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Let $S$ be a set of natural numbers such that $n$ belongs to $S$ whenever all numbers less than $n$ belong to $S$ (i.e., assume $\forall n(\forall m<n\ m\in S)\Rightarrow n\in S$, where the quantifiers range over all natural numbers). For indirect proof, suppose that $S$ is not the entire set of natural numbers $\mathbb{N}$. That is, the complement $\mathbb{N}\setminus S$ is nonempty. The well-ordering principle for natural numbers says that $\mathbb{N}\setminus S$ has a smallest element; call it $a$. By assumption, the statement $(\forall m<a\ m\in S)\Rightarrow a\in S$ holds. Equivalently, the contrapositive statement $a\in \mathbb{N}\setminus S \Rightarrow \exists m<a\ m\in \mathbb{N}\setminus S$ holds. This gives a contradition since the element $a$ is an element of $\mathbb{N}\setminus S$ and is, moreover, the \emph{smallest} element of $\mathbb{N}\setminus S$.
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Let $S$ be a set of natual numbers such that $n$ belongs to $S$ whenever all numbers less than $n$ belong to $S$ (i.e., assume $\forall n(\forall m<n m\in S)\Rightarrow n\in S$, where the quantifiers range over all natural numbers). For indirect proof, suppose that $S$ is not the entire set of natural numbers $\matbb{N}$. That is, the complement $\mathbb{N}\setminus S$ is nonempty. The well-ordering principle for natural numbers says that $\mathbb{N}\setminus S$ has a smallest element; call it $a$. By assumption, the statement $\forall m <a m\in S)\Rightarrow a\in S$ holds. Equivalently, the contrapositive statement $a\in \mathbb{N}\setminus S \Rightarrow \exists m<a m\in \mathbb{N}\setminus S$ holds. This gives a contradition since the element $a$ is an element of $\mathbb{N}\setminus S$ and is, moreover, the \emph{smallest} element of $\mathbb{N}\setminus S$.
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