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Math Help - Archimedean Property

  1. #1
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    Archimedean Property

    Is there a proof of the Archimedean Property that doesn't involve the words "least upper bound" or "completeness axiom"?

    Naturally, there should be since we use the Archimedean Property in the proof that R complete implies R has the least upper bound property.

    If we assume the least upper bound property, then it is easy to prove the Archimedean Property. However, it must be the case that by simply assuming that R is complete we can deduce the Archimedean Property. Otherwise, the usual proof that complete implies least upper bound would be circular.

    Any ideas on how to do this?

    Well, let's define R to be the set of equivalence classes of Cauchy sequences of rational numbers. With a little thought, we see that given this definition, we can approximate any real number by a sequence of rational numbers. If x < y are positive rational numbers, then by the previous comment we can find

    0 < (p/q) < x < y < (a/b)

    for some positive integers p,q,a,b. Then, set n = aq to obtain n(p/q) > a/b and therefore

    nx > n(p/q) > (a/b) > y ,

    as desired. This proof seems like it works and properly avoids mention of the least upper bound property. Any objections to this line of reasoning?
    Last edited by patrick; February 21st 2010 at 05:58 PM. Reason: Added some work
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by patrick View Post
    Is there a proof of the Archimedean Property that doesn't involve the words "least upper bound" or "completeness axiom"?

    Naturally, there should be since we use the Archimedean Property in the proof that R complete implies R has the least upper bound property.

    If we assume the least upper bound property, then it is easy to prove the Archimedean Property. However, it must be the case that by simply assuming that R is complete we can deduce the Archimedean Property. Otherwise, the usual proof that complete implies least upper bound would be circular.

    Any ideas on how to do this?

    Well, let's define R to be the set of equivalence classes of Cauchy sequences of rational numbers. With a little thought, we see that given this definition, we can approximate any real number by a sequence of rational numbers. If x < y are positive rational numbers, then by the previous comment we can find

    0 < (p/q) < x < y < (a/b)

    for some positive integers p,q,a,b. Then, set n = aq to obtain n(p/q) > a/b and therefore

    nx > n(p/q) > (a/b) > y ,

    as desired. This proof seems like it works and properly avoids mention of the least upper bound property. Any objections to this line of reasoning?
    WEll, unfortunately you can't have a proof without using the completeness of \mathbb{R} since the Archimedean principle in an equivalent concept.
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    Well, when I said "completeness axiom" I really meant the least upper bound property (sorry for the confusion).

    I think that the second thing that I wrote (the "proof") implicitly uses the completeness of \mathbb{R}, so it should work (though I could be missing something here).
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by patrick View Post
    Well, when I said "completeness axiom" I really meant the least upper bound property (sorry for the confusion).
    In this context they're synonyms.

    I think that the second thing that I wrote (the "proof") implicitly uses the completeness of \mathbb{R}, so it should work (though I could be missing something here).
    Before we get into that what may I ask is an equivalence class of Cauchy sequences? What is the equivalence relation? Is it that \{x_n\}\sim\{y_n\} if \lim\text{ }x_n=\lim\text{ }y_n (we of course have to define what this means).
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    Yes, the equivalence relation is that two sequences are equivalent if and only if the sequence of their differences is a null sequence.

    There is of course some work to do to show that this is well-defined, but the nice thing is that we get completeness through this definition (again, there are some details to write out). This is why I was differentiating between "completeness axiom" and "least upper bound property." I am assuming that we don't know that they are equivalent yet...
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by patrick View Post
    Yes, the equivalence relation is that two sequences are equivalent if and only if the sequence of their differences is a null sequence.

    There is of course some work to do to show that this is well-defined, but the nice thing is that we get completeness through this definition (again, there are some details to write out). This is why I was differentiating between "completeness axiom" and "least upper bound property." I am assuming that we don't know that they are equivalent yet...
    Ok. Let's restart. If we really want to do this right (get your questions answered) please answer three questions:

    1. What is your background?

    2. What is your goal?

    3. What subject are we looking at here. Are we looking at this from a topology or real analysis standpoint?
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  7. #7
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    Thanks for the reply. What I want to prove is the following:

    \mathbb{R} is complete implies \mathbb{R} has the least upper bound property.

    The way that I remember doing this when I took real analysis several years ago is the following:

    Let S be a non-empty subset of R bounded above by y. Then, choose x in S and set s0 = s-1. We have a non-empty interval [s0 , y]. Let a be the midpoint of [s0 , y]. If a is an upper bound for S, then let the second interval be [s0 , a]. Otherwise, let the second interval be [a , y]. Now continue inductively in this way to obtain a sequence of nested compact intervals

    I_{1}\supset I_{2}\supset \cdots .

    Now, the main point is that there exists a unique point which is in each of these intervals, which can be shown quite easily to be the desired least upper bound. However, in order to conclude that there is such a unique point, we need to know that the diameter of these nested intervals goes to 0.

    We clearly have \textrm{diam}\left(I_{n}\right) = 2^{-n}\textrm{diam}\left(I_{1}\right), so what we need to do is to show that \lim 2^{-n} = 0. We cannot conclude the latter fact without the Archimedean Property, and since we want to deduce the least upper bound property, we cannot use the LUB to prove the Archimedean Property. This is where my concern came in about being able to deduce the Archimedean property simply from the fact that \mathbb{R} is complete.

    Now, using the definition of R as such a set of equivalence classes, I have my "proof" above (which I'm not sure is correct but it seems OK).

    I am interesting in seeing whether my proof from before is correct and whether or not there is a slicker way of deducing the Archimedean Property directly from the completeness of \mathbb{R}. Analytical or topological approaches will be welcomed.

