How do we prove the limit of a sequence of real numbers is unique? I wasn't sure if this proof was correct or not.
Proposition. If $(a_n) \to a$ and $(a_n) \to b$, then $a = b$.
Proof. Suppose $(a_n) \to a$ and $(a_n) \to b$.
Then, $\lim \limits_{n \to \infty}(a_n)=a$ and $\lim \limits_{n \to \infty}(a_n)=b$.
So for every $\epsilon > 0$ there exists a $N \in N$ such that $n>N$ implies $|a_n -a| < \frac{\epsilon}{2}$ and a $M \in N$ such that $n>M$ implies $|a_n-b| < \frac{\epsilon}{2}$.
So $|(a_n-a)+(a_n-b)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.
So $|2a_n - (a + b)| < \epsilon$.
So $\lim \limits_{n \to \infty} 2a_n = a + b$.
So $2\lim \limits_{n \to \infty} a_n = a + b$.
So $\lim \limits_{n \to \infty} a_n = \frac{a + b}{2}$.
Case 1: $\lim \limits_{n \to \infty} a_n = a$
If $\lim \limits_{n \to \infty} a_n = a$,
then $a = \frac{a+b}{2}$.
So $2a = a + b$ and $a = b$.
Case 2: $\lim \limits_{n \to \infty} a_n = b$
If $\lim \limits_{n \to \infty} a_n = b$, then $b = \frac{a+b}{2}$.
So $2b = a + b$ and $a = b$.
Therefore $a = b$.
 A: I suggest that you adapt your proof using the following trick at the beginning:
$$|a-b| = |a-b+(a_n-a_n)|= |(a-a_n)+(a_n-b)| \leq |a-a_n|+|b-a_n|$$
In particular you will get for all $n>M$ that $|a-b|\leq \epsilon$ which implies $a=b$, since $\epsilon>0$ is arbitrarily small.
A: Here is a better proof: If $a \neq b$. Assume $a < b$. Let $\epsilon = b - a$, then there is $N_1$ such that $n \ge N_1 \implies |a_n-a| < \dfrac{\epsilon}{2}$, and also $N_2$ such that $n \ge N_2 \implies |a_n - b| < \dfrac{\epsilon}{2}$. So if $n \ge N_1+N_2$, then $\epsilon \le |a_n-a|+|a_n-b| < \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2} = \epsilon$, contradiction. So $a = b$.
A: Claim
Let $(X,d_{X})$ be a metric space.
If $(x_{n})_{n\in\mathbb{N}}$ is a convergent sequence of points in $X$, then its limit is unique.
Proof
We are going to prove the desired claim by contradiction.
Let $(x_{n})_{n\in\mathbb{N}}$ be a sequence of points in $X$ which converge to $a$ and $b$, respectively, where $a\neq b$.
On the one hand, the first convergence means that
\begin{align*}
(\forall\varepsilon\in\mathbb{R}_{>0})(\exists n_{1}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{1} \Rightarrow d_{X}(x_{n},a) < \varepsilon)
\end{align*}
On the other hand, the second convergence means that
\begin{align*}
(\forall\varepsilon\in\mathbb{R}_{>0})(\exists n_{2}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{2} \Rightarrow d_{X}(x_{n},b) < \varepsilon)
\end{align*}
Since $a\neq b$, we can take $3\varepsilon = d_{X}(a,b)$.
Consequently, there is $n_{\varepsilon} = \max\{n_{1},n_{2}\}$ such that for every $n\in\mathbb{N}$ satisfying $n\geq n_{\varepsilon}$ results that:
\begin{align*}
d_{X}(a,b) \leq d_{X}(x_{n},a) + d_{X}(x_{n},b) < 2\varepsilon = \frac{2d_{X}(a,b)}{3} \Rightarrow d_{X}(a,b) < 0,
\end{align*}
which is clearly a contradiction, because the metric is always non-negative.
Hopefully this helps!
EDIT
Claim
Based on the suggestion of @MarekKryspin, consider a Hausdorff topological space $(X,\tau)$.
If $(x_{n})_{n\in\mathbb{N}}$ is a sequence of points in $X$ which converges, then the limit is unique.
Proof
Suppose the sequence $(x_{n})_{n\in\mathbb{N}}$ of points in $X$ converges to $a$ and $b$, respectively, where $a\neq b$.
Due to the definition of limits in topological spaces, the first convergence means that
\begin{align*}
(\forall N(a))(\exists n_{1}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{1} \Rightarrow x_{n}\in N(a)).
\end{align*}
Based on the same definition, the second convergence means that\begin{align*}
(\forall N(b))(\exists n_{2}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{2} \Rightarrow x_{n}\in N(b)).
\end{align*}
Once the $(X,\tau)$ is Hausdorff, we can choose neighborhoods $N(a)$ and $N(b)$ s.t. $N(a)\cap N(b) = \varnothing$.
But this leads to a contradiction, because $x_{n}\in N(a)\cap N(b)$ whenever $n\geq\max\{n_{1},n_{2}\}$.
Hopefully this helps!
