# Limits are unique

$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$
Note that in the definition of convergence that the definite article is used for the limit, i.e., we have ‘the limit’ rather than ‘a limit’. A priori there is no reason to expect that a limit like that in the definition is unique. Well, it is and we prove that fact now.

First we need a lemma which should be intuitively obvious: if the difference between two numbers is less than all positive real numbers, then that difference is in fact zero and the numbers are equal.

Lemma
Suppose that $x$ and $y$ are real numbers such that $|x-y|\lt \varepsilon$ for all $\varepsilon >0$. Then $x=y$.

Proof
Suppose $x\neq y$. Then $|x-y|=\widetilde{\varepsilon}>0$ for some $\widetilde{\varepsilon}\in \R$. But then $|x-y|\lt \varepsilon$ is not true for $\varepsilon =\widetilde{\varepsilon}$ and this is a contradiction.

Now for the observation that limits are unique.
Theorem (Limits are unique)
Suppose that $a_n \to a$ and $a_n \to a’$. Then $a=a’$.

Proof
Let $\varepsilon >0$ be an arbitrary real number. Then there exists integers $N$ and $N’$ such that
$n>N \Longrightarrow |a_n-a|\lt \frac{\varepsilon }{2}$
and
$n>N’ \Longrightarrow |a_n-a’|\lt \frac{\varepsilon }{2} .$
(If the division by two is confusing, then see the remark following the proof.)

Hence, if $n>\max \{ N, N’ \}$, then
\begin{eqnarray*}
|a-a’|
&=& |a – a_n+ a_n -a’ | \\
&\leq & | a -a_n | + |a_n -a’| , {\text{ by the triangle inequality}}, \\
&=& |a_n -a| + |a_n -a’ | \\
&\lt & \frac{\varepsilon }{2} + \frac{\varepsilon }{2} \\
&=& \varepsilon .
\end{eqnarray*}
Since $\varepsilon$ was arbitrary, by the preceding lemma we deduce that $a=a’$.

Remark
The use of $\varepsilon /2$ in the proof is confusing for many students. The objection is that the definition of limit says that given $\varepsilon >0$, there exists an $N$ such that $n>N\Longrightarrow |a_n-a|\lt \varepsilon$. Why can we divide the $\varepsilon$ by $2$?

Well, imagine that you have a machine that takes $\varepsilon$’s and produces $N$’s. Now suppose that someone gives you an $\varepsilon$. You can put $\varepsilon /2$ into the machine (instead of $\varepsilon$) and out will come an $N$ with $n>N\Longrightarrow |a_n-a|\lt \varepsilon /2$. In other words we have the situation that given $\varepsilon >0$, there exists $N$ such that $n>N\Longrightarrow |a_n-a|\lt \varepsilon /2$.

In fact, for a convergent sequence, given any $\varepsilon$ we can find $N$ such that $n>N\Longrightarrow |a_n-a|\lt \varepsilon /3$, or less than $\varepsilon /4$, or $7\varepsilon$, or $\sqrt{\varepsilon}$, or $\varepsilon ^{18}/25000$ or whatever.