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Basic question regarind archimedian property

  1. Feb 9, 2005 #1
    I need to prove that if I := [0, 1/n], the element 0 belongs to all I(n) and the archimedian property to show that 0 is the only common point.

    so basically, I need to prove the intersection from 1 to infinite of I(n) = {0}.

    the book says to use the archimedian property that if t > 0, then tehre exsists n(t) such that 0 < 1/n(t) < t.

    i've been thinking all day, but still can't go anywhere.

    ok, so I guess to start off,

    let t > 0, this implies that that there is a n in N, such that t < n(t).

    so this means 1/n(t) < t.

    and I guess by looking at the archimedian property that 0 < 1/n(t) < t, this means that 1/n(t) is always less then any given t. so I guess if it is less then t, then obviously, t is not included with any 1/(t + 1), and it is always greater then 0, so I guess that's why 0 is the only thing in common w/ all of them.

    notice I said "I guess" a lot, because I really dont believe my proof......so can someone give me some tips? am I close?

    and then, part 2:

    same question, but with (0, 1/n) and need to prove that the intersection is the empty set. both are confusing
     
    Last edited: Feb 9, 2005
  2. jcsd
  3. Feb 9, 2005 #2
    Seems like you were on the right track.

    (I'll use )0, 1/n( to denote [0, 1/n], since the latter seems to mess up Latex).

    You want to show that [itex]\bigcap^{\infty}_{n = 1} )0, 1/n( = \{0 \} .[/itex]

    The hard part is [itex]\bigcap^{\infty}_{n = 1} )0, 1/n( \subseteq \{ 0 \}[/itex], the other inclusion is trivial.

    So, suppose [itex]x \in \bigcap^{\infty}_{n = 1} )0, 1/n([/itex] (that set will be referred to as "the set"). x can't be negative, so we have x >= 0. If x = 0, we are done. If x > 0, take an integer n such that nx > 1. Since x was in the set, we must have [itex]x \in )0, 1/n([/itex], so that 0 < x < 1/n. But then nx < 1, which contradicts nx > 1. Thus, there are no non-zero numbers in the set.
     
    Last edited: Feb 9, 2005
  4. Feb 9, 2005 #3

    matt grime

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    Hmm, review the definitions etc.

    1. the intersection is the set of all elements common to all the sets. Obviously 0 is in all the sets, and the intersection contains only positive numbers.

    2. Given any number, t, strictly greater than 0 there is an integer n, dependent on t, such that 0<1/n<t, ie t isn't in the inteval [0,1/n]

    3. If the intersection contained anything other then 0, it contains a strictly positive t, so by 2. we see no such t exists. Hence the intersection only contains 0.

    So, where's the problem? I just wrote out the information in your question and it's the proof you require.
     
  5. Feb 9, 2005 #4
    hm....still not understanding quite yet.

    I understand number 1..
    on number 2, is ok
    but I still dont quite follow you on number 3. what do you mean that t doesn't exist if it is in the intersection, and again....can you expalin why??
     
    Last edited: Feb 9, 2005
  6. Feb 9, 2005 #5

    matt grime

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    Suppose that t is strictly positive and in the intersection. Got it? ASSUME that is true....

    then since by the archimedian principle there is an n in N suc that 1/n <= t t is not in [0,1/n]

    This is a contradicion, so something we assumed is wrong. Let's look, what did we ASSUME....

    (apologies if the idea of proof by contradiction is alien, it can be done without contradiction)
     
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