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Rudin's Principles Theorem 1.11 (supremum, infimum)

  1. Feb 1, 2015 #1
    Mod note: Edited by removing [ sup ] tags.
    To the OP: Please don't fiddle with font tags, especially the SUP tag, which renders what you write in very small text (superscript).

    Hello everyone

    I have just started studying mathematics at university this summer and I have decided to supplement my scheduled classes and accompanied books by working through Rudin's analysis book for undergraduates. I am not completely certain, though, that I understand theorem 1.11 found on page five and I hope for some of you to help me a little.

    As I understand, a set, say S, has the least-upper-bound property if given any non empty subset of S, say E, which is also bounded above its supremum must lie in S.
    So since T={x is in Q: 2<x^2} is a subset of Q and the inf(T)=sqrt(2) is not in Q itself Q does not have the least-upper-bound property, right?

    So in my own words theorem 1.11 says something like:

    Let S be any set which has the least-upper-bound property and let B be a non empty subset of S which is bounded below. Since it is bounded below there must exists another set whose elements are all lower bounds of B and which consists and all possible lower bounds of B; although the set could theoretically consist of just one element. Call this set L. Conversely, every element of B must function as a upper bound of L. Because S has the least-upper-bound property every subset of S which is not empty and has an upper bound must also have an supremum, I.e. sup(L) exists. Call it £. But because B is defined as consisting of all upper bounds of L we know that £ must lie in B. Also, for every x>£, x must be in B and B only. If we negate it we get x<£ or x=£ and equivalently the opposite of x is only in B, namely x could be outside B. But since B is arbitrary the only way this be certain is for x to be in L. Hence £ is in L. But because every lower bound of B is in L, it's greatest lower bound must also be in L and the largest element in L is £ so it follows sup(L)=inf(B).

    I am truly sorry to use that much space, but I cannot formulate myself any better at this point.

    Thank you in advance
    Last edited by a moderator: Feb 2, 2015
  2. jcsd
  3. Feb 1, 2015 #2

    Stephen Tashi

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    If you are asking for an evaluation of your own words, if would help if you quoted the wording of the theorem from the book.
  4. Feb 2, 2015 #3


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    You seem to be confusing "upper bound" and "lower bound". Since sqrt(2) is a lower bound for T, it has nothing to do with upper bounds on subsets of T.

    Last edited by a moderator: Feb 2, 2015
  5. Feb 5, 2015 #4


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    Hard to be sure, but I think he is proving that the LUB property implies the INF property. I.e. he assumes that every non empty set which has an upper bound has a least upper bound, and he proves that every non empty set with a lower bound has a greatest lower bound, by showing that the greatest lower bound of the given set actually equals the least upper bound of the set of lower bounds.

    As Halls says, we are not concerned here with upper bounds of subsets of the given set, but with upper bounds of a related set, namely of the set of lower bounds of the given set.
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