Since each of these intervals is a countable union of closed intervals, then O is a countable union of closed sets. Let O 0 denote the collection of all open intervals.
An open subset of R is a subset E of R such that for every xin Ethere exists >0 such that B (x) is contained in E. For example, the open interval (2;5) is an open set. If S is an open set for each 2A, then [ 2AS is an open set. Theorem 2 Structure of Open Sets Every proper open subset of R is a countable, disjoint union of open intervals and open …

So the set [itex]\{I_x:x\in U\}[/itex] is countable and its elements are disjoint open intervals whose union is U. q.e.d. Note that q I 6= q J for I6= J, since Iand J must be disjoint, so the set fq I: I2Cgis in one-to-one correspondence with the elements of C. But this set is countable, being a subset of the rational numbers.

5. You can show that [itex]\mathbb R[/itex] is second countable, so an at most countable number of open intervals suffices.
In mathematics, particularly in topology, an open set is an abstract concept generalizing the idea of an open interval in the real line. The union of open sets is an open set. Given any base for a second countable space, is every open set the countable union of basic 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. In other words, the union of any collection of open sets is open.

In light of this, it seems to me that the only ingredient necessarily is separability: Let X be a separable topological space.

(By the way, notice that I_x is just the connected component of U containing x.) Any union of open intervals is an open set.

is open for every n2N, but \1 n=1 I n = f0g is not open.

If C is any closed subset of R, then its complement is open … We know every open set O of R is a countable union of pairwise disjoint open intervals.

Hence any open set in $\mathbb{R}$ can be written as countable union of open intervals. Let d be a metric on an infinite set M. Prove that there is an open set U in M such that both U and its complement are infinite. Both R and the empty set are open. Look at any set of open sets {Aα}.If x ∈ ∪αAα, then by definition of union, x ∈ Aα for some particular α.

The simplest example is in metric spaces, where open sets can be defined as those sets which contain a ball around each of their points; however, an open set, in general, can be very abstract: any collection of sets can be called open, as long as the union of an arbitrary number of open sets in the collection is open, the intersection of a finite number of open If any two of exploited open intervals overlap, merge them. Hence, O is F sigma. Clearly, collection of open intervals is a base for the standard topology.

5.1.1.

Proposition : Every non-empty open set is the disjoint union of a countable collection of open intervals. Any open interval is an open set.

Proof.

Show that every open interval (and hence every open set) in R is a countable union of closed intervals and that every closed interval in R is a countable intersection of open intervals.

, together with its limit 0 then the complement R−A is open.