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Is that proof correct?

  1. Nov 17, 2008 #1
    In an analysis book i found this proof for the uniqness of a limit of a function.

    This is not homework

    "It is easy to show that when a limit of a function f(z) exists at a point a,it is unique.To do this ,we suppose that

    lim f(z) =l and lim f(z)=m as z----> a (z goes to a).

    Then,for any positive number ε,there are positive numbers r,δ such that

    [tex]\left| f(z)-l\right|[/tex]< ε whenever 0<[tex]\left|z-a\right|[/tex]< r


    [tex]\left|f(z)-m\right|[/tex] < ε whenever 0<[tex]\left|z-a \right|[/tex]< δ.

    So if 0< [tex]\left|z-a\right|[/tex]< θ ,where θ denotes the smaller of the two Nos r and δ,we find that

    [tex]\left|m-l\right|[/tex] = [tex]\left|(f(z)-l)-(f(z)-m)\right|[/tex] =< [tex]\left| f(z)-l \right|[/tex] + [tex]\left|f(z)-m\right|[/tex] < ε+ε =2ε.

    But [tex]\left|m-l\right|[/tex] is a nonnegative constant, and ε can be chosen arbitrarily small.

    l-m =0 , or l=m."

    is that proof correct??
  2. jcsd
  3. Nov 17, 2008 #2
    I wouldn't have been so quick with the last line. The above lines only indicate that the limit as z approaches a of the expression |m - l| is 0. Since |m-l| is a constant, this then implies that this limit is equivalent to the value |m - l|, as constant functions are continuous. The rest then follows, assuming you have already proven that constant functions are continuous.
    Another method of concluding this proof is to note that |l - m| is a fixed number > 0, while [itex]\epsilon[/itex] is free to be any value > 0. Thus, pick [itex]\epsilon[/itex] < |l - m|/2 to derive a contradiction.
  4. Nov 18, 2008 #3
    So you are saying that the above proof is wrong??

    Please write down ,if you can, a complete correct proof ,i will be very thankfully
  5. Nov 18, 2008 #4
    It's not completely wrong. It's better to say it's incomplete. It lays out all the majors steps in the proof, but on the last one, it really skimps on an important detail. That detail is a contradiction which allows you to assert m = l.

    To complete the proof, you'd want to say

    "Suppose [tex]|l - m| > 0[/tex]. Then, by replacing [tex]\epsilon[/tex] with [tex]\frac{1}{2} |l - m| [/tex], we see that

    [tex]|l - m| < 2 (\frac{1}{2} |l - m|)[/tex]

    Which is a contradiction. Therefore, our assumption [tex]|l - m| > 0[/tex] is false, and so [tex]|l - m| = 0[/tex] and so [tex]l = m[/tex].
  6. Nov 18, 2008 #5


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    if a function has 2 different limits, say a distance e apart, then eventually all the values of the function are very near both limits. but small nbhds of two different points are disjoint, so you are saying eventually all the values are in one nbhd and also in the other disjoint nbhd, an impossibility.
  7. Nov 19, 2008 #6
    Here is a diagram of what mathwonk says:
    http://img237.imageshack.us/img237/6198/hausdorffmetricmp7.png [Broken]
    Last edited by a moderator: May 3, 2017
  8. Nov 19, 2008 #7

    State the geometrical axioms or theorems that do not allow the above to happen

    i am very interested to see this kind of proof
    Last edited by a moderator: May 3, 2017
  9. Nov 19, 2008 #8


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    That's basically just a drawing of the proof that you posted in the original post.
  10. Nov 20, 2008 #9
    In laymens terms, a point is a limit (in any standard sense of the word) when it can be approximated with arbitrary accuracy. If there were two distinct limit points, you simply approximate within a tolerance less than the distance between them, and it has to converge on one or the other.

    In topology, this is the hausdorff property: for any two points, there is a neighborhood of each such that both neighborhoods are disjoint.
  11. Nov 20, 2008 #10
    Let [itex]B_x[/itex] be the ball around x and [itex]B_y[/itex] be the ball around y.

    If [itex]a_i[/itex] -> x then there exists N such that [itex]a_n \in B_x[/itex] for all n>N.

    If [itex]a_i[/itex] -> y then there exists M such that [itex]a_n \in B_y[/itex] for all n>M.

    Then for some large k, k>N and k>M, [itex]a_k \in B_x[/itex] and [itex]a_k \in B_y[/itex]. Since [itex]B_x[/itex] and [itex]B_y[/itex] are disjoint, this is a contradiction (disjoint sets share no points).
  12. Nov 21, 2008 #11
    But thats proving the uniqueness of the limit of a sequence,isn't it?"
  13. Nov 21, 2008 #12
    Ahh yes. But that's OK since, if the function converges, you can always construct a convergent sequence. For example, let a_n = a + 1/n.

    That is a nicety of metric spaces - you can show all sorts of things with sequences that aren't possible in more general situations.
  14. Nov 22, 2008 #13
    No it is wrong.

    What the Author of that proof really proves here is:

    If limf(z) = l and limf(z) = m as z goes to a ,then given ε>0 and also given

    0< [tex]\left|z-a\right|[/tex]< θ ,where θ denotes the smaller of the two Nos r and δ,

    THEN l=m.

    ...............and not.

    If limf(z) = l and limf(z) = m as z goes to a, then l=m

    Note the Author in the last step of his proof ,he assumes the theorem:

    IF, for all ε>o [tex]\left|m-l\right|[/tex]<ε ,then l=m,which Tac-Tics proves by using contradiction.

    Many of analysis books give a wrong proof following ,more or less the lines of the above proof.
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