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text: 'A **well-ordered set** is a [-540] $(S, \\leq)$, such that for any nonempty subset $U \\subset S$ there is some $x \\in U$ such that for every $y \\in U$, $x \\leq y$; that is, every nonempty subset of $S$ has a least element.\n\nAny finite totally ordered set is well-ordered. The simplest [infinity infinite] well-ordered set is [45h $\\mathbb N$], also called [ordinal_omega $\\omega$] in this context.\n\nEvery well-ordered set is [4f4 isomorphic] to a unique [-ordinal_number], and thus any two well-ordered sets are comparable.\n\nThe order $\\leq$ is called a "well-ordering," despite the fact that "well" is usually an adverb.\n\n#Induction on a well-ordered set\n\n[mathematical_induction] works on any well-ordered set. On well-ordered sets longer than $\\mathbb N$, this is called [-transfinite_induction]. \n\nInduction is a method of proving a statement $P(x)$ for all elements $x$ of a well-ordered set $S$. Instead of directly proving $P(x)$, you prove that if $P(y)$ holds for all $y < x$, then $P(x)$ is true. This suffices to prove $P(x)$ for all $x \\in S$.\n\n%%hidden(Show proof):\nLet $U = \\{x \\in S \\mid \\neg P(x) \\}$ be the set of elements of $S$ for which $P$ doesn't hold, and suppose $U$ is nonempty. Since $S$ is well-ordered, $U$ has a least element $x$. That means $P(y)$ is true for all $y < x$, which implies $P(x)$. So $x \\not\\in U$, which is a contradiction. Hence $U$ is empty, and $P$ holds on all of $S$.\n%%',
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