Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Clarifications on the least upper bound property and the irrational numbers

  1. Oct 1, 2012 #1
    Hello everyone.
    I desperately need clarifications on the least upper bound property (as the title suggests). Here's the main question:

    Why doesn't the set of rational numbers ℚ satisfy the least upper bound property?

    Every textbook/website answer I have found uses this example:

    Let S={x∈ℚ : x≤√2}. This has an upper bound in ℚ, but it has no supremum in ℚ, since √2 is irrational, therefore ℚ does not satisfy the least upper bound property.

    Here's my major problem:
    A supremum is not necessarily an element of the set it's being a least upper bound of.
    By the definition of the least upper bound property:
    (i)A real number x is called an upper bound for S if x ≥ s for all s ∈ S.
    (ii)A real number x is the least upper bound (or supremum) for S if x is an upper bound for S and x ≤ y for every upper
    bound of S.
    Nowhere does this definition state that the supremum has to belong in that set.

    It seems to me that √2 should perfectly qualify as the supremum of S by this very definition. After all x≤√2 for all x∈S.
    But why do we dismiss it by saying it is not rational? Why does it HAVE to be? Why does the supremum of a subset of ℚ also HAS to be in ℚ, or else ℚ doesn't satisfy the least upper bound property?

    I'm probably missing something and this question looks idiotic to some, but frankly I don't see it. I hope someone can help me out by giving a detailed, comprehensible answer. Thanks in advance.
  2. jcsd
  3. Oct 1, 2012 #2
    You're using the wrong definition of supremum. Your definition only holds if the ambient set is the real numbers.

    Given an ordered set [itex](E,\leq)[/itex] and let [itex]S\subseteq E[/itex].
    We say that [itex]x\in E[/itex] is an upper bound of S if [itex]s\leq x[/itex] for all s in S.
    We say that [itex]x\in E[/itex] is a supremum of S if it is an upper bound of S and if for each other upper bound y of S holds that [itex]x\leq y[/itex].

    So, if you use this definition for [itex]E=\mathbb{Q}[/itex], then you see that the supremum must be in [itex]\mathbb{Q}[/itex].
  4. Oct 1, 2012 #3


    User Avatar
    Science Advisor

    It looks like a simple misunderstanding. A set has the least upper bound property if the least upper bound of any bounded subset is in the set. The rationals do not satisfy this property, since the least upper bound (as in the example) may not be rational.
  5. Oct 2, 2012 #4
    Thanks a lot!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook