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Uniqueness of limit by transitive property?

  1. Jan 8, 2010 #1
    [for all of the following, "lim" means the limit as n->∞]
    Let an be a sequence of real numbers.

    Theorem: if lim an = L and lim an = M, then L=M.

    (Incorrect) "Proof":
    lim an = L and lim an = M
    Thus, L = lim an = lim an = M (transitive property)
    Therefore, L=M.

    To me, every step in the proof seems to be justified and correct.
    Can someone please explain what is wrong with this proof?
     
  2. jcsd
  3. Jan 8, 2010 #2
    You just stated (not proved) that, if a = b and b = c, then a = c. BUT, if you're trying to prove the uniqueness of limits, then this won't do, because you are assuming that something, called "lim an", is equal to both L and M, and this is precisely what you want to prove, so you cannot assume it.
     
  4. Jan 8, 2010 #3

    Hurkyl

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    This is pattern is not uncommon. First, we give a meaning to the formal expression
    [tex]\lim_{n \rightarrow \infty} a_n = L[/tex]​
    but only to the entire expression -- we do not assign any meaning to any subcollection of symbols. [itex]n \rightarrow \infty[/itex] doesn't get a meaning. [itex]a_n = L[/itex] doesn't get a meaning. [itex]\lim_{n \rightarrow \infty} a_n[/itex] doesn't get a meaning. Only the expression in its entirety gets a meaning.


    Then, we prove uniqueness.

    Once when we have proven uniqueness can justify giving a meaning to the expression [itex]\lim_{n \rightarrow \infty} a_n[/itex] on its own.
     
  5. Jan 8, 2010 #4
    Indeed, in sufficiently screwed up topological spaces (i.e. non-Hausdorff), limits need not be unique.
     
  6. Jan 8, 2010 #5
    Parent(me)=my father.
    Parent(me)=my mother.
    Therefore, my mother=my father. No wonder I'm so screwed up...
     
  7. Jan 8, 2010 #6
    But if a,b,c are "real numbers", then a=b AND b=c => a=c. (this is definitely true)

    Now lim a_n = L and lim a_n = M => lim a_n exists and must be equal to a real number. Therefore, lim a_n, L, and M are all real numbers, and by the transitive property above, L=M ??????
     
  8. Jan 8, 2010 #7
    Think about my post. It's pretty stupid, but it basically explains where your reasoning is wrong.
     
  9. Jan 8, 2010 #8
    You're assuming what you're trying to prove.
     
  10. Jan 8, 2010 #9
    hmm...I don't think so.
    I am simply assuming the hypothesis of the theorem and using the transitive property of the real numbers.
     
  11. Jan 8, 2010 #10
    A priori, you know that the limit exists (for the sake of this argument, we assume this at least), but not that it is single valued, which is what you're assuming. Take the set A={1,2} with the trivial topology t={{},{1,2}}. Then the sequence a_k=1 converges to both 1 and 2.
     
  12. Jan 8, 2010 #11
    I disagree with the first two equalities.
    Equality is "if and only if" kind of statement. To use the equal sign, it has to be true both ways.
     
  13. Jan 8, 2010 #12
    Fine, then change it to "a parent of mine is my father" and so on. But the little details don't matter. You're missing the point that until you prove otherwise, there is no reason that a sequence can't have two different limits.

    You're saying the limit is so and so and the limit is also so and so, when you should be saying "a limit is" and then proving that it is indeed ok to say the limit. I know you never used those exact terms, but that's what's going on in your proof, only it's more subtle.
     
    Last edited: Jan 8, 2010
  14. Jan 9, 2010 #13
    The equality that we're all used to is an equivalence relation on [tex] \mathbb{R} \times \mathbb{R} [/tex]. If you wanted to use this notion of equality (and hence, use its transitivity property) in the expression "lim a_n = L", you'd have to know that [tex] \lim a_n \in \mathbb{R} [/tex]. You'd have to know that "lim a_n" denoted a SINGLE element of [tex]\mathbb{R}[/tex]. But that's precisely what you're trying to prove; hence, the circularity arises.

    Therefore, since this is not an equivalence relation in the normal sense, there is NO transitivity property. What if the sequence converged to more than one point (which, as has already been pointed out, can happen in some spaces), then would "lim a_n" be a set? If it were a set, what would "lim a_n = L" mean?
     
  15. Jan 9, 2010 #14
    Remember that lim an=L is just an abbreviation to the long defintion "for every e>0...."
    and not vice versa. Before you've proven it, you can't assume that limits act as a function you're used to, and that's exactly what you're trying to prove here.
     
  16. Jan 9, 2010 #15

    HallsofIvy

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    Yes, you are assuming what you want to prove. The "common term" in your argument "lim an= a, lim an= b" is lim an. To use the "transitive property of the real numbers", you must assume that "lim an" is a specific number- and that is what you are trying to prove.

    Would you consider
    "The number whose square is 4"= 2,
    "The number whose square is 4"= -2,
    therefore 2= -2.
    a valid argument? The error in it is exactly the error in your argument.
     
    Last edited by a moderator: Jan 9, 2010
  17. Jan 9, 2010 #16
    Well said, HallsofIvy.
     
  18. Jan 10, 2010 #17

    Landau

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    Some elaboration (sorry if this is redundant):
    Obviously the mistake in this argument is in the use of the word "the". When you say "the number whose square is 4" you implicitly say that there is a UNIQUE number with this property. Correct would be:

    "a number whose square is 4"= 2,
    "a number whose square is 4"= -2,
    but 2=-2 cannot be concluded from this.

    Likewise, in your false argument your are implicitly assuming that "lim an" is a UNIQUE number:

    "The number which equals lim a_n"=L
    "The number which equals lim a_n"=M
    therefore, L=M.

    Of course you know lim a_n is a number, but you don't know that it is unique! So, correct would be:

    "a number which equals lim a_n"=L
    "a number which equals lim a_n"=M
    but the conlusion L=M does not follow.

    The uniquess assumption is in fact precisely what you're trying to prove.
     
  19. Jan 11, 2010 #18
    Ok, I really wanted to place my hand in the pot on this one :) I really think the problem we see here is in the abuse of notation when we use equals in the following:

    lim n->inf a_n = L

    We see that once we have the equals sign in our notation, we feel that we can use all of the natural properties of equals in or calculations, including the transitivity property. However, this notation is saying nothing more than that L is a limit of the sequence a_n, defined by our usually epsilon-delta argument.

    The question at hand is really asking for a justification for using the equals sign, provided that the limit exists. Keeping that in mind, we see that we may not use any properties of the equals sign (like transitivity), while trying to prove the uniqueness of limits.

    Another way we can view this question is as such, we temporarly define two classes of (convergent) sequences, one that has multiple limits, and one that has only one limit. Show that the former (the set of all sequences that have multiple limit points) is empty. With this approach we are forced to assume that our sequence has multiple limit points, at which the use of equality with a real number is just meaningless... from which we would have to use set equality.

    So we can do this by;

    Suppose L is in lim a_n, and M is in lim a_n (note the use of set terminology, and that now lim a_n is considered as a set). Show that L = M. By which we can show that every convergent sequence has a singleton set for limits (ie, a unique limit point).

    I don't know if I went too far afield with this explanation, but i thought it really might make the proper distinctions to illustrate the question at hand.
     
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