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Here is a correct proof by contradiction.

Assume that the natural numbers are bounded, then there exists a least upper bound in the real numbers, call it x, such that n ≤ x for all n.

Consider x - 1. Since x is the least upper bound, then x - 1 is not an upper bound; i.e. there exists a specific n such that x - 1 < n.

But then, x - 1 < n implies x < n + 1, hence x is not an upper bound.

QED

This concludes the proof; i.e. there exists no upper bound in the real numbers for the set of natural numbers.

P.S. There exists sets in which the set of natural numbers are bounded, but these are not in the real number system.

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Q: Show that the natural set of numbers is an unbounded set?
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