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

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 $$\epsilon$$?

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 $$|l_1-l_2|=0$$.
• Thanks, but I am still confused. This question arised after reading the OP solution here math.stackexchange.com/questions/4423777/… OP's argument and Abbott's one seems the same to me, but OP's one is consided wrong. Maybe they differ because here $l_1$ and $l_2$ are fixed real numbers while in the linked question $f(x_1)$ and $f(x_2)$ depends on $\epsilon$ from the inequalities $x_1,x_2 \ge M_\epsilon$, so they don't obey to $\forall \epsilon |a-b|<\epsilon \implies a=b$ because $f(x_1)$ and $f(x_2)$ are not fixed?
• I would love to answer to that, but you've done it your self. In the proof from my answer, when I reach $|l_1-l_2|<\varepsilon$, the number $N$ is not mentioned anymore. But in that other question, you cannot jump from $|f(x_1)-f(x_2)|<\varepsilon$ to $|f(x_1)-f(x_2)|=0$ because $x_1$ and $x_2$ are not fixed numbers. They were chosen depending on $\varepsilon$. Commented Apr 25, 2022 at 11:53