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Completeness Axiom

  1. Dec 29, 2004 #1

    cepheid

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    If a nonempty set, S, of real numbers has an upper bound M ([itex] x \leq M [/itex] for all x in S), then S has a least upper bound b. (This means that b is an upper bound for S, but if M is any other upper bound, then [itex] b \leq M [/itex].) The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line.

    I am confused about this theorem. I apologize, for I have not been formally taught any number theory, or whatever branch of math into which this falls. I only came across this in my calc text when reviewing sequences and series. It was used to prove the Monotonic Sequence Theorem. However, I do not understand why the condition that the set S must have a least upper bound guarantees that the real number line has no holes or gaps. What about this situation (hypothetical "broken" real number line):

    <----------------Smax_______________M---------------------->

    So, at the end of the left section of the real line, we have Smax, the highest number in the set S. At the beginning of the right section, we have the upper bound M. Is M not, by necessity, a least upper bound? What part of the completeness axiom is not satisfied by the broken number line?

    Thank you.
     
  2. jcsd
  3. Dec 29, 2004 #2
    In your example, M is an upper bound, but not the least upper bound (the LUB happens to be Smax in your example).

    I don't see how this affects the fact that the real line has no holes though... First of all, the completeness axiom doesn't say anything about there not existing any numbers between an upper bound of a set, and the least upper bound of that set! And second, if a and b are real numbers with a < b (for example Smax and M...), then surely (a + b)/2 is another real number, and we have a < (a + b)/2 < b (which is easy to show). That seems to imply that the real line has no holes, but of course, I don't think I've ever heard a rigorous definition of "hole" when it comes to real analysis ;)
     
    Last edited: Dec 29, 2004
  4. Dec 29, 2004 #3

    cepheid

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    Oops. I forgot that in the definition of an upper bound M, it said that any number x in the set S is less than or equal to M. I was thinking they all had to be less than M. But, if it is less than or equal to, then doesn't that mean that the highest number in the set S is always the least upper bound? I know that this axiom is saying something very basic, but I'm still doubtful that it could be that simple.

    I'm afraid that the way you have worded this is a bit convoluted, but it seems to me that you are saying that the completeness axiom doesn't lead directly to the scenario I drew. I agree. I chose that hypothetical situation with a gaping hole in the real number line because it seemed to me that there was nothing in it that contradicted the axiom. I still don't see what does. You stated yourself that in the scenario I gave, Smax is the least upper bound. So the set of numbers has a least upper bound! The completeness axiom is satisfied...even with a number line that is discontinuous! Please help!!!
     
  5. Dec 29, 2004 #4
    Yes, if S has a largest element. What's the largest element of S = {1 - 1/n; n natural}?

    *edit*

    Oh, I see. The first paragraph in your first post was actually a quote, not something you'd written yourself. Well then, I'm not sure what to say. The last sentence seems faulty to me too.
     
    Last edited: Dec 29, 2004
  6. Dec 29, 2004 #5

    matt grime

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    "Oops. I forgot that in the definition of an upper bound M, it said that any number x in the set S is less than or equal to M."

    well that'd be the definition of an upper bound... this doesn't imply that there is a number in S equal to M, or anywhere near M at all.
     
  7. Dec 30, 2004 #6

    cepheid

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    Oh...cool. So the LUB is 1 in this instance? By making n arbitarily large...the numbers in the set can approach arbitrarily close to 1?

    Hmm...I took it verbatim from Stewart Calculus 5e section 12.1 page 745, and though I desperately want it to make sense.... :confused:

    Matt, thanks for clearing that up. I was aware that no number in the set need be equal to an upper bound M, or anywhere close to it, and Muzza's example shows that none of them need be equal to the least upper bound either! I never even thought of infinite sets!. But, it seems to me that for a finite set, (which necessarily has a largest number...or am I making too many assumptions here that call upon the axioms I am trying to understand?), the least upper bound is the highest number in the set. Glad we sorted that out. Now, only this problem of whether the completeness axiom guarantees no holes in the real number line, and what that means, remains.
     
