# Verification of a Proof: Uniqueness of Sequence Limit

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.

• 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? Jan 13 at 1:02
• @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. Jan 13 at 1:05
• 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 $\varepsilon>0$. Jan 13 at 1:09
• I see: misunderstanding of the quantifiers then. Thanks. Jan 13 at 1:10
• That's right. Your welcome! 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 $$\mathbb{R}$$.