). kathrynmath said: A set A is called a F set if it can be written as the countable union of closed sets. Indeed, there exists a very famous closed set called the Cantor set whose structure is much more interesting.

Consider the sequence of sets. a) Show that a closed interval [a,b] is a G set