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Suppose S is nonempty and unbounded. Prove that there exists a divergent sequence in S.

User Andreina
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Final answer:

A divergent sequence exists in a nonempty, unbounded set by constructing a sequence where each term is greater than its position in the natural numbers. This sequence diverges to infinity as the set is unbounded.

Step-by-step explanation:

The question is asking to prove the existence of a divergent sequence in a set S that is nonempty and unbounded. To construct a divergent sequence, we can define a sequence {a_n} by choosing a_n in S such that a_n > n for every n in the natural numbers. As the set S is unbounded, for each n, there exists an element a_n in S that is larger than n, and because n can be made arbitrarily large, the sequence will diverge to infinity.

Essentially, since S is unbounded above, we can always find a member of S that exceeds any given natural number. Hence, this sequence does not converge to any real number, and is therefore a divergent sequence in S.

User Lawrence Cooke
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