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Inf(S) = 0 proof

  1. Oct 12, 2013 #1
    This is from the haggarty book on analysis

    take the set ## S := \{ 2^{-m} + 3^{-n} + 5^{-p} | m,n,p \in \mathbb{N} \} ##

    claim [itex] inf(S) = 0 [/itex]

    proof by contradiction:
    suppose ## A > 0 ## where A is the largest lower bound for S

    then ## 1/2^m + 1/3^n + 1/5^p \geq A ##
    i.e. ## 2^m + 3^n + 5^p \geq A(2^m3^n5^p) ##
    let ## X = 2^m + 3^n + 5^p ## and ## Y = 2^m3^n5^p ## where ## Y,X \in \mathbb{N} ##
    then ## X \geq AY ## hence ## Y \leq \frac{X}{A} ## which is a contradiction to the archimedean postulate which states for any real x a natural number n >= x hence the lower bound is 0

    is this proof OK or have I gone wrong somewhere?
     
  2. jcsd
  3. Oct 12, 2013 #2

    LCKurtz

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    I don't understand your contradiction. You made a hypothesis and just wrote it several different ways ending in ## Y \leq \frac{X}{A} ##. You haven't explained how it contradicts the Archimedian postulate. You also haven't observed that the glb is greater or equal 0.
     
  4. Oct 12, 2013 #3

    pasmith

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    I would prove it directly: show that for all [itex]\epsilon > 0[/itex] there exist [itex]m \in \mathbb{N}[/itex] such that [itex]2^{-m} < \frac13\epsilon[/itex], [itex]n \in \mathbb{N}[/itex] such that [itex]3^{-n} < \frac13\epsilon[/itex], and [itex]p \in \mathbb{N}[/itex] such that [itex]5^{-p} < \frac13\epsilon[/itex], and then add the inequalities together. (This assumes you've shown that 0 is a lower bound for S.)
     
  5. Oct 12, 2013 #4
    1.
    well we know the X and Y are natural numbers, and A is any number greater than 0, the archimedian postulate states that if we have a real number x, then any natural number n will be greater or equal to x. Here we have Y (which is a natural number), which we showed to be less than or equal to a natural number over a real number A>0, which is a contradiction to the archimedian postulate.

    2.
    we know the glb is more than or equal to 0 as m,n,p are natural numbers, hence as m,n,p get large 1/2^m + 1/3^n + 1/5^p will become small, but most certainly not less than 0 (it's a summation of rational numbers).

    Should I add 2. to the start of the proof, and 1. to the end to show it's a contradiction? Would that be more clear?

    thank you
     
  6. Oct 12, 2013 #5
    How would I prove 0 is a lower bound for S? It seems obvious but have no idea how to prove it

    e.g. can I just say 0 is a lower bound as S is never negative?
     
  7. Oct 12, 2013 #6

    LCKurtz

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    No, it doesn't say any such thing.
     
  8. Oct 12, 2013 #7

    DrClaude

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    I'll let the experts help you with the proof, but I'd like to understand how this
    can mean that
     
  9. Oct 12, 2013 #8
    The book I'm reading states this exactly:

    "Another way of phrasing the Archimedean postulate is to say that, given any real number x, there exists an integer n with n ≥ x"

    here we have the integer Y, and the real number X/A but Y ≤ X/A
     
  10. Oct 12, 2013 #9

    pasmith

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    The relevant ordered field axioms are:

    - For all a and b, if a < b then for all c > 0, ac < bc.
    - For all a and b, if a < b then for all c, a + c < b + c.

    Can you prove from those axioms that if a > 0 then [itex]a^n > 0[/itex] for all [itex]n \in \mathbb{N}[/itex], and that if a > 0 and b > 0 then a + b > 0?

    "There exists" means that at least one integer is greater than or equal to x, not that all integers are greater than equal to x. For example, [itex]\pi[/itex] is a real number, and there are three strictly positive integers which are less than [itex]\pi[/itex]. On the other hand [itex]4 > \pi[/itex], and there is no contradiction.
     
    Last edited: Oct 12, 2013
  11. Oct 12, 2013 #10

    LCKurtz

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    Do you not understand the difference between the statements "any natural number n will be..." and "there exists a natural number n with..."? The distinctions matter when you are constructing a proof.
     
  12. Oct 12, 2013 #11
    so the proof is not valid?

    I really don't understand

    if the archimedean postulate states there exists an integer greater than or equal to a real number, and I just showed the an integer is less than or equal to a real number then it's not a contradiction?

    I will show you one other solution in the book where they used the same method:

    Q: http://gyazo.com/af55675c5d3919c8af02a4dffbff69e7
    A: http://gyazo.com/e019c33314ce8a189b798ba2661fd437

    they also go to use the same method for the inf
     
  13. Oct 12, 2013 #12
    Also, isn't it a contradiction as I reached the statement for all Y <= X/A which is not true as there is a Y > X/A? (by the archimedean postulate)
     
  14. Oct 12, 2013 #13

    LCKurtz

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    DrClaude pointed out to you in post #7 that you have an algebra error in your argument that the rest of us overlooked.

    Regardless of that mistake, I can not make heads or tails of your reasoning, even if the algebra had been correct. So, no, I don't consider it a correct proof.

    Finally, pasmith showed you in post #3 the obvious way to work the problem. I suggest you think about that.
     
  15. Oct 13, 2013 #14
    OK going about it a pasmith's way:

    I previously proved in the question before that for any ## a \geq 2 ## ## a^n > n ##. This question also hints to use for any "## \epsilon > 0 \exists n \in \mathbb{N} ## s.t. ## 1/n < \epsilon ##. So with this we get ##n > 1/\epsilon ## so take for 3^n ## 3^n > n > 1/\epsilon ## so ## \epsilon > 1/3^n ## which would be the same as saying ## \frac{1}{3} \epsilon > 1/3^n ## and by similar arguments you'd get the same for 2^-m and 5^-p and adding this you get ## \epsilon > \frac{1}{2^m} + \frac{1}{3^n} + \frac{1}{5^p} ##

    but I don't understand how this shows the inf is 0?
     
  16. Oct 13, 2013 #15

    LCKurtz

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    Why would ##\epsilon > \frac 1 {3^n}## be the same as saying ## \frac{1}{3} \epsilon > 1/3^n ##? They don't look the same to me.

    You need to read post #3 again and actually try what he says.

    With respect to not seeing how this shows the inf is 0, what are the two things you need to show that inf(S) = 0? It would be good to have them carefully in mind if you are trying to prove it.
     
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