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- Thread starter leinadle
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Office_Shredder

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Sorry for the vagueness, here is an example. Suppose we have an infinite sequence of functions {f_n(x):R -> R}, each of which is bounded by some M_n > 0; i.e. |f_n(x)| < M_n for all x in R. Can we say that there exists an M > 0 such that |f_n(x)| < M for all n and for all x in R?

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I suppose the answer to my question must be no, after reading your response again. If I consider the set S which consists of all positive reals then each element is individually bounded by the x+1 argument, but S is certainly not bounded since R is unbounded. However, this logic is really bugging me for some reason. It seems to suggest that since the set isn't bounded, how can we possibly state that there exists an element that isn't bounded when we know that x+1 is a bound for that number. What piece of logic am I missing here? I feel like Ouroboros.

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AlephZero

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I suppose the answer to my question must be no

True. A simple counterexample is ##f_n(x) = n## which makes the functions question exactly the same as Office Shredder's post.

I don't know what is bugging you about this, but it is pretty much the same as any divergent series. For example ##\sum_{k=1}^n \frac{1}{k}## is finite for every integer ##n##, but ##\sum_{k=1}^\infty \frac{1}{k}## diverges. When your proposition in the OP is false, the ##M_n## are a divergent series

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WWGD

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http://en.wikipedia.org/wiki/Uniform_boundedness_principle

But, like from your example , there is no guarantee; use the set {1,2,3,..}

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