Least upper bound, greatest lower bound

Discussion in 'Physics & Math' started by kingwinner, Nov 9, 2006.

  1. kingwinner Registered Senior Member

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    796
    A math question...

    For the following subsets of the real number, find (if it exists) the maximum, minimum, least upper bound, and greatest lower bound. Justify all of your answers with a proof.
    (i) A={x|x is rational and x^2 <=5}
    [no idea how to do this question, can someone give me a hint?]

    (ii) {x|x=1- (1/10)^n, n E N}
    [very clealy I can see no max, min=9/10,lub=1,glb=9/10,but can I prove something obvious like this?]

    Can someone give me some hints on how to prove things like this? Any help is appreciated!
     
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  3. Absane Rocket Surgeon Valued Senior Member

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    i - doesn't exist for either. Can you justify this?

    ii - If it were me, I'd show any other lower bound isn't a member of the set.
     
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  5. jdheiden Registered Member

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    For i, it certaintly has a LUB, which is root(5), however it is not contained in the set since the set is only rational numbers, so it is not a max. So there is no maximum for the set.
     
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  7. Absane Rocket Surgeon Valued Senior Member

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    Not by the definition of the set. Sqrt(5) is an upperbound, but not the least.
     
  8. funkstar ratsknuf Valued Senior Member

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    I don't agree. sqrt(5) is the least upper bound, as no other real exists that is greater than (or equal) to every element of the set, and strictly less than sqrt(5). Remember, the set is considered to be a subset of the reals, so the lub can be an irrational real - it doesn't have to be rational.

    (Proof: Assume that b is a least upper bound for A strictly less than sqrt(5), i.e. b < sqrt(5) and for every x in A, x <= b. Since the rationals are dense in R, a rational q exists between b and sqrt(5), such that b < q < sqrt(5). Then q is in A by definition, which contradicts that b was a least upper bound for A. Hence sqrt(5) is the least upper bound for A.)
     
  9. Zephyr Humans are ONE Registered Senior Member

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    I think Funkstar has it ... The least upper bound (supremum) exists for every bounded set. If the supremum is part of the set, it's also the maximum; otherwise the maximum of that set doesn't exist.
     
  10. D H Some other guy Valued Senior Member

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    Not true. Every bounded set does not have a least upper bound.

    This is true for the reals because they are complete. The rationals, on the other hand, are not complete. The least upper bound of the problem in the OP does not exist. One cannot say that the least upper bound is sqrt(5) because sqrt(5) is not rational.

    From http://en.wikipedia.org/wiki/Least_upper_bound_axiom (key text bolded for emphasis):
    A perfect example is S={x in Q | x<sup>2</sup> < 2}. 2 is certainly an upper bound for the set. However, this set has no least upper bound — for any x in S, we can find a y in S with y > x.​
     
  11. §outh§tar is feeling caustic Registered Senior Member

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  12. shmoe Registred User Registered Senior Member

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    These are being considered as subsets of the real numbers, hence they both have least upper bounds (as they are bounded from above). The least upper bound doesn't need to be in the set.

    In the Wiki example, that set is being considered a subset of the rationals. The point being to show the rationals don't satisfy the least upper bound property.
     
  13. Zephyr Humans are ONE Registered Senior Member

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    Yes, it only holds for complete ordered (I think those are the conditions) sets, like the reals. But I sort of assumed we were talking about reals

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    That set has no least upper bound in the rationals, but it does have one in the reals. (Squareroot of 2)
     
  14. funkstar ratsknuf Valued Senior Member

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    It does say, rather explicitly, that the sets are considered as subsets of the reals. So one can indeed say that sqrt(5) is the lub of the set, because it only has to be a real, not necessarily a rational.

    But I'm repeating myself. [Edit: And others, I see.]
     
  15. Absane Rocket Surgeon Valued Senior Member

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    You're right. I missed one condition..
     
  16. §outh§tar is feeling caustic Registered Senior Member

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    Did my post get deleted?
     
  17. Absane Rocket Surgeon Valued Senior Member

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    I see one post of yours.. "^ pedant."
     
  18. D H Some other guy Valued Senior Member

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    Posts seem to appear, disappear, and then reappear lately. I think something may be wrong with the server.
     
  19. kingwinner Registered Senior Member

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    Hi everyone,

    For (i) A={x|x is rational and x^2 <=5}

    How can I prove that there are no maximum and no minimum?

    And funkstar, your proof about the glb A=sqrt5 includes "Since the rationals are dense in R...", is there a more rigorous way to prove it or think about it??

    Thanks!
     
  20. §outh§tar is feeling caustic Registered Senior Member

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    I had an entire paragraph right below that..
     
  21. §outh§tar is feeling caustic Registered Senior Member

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    A is the set of all rationals in the open interval (-sqrt 5, sqrt 5). That should help you prove that there is no max or min (probably once you understand why the rationals are dense).

    I think you know how to show that in between any two distinct rationals, there is a rational number. Showing that this extends to any two distinct real numbers involves some cleverness. See if you can find the proof in your book. I remember I wasn't able to figure it out myself but after I did, it became a piece of cake to understand.
     
  22. funkstar ratsknuf Valued Senior Member

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    Well, you've seen the proof that the lub is sqrt(5) above. If there is a maximum, then what correspondence must it have to the lub? Wrt. the minimum, assume k is the minimal element and k < 0 (establish first that the minimum, if it exists, must ne negative. This is really easy: Is -1 a member of A? If it is, then is the glb (if it exists) negative, and why?). Now, is k*2 < k and is k*2 a member of A? What does this tell you about k as a minimal element of A?
    It's the lub not the glb. But anyway: Not really. It's a rather basic property of the rationals that there has to be another rational (strictly) between any two given (different) rationals. That's what "dense" means, basically. You can't really get around using this property, because the problem very much concerns the rationals distribution in the reals (think about it).
    You're welcome. Btw, as much as we enjoy these little problems, it is better if you give us some sort of idea as to how far you've come with each problem, and what particularly is giving you grief. Copying solutions from here will not aid you in the long run...
     
  23. §outh§tar is feeling caustic Registered Senior Member

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    There is actually a way to prove that the rationals are dense. I just haven't found any real use for it. It's like being able to prove that any two (non-empty) closed intervals are equivalent. It's nice to know but I don't see how it really helps you.

    EDIT: Except if you mean the 'dense' property of the real numbers comes from its being complete. But I think that too can be proved (by way of construction with Dedekind cuts). One of the beautiful things about mathematics is the drive for minimalism when it comes to assumptions. I was reading Gauss' biography yesterday and I was impressed that he was able to come up with a non-Euclidean geometry at such a young age. Of course, one's age isn't much in mathematics but still..

    I wonder how one goes about constructing such a thing. I suppose if you knew how to construct a 17-gon then doing so would be child's play.
     
    Last edited: Nov 12, 2006

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