Prove that if $(a_n)$ and $(b_n)$ are sequences such that $(|a_n - b_n|) \rightarrow 0$ then either: 1) both sequences are convergent to the same limit or 2) both sequences are divergent.

Is my following attempt at this proof correct?

The above statement can be logically rephrased as

$$(|a_n - b_n|) \rightarrow 0 \implies \left[(a_n) \rightarrow c \iff (b_n) \rightarrow c\right]$$

Let $\epsilon > 0$. Now, $(|a_n - b_n|) \rightarrow 0$ implies that there exists $N_1 \in \mathbf{N}$ such that $|a_n - b_n| < \epsilon /2$ for all $n \ge N_1$. Now assume $(a_n) \rightarrow c$ which implies that there exists $N_2 \in \mathbf{N}$ such that $|a_n - c| < \epsilon /2$ for all $n \ge N_2$. Set $N = \max\{N_1, N_2\}$, then whenever $n \ge N$ it follows that $|b_n - c| = |b_n - c -a_n+a_n| \le |a_n - b_n| + |a_n-c| < \epsilon/2 + \epsilon/2 = \epsilon$, hence $(b_n) \rightarrow c$. Now if we assume $(b_n) \rightarrow c$, then all we do is swap the $a_n$ and $b_n$ in the previous argument.

  • $\begingroup$ It's close. Your rephrasing isn't entirely correct, it states that if $|a_n-b_n| \to 0$ then $a_n \to c$ is equivalent to $b_n \to c$, but you haven't done anything about the divergent part. $\endgroup$ – B. Mehta May 26 '17 at 11:28
  • $\begingroup$ What would be the correct logical rephrasing? Because I thought after proving the iff part, the divergent part follows naturally from its negation. $\endgroup$ – SwiftMo May 26 '17 at 11:39
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    $\begingroup$ @B.Mehta Well, if $(a_n)$ is divergent, then $a_n\not\to c$ for all $c$, hence (under the given condition) $b_n\not\to c$ for all $c$, hence $b_n$ is divergent (and vice versa). $\endgroup$ – Hagen von Eitzen May 26 '17 at 11:52

Your proof is entirely correct, and more than that, it's well written.

  • $\begingroup$ It's true if you assume $c\in\mathbb R \cup\{-\infty,\infty\}$ right? $\endgroup$ – Oussama Boussif May 26 '17 at 11:29
  • $\begingroup$ @OussamaBoussif no, one of the $c$'s should (technically) be $d$. $\endgroup$ – 5xum May 26 '17 at 11:30
  • $\begingroup$ I'm sorry I don't get it, what's $d$? $\endgroup$ – Oussama Boussif May 26 '17 at 11:34
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    $\begingroup$ @OussamaBoussif The point is that the original statement just speaks of $a_n$ and $b_n$ as being convergent, not that they converge to the same point. So, it could be that they converge to two different points, $c$ and $d$. It's a minor technical detail (because it's actually easy to show they cannot converge to two different points) $\endgroup$ – 5xum May 26 '17 at 11:35
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    $\begingroup$ @5xum Sorry if I'm misunderstanding, but the original statement says "...are convergent to the same limit"? $\endgroup$ – SwiftMo May 26 '17 at 11:37

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