  8. Dec 30, 2004 #7

    mathwonk

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    i agree there are sets with holes that have the least upper bound axiom. e.g. the reals minus the open interval (0,1). also any finite set of reals. so lets see if we can prove the reals are connected. that will prove there are no holes. obviously we will need to use more properties than just the least upper bond property.

    ok. how DO we prove the reals are connected? of course when in doubt we always start from a definition of connectedness. here's a sneaky one that might satisfy you: the reals are connected if there is no continuous surjection from the reals to the set {0,1}. no wait let's just prove the intermediate value theorem. it is even stronger.

    so assume we have a continuous function f from the reals to the reals such that f(a) < 0 and f(b)>0, nd a<b. we claim some real number c with a<c<b maps to 0.

    consider the set S of all reals x between a and b such that f(x) is negative. Since f is continuous, all reals near a, map to negative numbers. but how do we know there ARE any such numbers between a and b? Here we use the fact that the reals contain the rationals and are a field. so there do exist numbers of form a+1/n for all large n, and these numbers are as close to a as desired. this step in the proof is often skipped over. also all numbers close to b map to positive numbers.

    anyway now we have that S is non empty. S is obviously bounded above by b. so S has a least upper bound c < b. then if f(c) < 0, then also f(x) < 0 for all x near c. and again this is a contradiction to c being an upper bound for those x with negative images, since any number of form c+1/n is also less than b and maps to a negative number.

    similarly one gets contradiction from assuming f(c)>0. hence f(c) = 0.

    thus we have proved property of the reals which is not true for the sets you proposed. the difference is that even though your sets did have the lub property, they bwere not ordered fields, i.e. did not contain those needed points of form a+1/n, and c+1/n used in the proof.

    so maybe we sould say that since the reals are closed under addition and contain the rationals, AND have the lub proeprty, then they have no holes.
     
  9. Dec 30, 2004 #8

    Hurkyl

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    This isn't a freebie: you've assumed the Archmedian property of the real numbers!
     
  10. Dec 31, 2004 #9

    matt grime

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    A finite set can be ordered, and the maxmimum can be taken, and it will be in the set.
     
  11. Jan 1, 2005 #10

    mathwonk

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    of course hurkyl knows the archimedean proeprty of the reals is a corollary of the least upper bound property which we were given.

    for novices, the archimedean property says the integers are not bounded above, but if they were, then they would have a leat upper bound b, and then b-1 is not an upper bound for the integers, so some integer n is larger than b-1, but then n+1 is larger than b, a contradiction.
     
    Last edited: Jan 1, 2005
  12. Jan 1, 2005 #11

    Hurkyl

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    The reason I'm picky is that I hadn't worked out of your weaker version is right. i.e.:

    If G is an ordered group containing the rationals, and G has the LUB property, then G has the Archmedian property.

    (Of course, I know if you replace group with field it's true)


    But it's clear, now that I think of it -- it's fairly straightforward to embed R into G via Dedekind cuts, the LUB property guarantees there are no infinite numbers nor nonzero infinitessimals... and if G had something that wasn't in R, one could produce a nonzero infinitessimal number.
     
  13. Jan 2, 2005 #12

    mathwonk

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    i guess i am missing something. consider the following argument, as before:

    "the archimedean property says the integers are not bounded above, but if they were, then they would have a leat upper bound b, and then b-1 is not an upper bound for the integers, so some integer n is larger than b-1, but then n+1 is larger than b, a contradiction."

    this appears to me to use only least upper bounds, addition, and ordering, hence holds in any setting where those are true. now ,to pass from arbitrarily large integers to arbitrarily small rationals, needs the rationals to be there, but nothing else i can see. why introduce all that extra complexity of dedekind cuts, etc...?
     
  14. Jan 2, 2005 #13

    Hurkyl

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    Because I've been trying to work out some of the basics of real closed fields for the past couple months, so my mind was already in this gear.
     
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