Prove that every open set in Ris the union of an at most countable collection of disjoint segments. 1;and X are both open and closed. Proof. Prove that m([1 P k=1) 1 k=1 m(E k). Hint in book Proof: Suppose G ⊂ R 1 is an open set. of R. (Note that these are the connected open subsets of R.) Theorem. Since R is un-countable, R is not the union of two countable sets. Homework Equations The Attempt at a Solution This sounds like something requiring proof by contradiction. A nite intersection of open sets is open, but an intersection of infinitely many open sets needn’t be open. An arbitrary union of open sets is open; one can prove that every open set in R is a countable union of disjoint open intervals.

Haiyu Huang 2.29 Problem. Let A denote the set of algebraic numbers and let T denote the set of tran-scendental numbers. 3 The intersection of a –nite collection of open sets is open.

Problem 3 (Chapter 1, Q56*). Prove that the collection of isolated points of a set S ⊂ Rn must be countable. The end points of the intervals do not belong to U. For each x 2U we will nd the \maximal" open interval I x s.t. Chapter 2, problem 29. HW4.5 Rudin Chap 2, 29. Hence T is uncountable. Prove that every closed subset of R is the intersection of a countable collection of open sets. The set {x in R | x d } is a closed subset of C. Each singleton set {x} is a closed subset of X. But, since others have already shown that method, I’ll show a slightly different proof. 1 Already done. Homework Statement Show that the set of rational numbers in the interval (0, 1) cannot be expressed as the intersection of a countable collection of open sets. Assume that Ois nonempty. Prove that any open set in R is the union of a collection of pairwise disjoint open intervals which is at most countable. Every open subset Uof R can be uniquely expressed as a countable union of disjoint open intervals. Let OˆR be open. Let fbe a real-valued function de ned on R. Show that the set of points at ... k=1 be a countable collection of sets in A. Hence Q kis dense in Rk and R is separable. For each q2O\Q, let R q = fr>0j(q r;q+ r) ˆOg. Note that R = A∪ T and A is countable. Proof. 2 The union of an arbitrary (–nite, countable, or uncountable) collection of open sets is open. Any element [math]x\in S[/math] is contained in an interval [math]I_x\subseteq S[/math], which in turn contains a rational number [math]r_x[/math]. Note { the pairwise was added afterwards, since a few people were confused by the meaning of ‘disjoint’ otherwise. Let U ˆR be open. If T were countable then R would be the union of two countable sets. Prove that every open set in R 1 is the union of an at most countable collection of disjoint segments.