« Section 31: Problem 8 Solution

Section 31: Problem 9 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
9. Let be the set of all points of of the form , for rational; let be the set of all points of this form for irrational. If is an open set of containing , show there exists no open set containing that is disjoint from , as follows:
(a) Let consist of all irrational numbers in such that is contained in . Show is the union of the sets and countably many one-point sets.
(b) Use Exercise 5 of §27 to show that some set contains an open interval of .
(c) Show that contains the open parallelogram consisting of all points of the form for which and .
(d) Conclude that if is a rational number with , then the point of is a limit point of .
(a) For every irrational point there is a basis neighborhood contained in , therefore, if then . The rest of , i.e. the rational points, is a countable set of points.
(b) According to the result of the Exercise 5 of §27, since has non-empty interior in being compact and Hausdorff, one of the closed sets in the union must have a non-empty interior in as well. Therefore, there is such that has non-empty interior, and contains a non-empty interval .
(c) I show a little different result: for found in (b) there is a non-empty interval and such that for every , . Take where and are such that (found in (b)). Let . is not empty. For each and there is a point . Therefore, .
(d) Given (c), for every point and every neighborhood of there is such that .