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Unbounded sequence

  1. Jan 23, 2008 #1
    if the real sequence {[tex]x_n[/tex]} satisfies that for every m>n, [tex]|x_n-x_m|>\frac{1}{n}[/tex], can we prove it's unbounded?

    here is what i thought:
    let's suppose the sequence is bounded, then there is its sub-sequence who converges to some real number L. now the problem is converted to this: can a sequence converge if for every m>n, [tex]|x_n-x_m|>\frac{1}{n}[/tex]? i guess such sequence doesn't converge, yet i don't know how to put it in solid proof.
    can anyone help?
  2. jcsd
  3. Jan 24, 2008 #2
    well such a sequence that [tex]|x_n-x_m|>\frac{1}{n}[/tex]
    does not converge, it is sufficient to take epsylon = 1/n
    From the Cauchy theorem for the convergence of a sequence we have that:
    A sequence [tex](x_n)[/tex] is a convergent sequence if and only if for every epsylon >0, there exists some number N(epsylon)>0 such that for every two numbers n,m>N we have
    [tex]|x_n-x_m|<e[/tex], let e=epsylon,
    now if we take m>n, or n>m, whichever we want, we get a more convinient theorem, or whatever, for the convergence of a sequence. it goes like this
    A sequence [tex](a_n)[/tex] converges if and only if for every e>o there exists some N(e)>0 such that for every n>N, and for every p from naturals the following is fullfilled:
    here we basically have only taken m=n+p, or we could take n=m+p.
    Now the reason why a sequence of the form [tex]|x_n-x_m|>\frac{1}{n}[/tex]
    cannot converge is that according to cauchy's theoreme a convergent sequence should fulfill these requirements for any e, and also for [tex]e=\frac{1}{n}[/tex]. But since in you case it does not hold for e=1/n , consequently the sequence does not converge. Because from this notation [tex]|x_n-x_m|>\frac{1}{n}[/tex] we can read that we indeed cannot choose the terms of this sequence [tex](x_n)[/tex] to be arbitrarly close to one another, in other words we cannot make them as close as we want. This in particular is somehow the contrary(negation) of a cauchy sequence(of what cauchy theorem states) so the sequence diverges, and hence is not bounded.
    i hope i was clear enough.
  4. Jan 24, 2008 #3
    thanks, sutupidmath.
    but there is a question about the Cauchy's theorem for convergence. as you've mentioned, N(e) is determined by an arbitrary e>0, which we take here e=1/n, so N(e)+1 need not to be necessarily smaller than 1/e=n.
    thus the fact [tex]|x_{N(e)+p+1}-x_{N(e)+1}|>\frac{1}{N(e)+1}[/tex] does not necessariy contradicts the fact [tex]|x_{N(e)+p+1}-x_{N(e)+1}|<e[/tex], since it's possible that [tex]\frac{1}{N(e)+1}<e=\frac{1}{n}[/tex].
    so I still can't prove anything.
    can you tell me where i've got wrong? thanks.
  5. Jan 24, 2008 #4
    well, i don't actually know if i am getting you right, but this last part
    [tex]\frac{1}{N(e)+1}<e=\frac{1}{n}[/tex] defenitely cannot hold, and the reason is this, because the very first requirment is that n be rigorously greater than N(e).It is true that N(e)+1, need not necessarly be smaller than n=1/e, but it cannot defenitely be greater than it.
    Look at the definition of cauchy's sequence:
    A sequence converges if and only if for every e>o there exists some N(e)>0 such that for every n>N(e), and for every p from naturals the following is fullfilled:
    [tex]|x_(_n+_p)-x_n|< e [/tex]. So, from here n can only be equal or greater than N(e)+1, hence:[tex]\frac{1}{N(e)+1}>=e=\frac{1}{n}[/tex]. Now you might as well say, what if we take N(e)+100, than will this be true: :[tex]\frac{1}{N(e)+100}<e=\frac{1}{n}[/tex]?? I again argue that this cannot hold, since it should hold for all values of n greater than N(e), so it must also hold for lets say n=N(e)+200 which is not true. I do not know if i am gettin you right. But if the answer to this problem is really crucial to you, i mean if you really need it for an exam or something that plays an importan role, wait untill somebody else who has more experience and knowledge here comments on this.

    P.S. This is how i would go about this.
    Last edited: Jan 24, 2008
  6. Jan 26, 2008 #5
    In such a proof of a contradiction of the Cauchy theorem, you are free to chose any e you want, i.e. your choice of 1/n is valid one. But now you have to show also that for all N, in particular very large ones, there is a pair k, l>N such that [itex]|x_k-x_l|>1/n[/itex]. But for large N the condition on the sequence [itex]|x_k-x_l|>1/k[/itex] is too weak to beat 1/n.
    Last edited: Jan 26, 2008
  7. Jan 26, 2008 #6
    Here is my sketch of a proof. I leave out some technical details which (I think) I have correctly on paper, but they are too tedious to type in here.

    First we define the intervals [itex]u_n:=[x_n, x_n+1/n)[/itex] and [itex]d_n:=(x_n-1/n, x_n][/itex] where the round brackets denote an open side of an interval, and u and d stand for up-interval and down-interval.

    The techical bit left out is to show that each up-interval intersects at most with one down-interval as well as the other way around, independent of the order of the n and m indices involved (of course [itex]n\neq m[/itex]).

    Now we want to show that despite some possible overlap of the intervals, their union is not bounded. To do so, we construct a sequence of values [itex]v_i[/itex] the sum of which is no larger than the "area" covered by the intervals. And we will see that this sum does not converge.

    We start by initializing all [itex]v_i[/itex] with zero. For each [itex]i[/itex] we perform the following for [itex]l_i\in\{d_i, u_i\}[/itex], i.e. we iterate over the two intervals indexed with i:
    • If [itex]l_i[/itex] does not overlap with any interval, add [itex]1/i[/itex] to [itex]v_i[/itex].
    • If [itex]l_i[/itex] overlaps with a smaller [itex]l_k[/itex], i.e. [itex]i<k[/itex] then add [itex]1/(2i)[/itex] to both, [itex]v_i[/itex] and [itex]v_k[/itex]. Note that we distributed no more than the size of the interval to the two v. See below that we don't distribute anything from the overlapping partner to not overrepresent the union of the two intervals. Together they may cover much more "area", but they certainly cover not less then [itex]1/i[/itex].
    • If [itex]l_i[/itex] overlaps only with a larger other interval, leave [itex]v_i[/itex] alone. We know that it got already a value assigned that is greater than [itex]1/(2i)[/itex].

    Now we have a sequence [itex]v_i[/itex] bounded from below by [itex]1/(2i)[/itex] and the sum of the sequence is certainly smaller than the "area" covered by the union of all up- and down-intervals. But the sum is [itex]1/2\sum_{i=1}^{\infty} 1/i[/itex] which is not bounded.

    Consequently the union of the up- and down-intervals is not bounded. I consider it another technicality to show that this suffices for [itex]x_i[/itex] to be unbounded.

  8. Jan 26, 2008 #7
    Can't that pair be n+p, and n, where p is from naturals?
  9. Jan 26, 2008 #8
    Does this proof include a little bit of topology (or sth that is done in real analysis) or i just got a wrong impression?

    P.S. I am gonna have a close look at this, cuz i just skipped it and did not quite understand it. (my ignorance what can i do!).
    Last edited: Jan 26, 2008
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