Section 10: Problem 7 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

Let $J$ be a well-ordered set. A subset $J_{0}$ of $J$ is said to be inductive if for every $\alpha\in J$ , $(S_{\alpha}\subset J_{0})\Rightarrow\alpha\in J_{0}.$ Theorem (The principle of transfinite induction). If $J$ is a well-ordered set and $J_{0}$ is an inductive subset of $J$ , then $J_{0}=J$ .

Let $A$ be a set of all elements not in $J_{0}$ , and suppose it is not empty. Then it has the smallest element $\alpha$ , and $S_{\alpha}\subset J_{0}$ (the section can be empty, it does not matter). Therefore, $\alpha\in J_{0}$ . Contradiction. We conclude that $A$ is empty, and $J_{0}=J$ .

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Section 10: Problem 7 Solution

Section 10: Problem 7 Solution