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Upper Bound Confusion

  1. Sep 12, 2009 #1

    jgens

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    The author of my calculus book defines an "almost upper bound" as follows: A number [itex]x[/itex] is an almost upper bound for the set [itex]A[/itex] if there are only finitely many number [itex]y \in A[/itex] with [itex]y \geq x[/itex].

    He then asks the reader to prove that if [itex]A[/itex] is a bounded infinite set, then the set [itex]B[/itex] of all almost upper bounds for [itex]A[/itex] is non-empty and bounded below. This seemed simple enough but I got confused when I thought about it in terms of a concrete example. Here are my thoughts . . .

    Let [itex]A = (0,1)[/itex] and [itex]x[/itex] be an almost upper bound for [itex]A[/itex]. Clearly [itex]x[/itex] can be written in the form [itex]x = 1 - \varepsilon[/itex] where [itex]0 < \varepsilon < 1[/itex]. Since [itex]x[/itex] is an almost upper bound, there should only be finitely many numbers [itex]y \in A[/itex] with [itex]y \geq x[/itex]. However, since the infinite sequence of numbers [itex]1-\varepsilon/2, 1-\varepsilon/3, 1-\varepsilon/4, . . . [/itex] are all in [itex]A[/itex] and greater than [itex]x[/itex] this is a contradiction.

    Could someone help me figure out where I went wrong in my thinking? Thanks.
     
  2. jcsd
  3. Sep 12, 2009 #2
    I was very confused.when i first came acrosse this.best tthing to do is try and remeber and "swat for yur exmas" it worked for me
     
  4. Sep 12, 2009 #3
    Why must x be in A? Wouldn't 1 be an almost upper bound of that set, since there are 0 elements in A greater than or equal to 1?
     
  5. Sep 12, 2009 #4

    tiny-tim

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    Last edited by a moderator: Apr 24, 2017
  6. Sep 12, 2009 #5

    jgens

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    Tiny-tim: This is from Spivak's Calculus

    Thanks for the explanation guys! I guess I just wasn't thinking about this the right way.
     
  7. Sep 12, 2009 #6

    HallsofIvy

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    For an interval like that, an "almost upper bound" is exactly the same as an upper bound.

    If the set were, say [itex](0, 1)\cup \{2\}[/itex] any number larger than or equal to 2 would be an upper bound but any number larger than or equal to 1 would be an "almost upper bound". Both sets are non-empty and bounded below.

    For a finite set finite set all numbers are "almost upper bounds"- which is why that theorem specifies that the set must be infinite.
     
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