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Mathematics
Calculus
Showing that a sequence of supremums of a sequence has these two properties
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[QUOTE="Office_Shredder, post: 6475089, member: 53426"] I did indeed. Yes. Since ##A_n## is an upper bound for a set which ##A_{n+1}## is the supremum, it must be at least as large as ##A_{n+1}## by definition of the supremum. That gives you part 1. For part 2, your proof is overly complicated and starts by assuming ##A_n## is unbounded (though I don't think it's necessary for the rest of your proof). Try something simpler: if ##A_k## is decreasing, and there is some ##n## for which ##A_n < \lambda-\epsilon##, then every other ##A_k## for ##k>n## must be below ##A_n## which gives you a contradiction to the fact that ##\lambda## is the limit. You don't need to prove that it has some other limit, just that it doesn't match the statement you were given. [/QUOTE]
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Showing that a sequence of supremums of a sequence has these two properties
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