    Any insight would be much appreciated. Thanks for your time .
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by patrick View Post
    so what we need to do is to show that \lim 2^{-n} = 0. We cannot conclude the latter fact without the Archimedean Property,
    Hmm...I feel as though they're is a very convoluted way to do this using topology (which actually probably assumes it too). Also, I'm sure you'll get to a point where you need the LUB propertry but have you tried show that it's Cauchy and concluding that since a convergent sequence's limit is unique in a Hausdorff space and zero is...oh there is where you'd need it haha

    Oh, btw. Your proof (if I am understanding it correctly) is incorrect because saying there exists \frac{p}{q} such that x<\frac{p}{q} says that...what?
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    Quote Originally Posted by Drexel28 View Post
    Hmm...I feel as though they're is a very convoluted way to do this using topology (which actually probably assumes it too). Also, I'm sure you'll get to a point where you need the LUB propertry but have you tried show that it's Cauchy and concluding that since a convergent sequence's limit is unique in a Hausdorff space and zero is...oh there is where you'd need it haha
    That seems very strange. So essentially, there is no way to show that complete implies LUB? I mean, we can't use LUB to prove LUB.

    This is what bothers me about introductory real analysis texts: They seem to always assume the LUB and then prove completeness, which is easy, but it never seems to be shown rigorously that completeness implies LUB (in my opinion, if we just say that \lim 2^{-n}=0 without justification, it isn't rigorous).

    Oh, btw. Your proof (if I am understanding it correctly) is incorrect because saying there exists \frac{p}{q} such that x<\frac{p}{q} says that...what?
    There exists p/q and a/b rational such that 0 < (p/q) < x < y < (a/b). The rationals clearly have the Archimedean Property, so setting n = aq we get

    nx > n(p/q) > (a/b) > y ,

    hence for each a,b positive with a < b, there exists a natural number n such that na > b, so {n} is unbounded. Then, as 2^{n} > n, the sequence 2^{n} is unbounded and therefore

    \lim 2^{-n} = 0.

    I *think* that the main point is that by the completeness of \mathbb{R}, we get that there exists p/q satisfying 0 < (p/q) < a, so we can apply the Archimedean Property of the rationals to get the Archimedean Property of the reals...
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by patrick View Post
    That seems very strange. So essentially, there is no way to show that complete implies LUB? I mean, we can't use LUB to prove LUB.

    This is what bothers me about introductory real analysis texts: They seem to always assume the LUB and then prove completeness, which is easy, but it never seems to be shown rigorously that completeness implies LUB (in my opinion, if we just say that \lim 2^{-n}=0 without justification, it isn't rigorous).
    No! There are other ways, and in fact there is probably a way to prove that \lim\text{ }\frac{1}{2^n}=0 without it...but I'm lazy right now[/tex]

    There exists p/q and a/b rational such that 0 < (p/q) < x < y < (a/b). The rationals clearly have the Archimedean Property, so setting n = aq we get

    nx > n(p/q) > (a/b) > y ,

    hence for each a,b positive with a < b, there exists a natural number n such that na > b, so {n} is unbounded. Then, as 2^{n} > n, the sequence 2^{n} is unbounded and therefore

    \lim 2^{-n} = 0.

    I *think* that the main point is that by the completeness of \mathbb{R}, we get that there exists p/q satisfying 0 < (p/q) < a, so we can apply the Archimedean Property of the rationals to get the Archimedean Property of the reals...
    My point is that the REAL Archimedean Property says that given any x\in\mathbb{R} there exists some n\in\mathbb{N} such that x<n. If we assume there exists some \frac{p}{q}\in\mathbb{Q} then x<\frac{p}{q}<p (well not always...but you get the idea)
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    My point is that the REAL Archimedean Property says that given any x\in\mathbb{R} there exists some n\in\mathbb{N} such that x<n. If we assume there exists some \frac{p}{q}\in\mathbb{Q} then x<\frac{p}{q}<p (well not always...but you get the idea)
    I see your point; this seems so obvious that there is nothing to prove. I just don't like assuming these little points without justification.

    I guess if we define the reals to be the set of equivalence classes of Cauchy sequences (with the aforementioned equivalence relation), then for any x\in R, we can approximate x arbitrarily close by a rational number. Suppose that (p/q) is the rational number within distance e > 0 of x. If (p/q) > x, then we're done. If not, then

    y := 2x - (p/q)

    is strictly greater than x, but it is irrational. We then let d = ( x - (p/q) )/2 and we find a rational number (a/b) which is within d distance of y. Then, x < a/b <= a, and we have the desired property. Does *this* work?
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  12. #12
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    Quote Originally Posted by Drexel28 View Post
    In this context they're synonyms.



    Before we get into that what may I ask is an equivalence class of Cauchy sequences? What is the equivalence relation? Is it that \{x_n\}\sim\{y_n\} if \lim\text{ }x_n=\lim\text{ }y_n (we of course have to define what this means).
    Quote Originally Posted by patrick View Post
    Yes, the equivalence relation is that two sequences are equivalent if and only if the sequence of their differences is a null sequence.
    I don't think the "yes" is appropriate here because the fact that a sequence is a Cauchy sequence does NOT imply that it converges- except of course, in the real numbers. We say that \{a_n\}~\{b_n\} if and only if \{a_n- b_n\} converges. IF either of the sequences converges, THEN you can show that the other also converges and that they converge to the same thing.

    The distinction is particularly important if we are talking about constructing the real numbers by taking equivalence classes of rational sequences- the only reason we get something other than the rational numbers is that not all Cauchy sequences converge in the rational numbers.
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