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I found the following statement in Munkres' Topology:

Theorem 4.2 (Strong induction principle). Let $A$ be a set of positive integers. Suppose that for each positive integer $n$, the statement $S_n \subset A$ [here $S_n = \{1, 2, \dots, n\}$] implies the statement $n \in A$. Then $A = \mathbb{Z}_+$.

Now, I think I understand strong induction. But what I don't understand here is, doesn't $S_n \subset A$ always imply $n \in A$? It's pretty much by definition of a subset, that if $\{1,2,\dots,n\} \subset A$ then $n \in A$. Did the author mean "Suppose that for each positive integer $n$ the statement $S_n \subset A$ is true"?

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Unless, I'm missing something, I don't see how the statement is true unless $1 \in A$. – user78680 May 20 '13 at 21:15
Since $S_1=\emptyset\subseteq A$, it follows that $1\in A$ by hypothesis. – Cameron Buie May 20 '13 at 21:21
up vote 5 down vote accepted

If you check the previous page, you should see that $S_n$ is the set of positive integers that are less than $n$. $S_n$ here denotes the section of $\Bbb Z_+$ by $n$. See also the more general definition immediately before Lemma 10.2.

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Oh, that makes a lot of sense. My bad. Thanks! – Javier May 20 '13 at 21:19

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