Section 10: Problem 4 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) Let $\mathbb{Z}_{-}$ denote the set of negative integers in the usual order. Show that a simply ordered set $A$ fails to be well-ordered if and only if it contains a subset having the same order type as $\mathbb{Z}_{-}$ .

(b) Show that if $A$ is simply ordered and every countable subset of $A$ is well-ordered, then $A$ is well-ordered.

(a) If $B\subset A$ has the order type of $\mathbb{Z}_{-}$ , then $B$ has no smallest element (as there is a bijection into $\mathbb{Z}_{-}$ preserving the order). If $A$ is not-well ordered then there exists a nonempty $B\subset A$ that has no smallest element. Let $b_{1}\in B$ . For $n\in\mathbb{Z}_{+}$ , there is $b_{n+1}\in B$ less than $b_{n}$ . By induction, there is a sequence $(b_{1},b_{2},\ldots)\in B^{\omega}$ such that $b_{1}>b_{2}>\cdots$ , and the subset $\{b_{1},b_{2},\ldots\}\subset B\subset A$ has the same order type as $\mathbb{Z}_{-}$ ( $f(b_{n})=-n$ ).

(b) It follows from (a). If $A$ is not well-ordered, then there is a countable subset having the same order type as $\mathbb{Z}_{-}$ , which is not well-ordered.

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Section 10: Problem 4 Solution

Section 10: Problem 4 Solution