« Section 24: Problem 11 Solution

Section 24: Problem 12 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
(a) If is a bijection preserving the order, then is a bijection preserving the order, similar for . Vice versa, if there are two order preserving bijections from and to and then the joint function is the order preserving bijection needed.(b) and has the order type of implies, by applying (a) two times, that has the order type of . Vice versa, we construct bijective order preserving functions and the resulting function is an order preserving surjective function onto .(c) Transfinite induction: we show that if it holds for any then it holds for . If has an immediate predecessor then is either empty or has the order type of , and has it as well. By (a), the property holds for . If has no immediate predecessor, then the set of all elements lower than has it as the supremum (for any there is ) and we can construct a sequence of points converging to . This is not a trivial fact, I think (and I don’t remember it was covered anywhere before). Indeed, if we consider , for example, then is a limit point of but there is no sequence converging to it (any sequence is a countable subset and has an upper bound in ). However, in the case of we prove it as follows: is countable; let us enumerate all its elements with positive integer indexes: ; then let be such that , let be the minimum index greater than such that , let be the minimum index such that and , etc. (we can always do this as there is no largest element in , therefore, there is infinite number of elements greater than any ); note that for every , and for all ; if then for some and for all . Now, once we have the sequence converging to , by using (b), has the order type of . By the principle of transfinite induction, this holds for any .(d) We show that any point is path connected to where is the immediate successor of . Indeed, for is path connected to ( is homeomorphic to ) and for any point , , is path connected to which is path connected to (by (a) and (c), has the same order type as and the order topology on a convex subset is the same as the subspace topology). Therefore, is path connected as the union of path connected subsets having a common point.(e) An open interval in has the order type of an open interval in . Indeed, by (c), has the order type of . Since for convex subsets the order topology is the same as the subspace topology, has the order type of . Therefore, for any there is an open neighborhood with the order type of . Moreover, there is a bijective order preserving function from one set onto the other, and by 7(a), it is a homeomorphism.(f) If could be embedded into then would be homeomorphic to a subspace of which has a countable basis for the topology. Therefore, there must be a countable basis of the topology of . We show that there is no. For any : is open and contains but does not contains for . Therefore, there is a basis element such that it contains but does not contain for . Call it . Note that for : . Since is uncountable, there must be uncountable number of distinct sets in the basis.