I've been reading Abbott's Understanding Analysis with Tao's Analysis I during my university's quarantine period (finally starting freshman year six months late! our class's ten month summer has come to an end. Thanks COVID). I'm still rather early on; an exercise from Abbott's asks the reader to prove the uniqueness of a limit of a sequence, and, after a good bit of searching to quell my lack of confidence in my work, I've noticed that my attempt is significantly different from the "canonical" approach by contradiction + triangle inequality, etc. Is this wrong? Is it necessarily weaker?

Take $a,b$ and a sequence $a_n$ such that $\lim_{n \to \infty} a_n = a$ and $\lim_{n \to \infty} a_n = b$. By definition, it follows that, for all $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that, $\forall n > N$, $|a_n - a| < \epsilon$ and $|a_n - b| < \epsilon$. From Theorem 1.2.6,** this is equivalent to saying that $\lim_{n \to \infty}a_n = a$ and $\lim_{n \to \infty}a_n = b$. It quickly follows that $a=b$. $W^5$

Theorem 1.2.6 from Abbott (paraphrased, slightly): $a,b \in \mathbb{R}$ are equal $\iff$ $\forall \epsilon \in \mathbb{R}$, such that $\epsilon > 0$, $|a-b| < \epsilon$.

If anything, I suppose I'm just worried that I'm misinterpreting and/ or misusing 1.2.6; however, the symmetry between that and the convergence definition made it seem obvious. Thanks in advance for your time and feedback.

  • $\begingroup$ Yes, if for any two real numbers $x$ and $y$ we have that $|x-y|$ is less than every positive number, then $x=y$. Now, if $\lim_n a_n = a$ and $\lim_n a_n = b$, given $\varepsilon>0$ we have that there exists some positive integer $N$ (that depends on $\varepsilon$) such that the following holds: for $n \geq N$, $|a_n-a| < \varepsilon$ and $|a_n-b| < \varepsilon$. How is exactly that you can conclude that $|a_n-a|$ and $|a_n-b|$ (also, what is $n$ here?) are less than every positive number? $\endgroup$
    – azif00
    Jan 13 at 1:02
  • $\begingroup$ @azif00 "also, what is $n$ here?" >> some index greater than $N$. Essentially copying down the limit/ convergence definition there. As for the rest, that's basically why I'm here; I think I'm starting to see that this is a misunderstanding of the $\epsilon$ selection portion of the definition, but I'm still unconfident. $\endgroup$
    – mmmeeeker
    Jan 13 at 1:05
  • $\begingroup$ So, if $n$ is some index $\geq N$, $|a_n-a|$ and $|a_n-b|$ are less than every positive number? No, that's not correct, $|a_n-a|$ and $|a_n-b|$ are just less than $\varepsilon$, note that the existence of $N$ depends on $\varepsilon$, so you cannot conclude that $|a_n-a|$ and $|a_n-b|$ are less than $\varepsilon$ for every $\varepsilon>0$. $\endgroup$
    – azif00
    Jan 13 at 1:09
  • $\begingroup$ I see: misunderstanding of the quantifiers then. Thanks. $\endgroup$
    – mmmeeeker
    Jan 13 at 1:10
  • $\begingroup$ That's right. Your welcome! $\endgroup$
    – azif00
    Jan 13 at 1:11

Hint (1): $\lvert x - y \rvert \le \lvert x - z \rvert + \lvert z - y \rvert $

Hint (2): If $(x_n)$ is a sequence in $\mathbb{R}$ such that $x_n \to x$ then $\forall \epsilon > 0,\exists N \in \mathbb{N}, \forall n \ge N$ we have $\displaystyle \lvert x_n - x \rvert < \frac{\epsilon}{2}$.

These two hints and the application of the theorem you have listed can be used to give the uniqueness of limit of a convergent sequence in $\mathbb{R}$.


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