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Bounds on Sets

  1. Jul 29, 2014 #1
    Hey guys,

    I'm puzzling a bit over an example I read in Rudin's Principles of Mathematical Analysis. He has just defined least upper bound in the section I am reading, and now he wants to give an example of what he means. So the argument goes like this:

    Consider the set A, where A = {p} s.t. p2 < 2 and p [itex]\in Q+[/itex] the set B, where B = {p} s.t. p2 > 2 and p is the same as above.

    Now let A [itex]\subset Q[/itex] and B [itex]\subset Q[/itex], where Q is the ordered set of all rational numbers. He says that A has no least upper bound and B has no greatest lower bound.

    I do not see why.

    If I consider A by itself a subset of Q, then I think 2 = sup A, and B by itself a sub set of Q, 2 = inf B.

    I could see that if we are talking about the set A AND B, then there is no sup A, if A [itex]\subset A AND B[/itex], because he just proved that there is no least element of B and no greatest element of A, and so it follows there is could be neither sup A nor inf B in this case.

    But he states to consider A and B as subsets of Q.

    Any help clarifying this matter would be greatly appreciated.

    Also, sorry if this is in the wrong place; not sure where it goes, so I figured general math would be best.
     
  2. jcsd
  3. Jul 29, 2014 #2

    Fredrik

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    He defines ##A=\{p\in\mathbb Q|p^2<2\}## and says that this set has no largest element. He then says that A has no least upper bound in ##\mathbb Q##. You say that 2 should be the least upper bound, but that's clearly not true, since (for example) 1.5 is an upper bound of A that's less than 2. The least upper bound of A in ##\mathbb R## is the irrational number ##\sqrt 2##.

    What Rudin is trying to explain is that no rational number can be a least upper bound of A. He defines ##B=\{p\in\mathbb Q|p^2>2\}## and points out two things: 1. A rational number M is an upper bound of A, if and only if it's an element of B. 2. B doesn't have a smallest element. These two things together imply that no rational number can be the least upper bound of A.

    Now regarding the notations of "upper bound in X" vs. "upper bound in Y", where X is a subset of Y, recall that the definition of "upper bound" involves a specific ordering relation. If S is a subset of X (which is still a subset of Y), no element of Y is "an upper bound of S, period". It can be an upper bound of S with respect to the ordering relation on X, or an upper bound of S with respect to the ordering relation on Y. Only an element of X can be an upper bound of S with respect to the ordering relation on X.

    In the case of ##\mathbb Q## and ##\mathbb R##, we can't just say "with respect to ≤", because that symbol is used both for the ordering relation on ##\mathbb Q## and the ordering relation on ##\mathbb R##. So we use phrases like "upper bound in ##\mathbb Q##" to shorten the phrase "upper bound with respect to the ordering relation on ##\mathbb Q##".
     
    Last edited: Jul 29, 2014
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