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If all elements of a set are individually bounded, is the set bounded?

  1. Jan 30, 2014 #1
    This is a concise question, so the title pretty much says it all. Also, this is not a HW question, but the idea has subtly popped up in two homework problems that I have done in the past. I cannot justify why the entire set would be bounded, because we know nothing of the nature of the boundedness of each element; i.e. the bounds themselves could grow without bound, e.g. if each element a_n is bounded by n, then surely the set is not bounded.
     
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  3. Jan 30, 2014 #2

    Office_Shredder

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    Your question needs to be made a lot more precise, if S is any subset of the reals each individual element of S is trivially bounded , as if x is a real number x+1 is an upper bound for x
     
  4. Jan 30, 2014 #3
    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?
     
  5. Jan 30, 2014 #4
    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.
     
  6. Jan 30, 2014 #5

    AlephZero

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    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
     
  7. Jan 30, 2014 #6

    WWGD

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