Question about a proof in uniqueness of limit in Abbott's "Understanding Analysis"

Question

In Abbott's textbook "Understanding Analysis", first edition, exercise 2.3.4 (page 49), it is asked: "Show that limits, if they exist, must be unique. In other words, assume $\lim a_n=l_1$ and $\lim a_n=l_2$ and prove that $l_1=l_2$."

In its solutions manual, the proof is the following: "Let $\epsilon>0$, we know that $\lim a_n=l_1$ so there exists $N_1 \in \mathbb{N}$ such that $n \ge N_1 \implies |a_n-l_1|<\epsilon/2$. Similarly, since $\lim a_n=l_2$ there exists $N_2 \in \mathbb{N}$ such that $n \ge N_2 \implies |a_n-l_2|<\epsilon/2$. Setting $N=\max\{N_1,N_2\}$, gives us that for $n \ge N$ it is $|l_1-l_2|\le|a_n-l_1|+|a_n-l_2|<\epsilon/2+\epsilon/2=\epsilon$, thus $|l_1-l_2|<\epsilon$. From the fact that two real numbers $a$ and $b$ are equal if and only if for every real number $\epsilon>0$ it follows that $|a-b|<\epsilon$, it is $l_1=l_2$."

My question is the following: the indexes $N_1$ and $N_2$ are actually $N_1=N_1(\epsilon)$ and $N_2=N_2(\epsilon)$, and so it is $N=\max\{N_1,N_2\}$, that is $N=N(\epsilon)$. I always seen the aforementioned result on $a$ and $b$ as something like "when $\epsilon$ become smaller and smaller, the distance between $a$ and $b$ becomes arbitrarily small and so they must coincide"; however, in this case, the estimations on the distance $|l_1-l_2|$ hold only when $n \ge \max\{N_1,N_2\}$, the latter depending on $\epsilon$; so how can Abbott be sure that, as $\epsilon$ varies to become smaller and smaller (and so does $N$) the condition $n \ge \max\{N_1,N_2\}$ is still verified since $\max\{N_1,N_2\}$ varies with $\epsilon$?

Answer 1

What Abbott proves is that, for each $\varepsilon>0$, $|l_1-l_2|<\varepsilon$. In order to do this, he picks $N_1,N_2\in\Bbb N$ such that $n\geqslant N_1\implies|a_n-l_1|<\frac\varepsilon2$ and that $n\geqslant N_2\implies|b_n-l_2|<\frac\varepsilon2$. And, yes, the numbers $N_1$ and $N_2$ depend upon $\varepsilon$. Then, if $N=\max\{N_1,N_2\}$,$$n\geqslant N\implies|a_n-l_|,|b_n-l_2|<\frac\varepsilon2\implies|l_1-l_2|<\varepsilon.$$Again, yes, $N$ depends upon $\varepsilon$. But that's irrelevant. What matters is that Abbott has proved that, for each $\varepsilon>0$, $|l_1-l_2|<\varepsilon$. At this point of the proof, $N_1$, $N_2$, and $N$ are gone. And now, since $|l_1-l_2|\geqslant0$ and since there is nor real number greater than $0$ which is smaller than any other real number greater than $0$, we must have $|l_1-l_2|=0$.