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Real Analysis: show sequences have the same limit if |Xn-Yn| approaches 0

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose {Xn}, {Yn} are sequences in ℝ and that |Xn-Yn|→0. Show that either: a) {Xn} and {Yn} are both divergent or b) {Xn} and {Yn} have the same limit.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I first prove that lim(Xn-Yn)=lim(Xn)-lim(Yn). I am not completely clear on the validity of this proof.
    Since |Xn-Yn|→0, |Xn-Yn-(x-y)|<ε. Now |Xn-Yn-(x-y)|=|(Xn-x)-(Yn-y)|≥|Xn-x|-|Yn-y| (from the triangle inequality).
    So, |Xn-x|-|Yn-y|<ε. Therefore, the lim(Xn-Yn)=lim(Xn)-lim(Yn).

    So, we have that lim(Xn-Yn)=lim(Xn)-lim(Yn). Then, if I assume Xn→x, then 0=x-lim(Yn). So, lim(Yn)=x. So, they have the same limit. If Xn does not have a limit, then both {Xn} and {Yn} must be divergent.

    I would like to know if this is the correct approach to take and if the proof is correct. Thank You.
     
  2. jcsd
  3. Oct 11, 2011 #2

    SammyS

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    What you really need to prove are the following:

    1) If one of them, say {xn}, converges, then so does the other.

    2) If one of them diverges, then so does the other.

    You can alternatively prove this by contradiction as follows. Assume one of them converges, and one of them diverges. Then show that this,
    together with limn→∞|xn - yn|=0, leads to a contradiction.
     
    Last edited: Oct 11, 2011
  4. Oct 11, 2011 #3
    But you still have to prove that if both of them converge that they converge to the same value?
    I don't quite understand what is implied by limn→∞|xn-yn|=0. Does this mean the lim(xn)→lim(yn)?
     
  5. Oct 11, 2011 #4

    SammyS

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    No, you do.

    limn→∞(xn-yn)=0 means |Xn-Yn|→0
     
  6. Oct 11, 2011 #5
    I meant you as a general you, meaning whoever is doing the proof, meaning me. Anyway, I'm not looking for answers so calm down. I'm just trying to understand the steps of the proof. Does limn→∞(xn-yn)=0 mean that xn and yn converge to the same value (namely, that x=y)?
     
  7. Oct 11, 2011 #6
    Also, I don't understand what would be wrong with proving lim(xn-yn)=lim(xn)-lim(yn) [which I now figured out how to prove correctly], and then say that assuming xn→x then 0=x-lim(yn) and lim(yn)=x. Otherwise, if xn does not approach x, then of course yn is also divergent.
     
  8. Oct 11, 2011 #7
    nyr:

    Assume for a moment that the difference xn-yn is always positive, or nonnegative. What can you then say about Limn→∞|xn-yn|? Consider then all other possible cases re the difference xn-yn
     
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