Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    11
    ReputationRep:
    I have been reading through the first chapter of Rudin's " Principles of mathematical analysis" and I got a bit confused with one of the definitions. I just need some clarification.

    He states ,as a definition, that "An ordered set S is said to have the least upper bound property if the following is true:

    **If E is a proper subset of S, E is not empty, and E is bounded above**, then Sup E exists in S"

    I can't figure out why this must be true. I first thought that if we consider E to be the proper subset of S containing the largest element of S, then provided that ** implies that Sup E exists in S then it must be true that S has the least upper bound property.
    But I am really uncertain about this, can some clarify why the existence of Sup S is defined in this way?






    Posted from TSR Mobile
    • Study Helper
    Online

    13
    (Original post by arrow900)
    I have been reading through the first chapter of Rudin's " Principles of mathematical analysis" and I got a bit confused with one of the definitions. I just need some clarification.

    He states ,as a definition, that "An ordered set S is said to have the least upper bound property if the following is true:

    **If E is a proper subset of S, E is not empty, and E is bounded above**, then Sup E exists in S"

    I can't figure out why this must be true. I first thought that if we consider E to be the proper subset of S containing the largest element of S, then provided that ** implies that Sup E exists in S then it must be true that S has the least upper bound property.
    But I am really uncertain about this, can some clarify why the existence of Sup S is defined in this way?
    What you quoted - in red - is not defining the existance of Sup S.

    That's covered by point 1.8 at the top of that page (in the 3rd edition)

    You should also note that S may not have a largest element, .e.g the natural numbers, or the reals, etc.

    What's quoted is saying:

    "A set, S, has the least-upper-bound property" means "If E is a proper subset of S, E is not empty, and E is bounded above**, then Sup E exists in S"

    If, and only if, the bit is green is true then, S has the least-upper-bound property.

    I may have misunderstood where you're coming from, but hope that helps.
    • Study Helper
    Online

    13
    Rephrasing the definition slightly may help.

    A set, S, has the least upper bound property, iff every non empty proper subset of S that is bounded above, has a least upper bound that is in S.
    Offline

    11
    ReputationRep:
    (Original post by arrow900)
    I have been reading through the first chapter of Rudin's " Principles of mathematical analysis" and I got a bit confused with one of the definitions. I just need some clarification.

    He states ,as a definition, that "An ordered set S is said to have the least upper bound property if the following is true:

    **If E is a proper subset of S, E is not empty, and E is bounded above**, then Sup E exists in S"

    I can't figure out why this must be true.
    It's neither true nor false. It's a definition of the least upper bound property.

    I first thought that if we consider E to be the proper subset of S containing the largest element of S, then provided that ** implies that Sup E exists in S then it must be true that S has the least upper bound property.
    It's hard to see what you're saying here. E is not a *specific* subset of S. It's any non-empty proper subset. And also your statement "if we consider E to be the proper subset of S containing the largest element of S," indicates confusion. S may not have a "largest element". For example, \mathbb{N} has no largest element, neither does \mathbb{R}

    Here are two examples.

    1) \mathbb{Q} does not have the least upper bound property. For example, consider:

    E =\{ x \in \mathbb{Q} : x^2 < 2 \}

    Then E is bounded above (by 10, for example), is non-empty, but it does not have a supremum in \mathbb{Q}. (The only possible candidate is \sqrt{2} but that is not rational).

    2) It is taken as an axiom that \mathbb{R} does have the least upper bound property since this then gives it the properties required for doing calculus. For example, if the l.u.b property holds for \mathbb{R} then you can prove that Cauchy sequences converge to something in \mathbb{R}. (Or alternatively, if you assume that \mathbb{R} is Cauchy-complete i.e. that all Cauchy sequence converge to something in \mathbb{R}, then you can prove that it has the l.u.b property)

    Now consider:

    E =\{ x \in \mathbb{R} : x^2 < 2 \}

    Then by the l.u.b. property, \sup E \in \mathbb{R} and it can be shown that \sup E = \sqrt{2}

    Note that the supremum of a set may or may not be an element of that set. In the case above, \sup E \notin E
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by ghostwalker)
    Rephrasing the definition slightly may help.

    A set, S, has the least upper bound property, iff every non empty proper subset of S that is bounded above, has a least upper bound that is in S.
    Yes that makes perfect sense. Its just that the definition in the textbook is phrased almost as if it is a theorem which caused the confusion. Thanks a lot.

    Posted from TSR Mobile
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.