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Is this proof correct?

  1. Jun 6, 2014 #1
    1. Question I 3.12 from Apostle's calculus volume 1
    If x is an arbitrary real number, prove that there are integers m and n such that m<x<n




    2. Theorem I.27:Every nonempty set S that is bounded below has a greatest lower bound; that is, there is a real number such that L= infS



    3. Suppose x[itex]\in[/itex]R and belongs to a nonempty set S with no maximum element. It follows that ∃B[itex]\in[/itex]Z[itex]\stackrel{+}{}[/itex] such that B is an upper bound for S. Let n=B, then we have n>x. Similarly, let -S denote the set of negatives of numbers in S. Suppose that -B[itex]\in-Z[/itex][itex]\stackrel{+}{}[/itex] of -S, then by theorem I.27 -B is a lower bound of -S. Let m=-B. , such that m<x<n.
     
    Last edited: Jun 6, 2014
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  3. Jun 6, 2014 #2

    Fredrik

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    You have stated this as if you're making two assumptions about x. You should only make one: ##x\in\mathbb R##. If you need to use some other statement about x, you need to be able to prove that it follows from your one and only assumption about x.

    It doesn't. Consider e.g. x=1 and S the set of integers (which doesn't have a maximum element).
     
  4. Jun 6, 2014 #3
    I thought an inductive set could have an upper bound but not a greatest upper bound / supremum because for every x in the set x+1 is also in the set?

    and to fix my assumption issue could i do something along the lines of- Suppose S is a nonempty set of real numbers and suppose there is a number B such that x[itex]\leq[/itex]B so B could be an upper bound and then go an to show the lower bound as well equating m and n, the integers, as a subset of the real numbers?
     
  5. Jun 6, 2014 #4

    Fredrik

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    The set of natural numbers is an inductive set, but it doesn't have an upper bound in ##\mathbb R##. This would contradict the result that you're trying to prove.

    If you let S be an arbitrary non-empty subset of ##\mathbb R## and let B be an arbitrary real number such that ##x\leq B##, then what set would B be an upper bound of? Certainly not S. S is arbitrary, so we could have ##S=\{B+1\}##. B is an upper bound of {x}, but that's not very useful.

    Edit: I'll give you a hint about one way to solve the problem. The set ##S=\{n\in\mathbb Z|n<x\}## is useful.
     
    Last edited: Jun 6, 2014
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