Every open set in R is a countable union of open intervals. We claim that b=2G. Let d be a metric on an infinite set M. Prove that there is an open set U … 2.16Prove that every open set in R is the union of an at most countable collection of disjoint segments. 1.5.3 (a) Any union of open sets is open. Since every open set in R is an at most countable union of open intervals, we must have σ(O 0)=B.LetD denote the collection of all intervals of the form (−∞,a], a ∈ Q.Let(a,b) ∈O 0for some b>awith a, b ∈ Q.Let a n= a+ 1 n so that a 29 Proof Let Gbe an open set in R. For each x2G, there are yand z, with z

Indeed, and a common way to prove it requires you to know that $\mathbb{N}\times\mathbb{N}$ is countable. CantorSet, what micromass didn't mention is that when you write an open set as a union of open cubes with rational endpoints, then the result is a union of countably many sets.This is because there are only countably many cubes with rational endpoints (something you can prove for yourself probably). You can show that $\mathbb R$ is second countable, so an at most countable number of open intervals suffices.

(Royden/Fitzpatrick, 4th edition) What is the most general setting in which every open set is a disjoint union of countable collection of open balls (or bases)? But, part of this result can be generalized to Rn. Prove. In R, every nonempty open set is the disjoint union of a countable collection of open intervals. Then Q\G is a countable dense set in G by the Archimedean prop- x2BˆU. An open set in $\mathbb R$ is by definition a (not necessarily disjoint) union of open intervals. Let Ube a collection of disjoint open subsets of Rn.

12. CantorSet, what micromass didn't mention is that when you write an open set as a union of open cubes with rational endpoints, then the result is a union of countably many sets.This is because there are only countably many cubes with rational endpoints (something you can prove for yourself probably). 0 denote the collection of all open intervals. 1.5.3 (a) Any union of open sets is open. Recall from The Union and Intersection of Collections of Open Sets page that if $\mathcal F$ is an arbitrary collection of open sets then $\displaystyle{\bigcup_{A \in \mathcal F} A}$ is an open set, and if $\mathcal F = \{ A_1, A_2, ..., A_n \}$ is a finite collection of open sets then $\displaystyle{\bigcap_{i=1}^{n} A_i}$ is an open set.

Let Bbe an open ball s.t. 8. Let b= supfy: (x;y) ˆGgand a= inffz: (z;x) ˆGg. Even in R2 it is easy to draw open sets that are not the disjoint union of open balls.