Ok so I know I've seen this before phrased differently but I can't quite put my finger on the solution. The question goes as follows:

Let $a_n, b_n$ be two converging sequences.

Also, let $\{n: n\in\Bbb{N}, a_n \leq b_n\} $ and $\{n: n\in\Bbb{N}, b_n \leq a_n\} $ be two unbounded sets.

Prove that $\lim\limits_{n\to \infty}a_n = \lim\limits_{n\to \infty}b_n$

I'm completely drawing a blank. I tried several approaches I know, such as Cantor's intersection theorem (which doesn't seem to fit) or other properties of converging sequences I've learned, but the question - as posed - doesn't provide enough detail (that I can see) to use any of those as far as I can tell.

Can anyone shed some light on this for me?


Edit (possible solution?) After some page-turning I found a similar proof in my professor's slides and adapted it to this question. Let me know if you think it's valid:

First we'll show that if $a_n \leq b_n $ for every $n$, then $\lim\limits_{n\to \infty}a_n \leq \lim\limits_{n\to \infty}b_n$:

Assuming by contradiction that if if $a_n \leq b_n $ for every $n$, then $\lim\limits_{n\to \infty}a_n > \lim\limits_{n\to \infty}b_n$, there then exists an $n$ in $a_n$ for which $a_n>b_n$, in contradiction to the fact that $a_n \leq b_n $ for every $n$, and therefor if $a_n \leq b_n $ for every $n$, then $\lim\limits_{n\to \infty}a_n \leq \lim\limits_{n\to \infty}b_n$.

We use the same process to prove the opposite direction, and since we have two infinite sets of indices for which $a_n\geq b_n$ and $b_n \geq a_n$, therefor $\lim\limits_{n\to \infty}a_n \leq \lim\limits_{n\to \infty}b_n$ and $\lim\limits_{n\to \infty}b_n \leq \lim\limits_{n\to \infty}a_n$ - which means $\lim\limits_{n\to \infty}a_n = \lim\limits_{n\to \infty}b_n$ - Q.E.D.

  • $\begingroup$ What do you know about the subsequences of a convergent sequence? $\endgroup$ – Daniel Fischer Apr 5 '15 at 8:47
  • $\begingroup$ Technically we haven't covered that topic yet, but I know from previous courses that all sub-sequences of a convergent sequence have the same sub-limit, except I can't use that because we haven't even touched the term 'subsequence' yet in class. $\endgroup$ – Elad Avron Apr 5 '15 at 8:49
  • $\begingroup$ It just occured to me that the mentioned sets are not of members in the sequences but of INDICES. Could this help me in any way? I can't seem to think how. $\endgroup$ – Elad Avron Apr 5 '15 at 8:50
  • $\begingroup$ But if you could use subsequences, would you see which subsequences you could take to reach the desired conclusion? If so, you can translate that idea into a proof not mentioning subsequences. $\endgroup$ – Daniel Fischer Apr 5 '15 at 8:51
  • 1
    $\begingroup$ As you emphasised above: indices of subsequences. For subsets of $\mathbb{N}$, "unbounded" and "infinite" are equivalent, so if one of the two sets were bounded, we wouldn't have the corresponding subsequence. $\endgroup$ – Daniel Fischer Apr 5 '15 at 8:57

Let the limits be $a$ and $b.$

Fix $\epsilon > 0.$ There exist $N_{\epsilon}$ such that for $n>N_{\epsilon}$ $$ |a_n- a | < \epsilon, \; |b_n-b| < \epsilon. $$ or alternatively $$ a-\epsilon < a_n < a+\epsilon, \; b-\epsilon < b_n < b+\epsilon. $$ There is some such $n$ with $a_n < b_n.$ So $$ a-\epsilon < b + \epsilon. $$ Similarly, $$ b-\epsilon < a + \epsilon. $$ So $$ |a-b | < 2\epsilon. $$ But $\epsilon$ was arbitrary, so $a=b.$

I call this pattern arbitrary closeness. (see my book "proof patterns")


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