Prove the limit of a sequence is unique in a metric space Let $(X,d)$ be a metric space, and let $(x^{(n)})_{n=m}^{\infty}$ be a sequence in $X$. Suppose that there are two points $x,x'\in X$ such that $(x^{(n)})_{n=m}^{\infty}$ converges to $x$ with respect to $d$, and $(x^{(n)})_{n=m}^{\infty}$ also converges to $x'$ with respect to $d$. Then we have that $x = x'$.
MY ATTEMPT
Suppose otherwise that $x\neq x'$. Consequently we can choose $\varepsilon = d(x,x')/3 > 0$. 
Thus, according to the definition of limit, for such $\varepsilon$, there are $N_{1}\geq m$ and $N_{2}\geq m$ such that
\begin{align*}
\begin{cases}
n\geq N_{1} \Rightarrow d(x^{(n)},x) \leq \varepsilon\\\\
n\geq N_{2} \Rightarrow d(x^{(n)},x') \leq \varepsilon 
\end{cases}
\end{align*}
Thus, if we take $n\geq N = \max\{N_{1},N_{2}\}$, the next result follows from the triangle inequality
\begin{align*}
d(x,x') \leq d(x^{(n)},x) + d(x^{(n)},x') \leq 2\varepsilon & \Rightarrow d(x, x') \leq \frac{2d(x,x')}{3}\\\\
& \Rightarrow d(x,x') \leq 0 \Rightarrow d(x,x') = 0
\end{align*}
whence we conclude that $x = x'$, which contradicts our assumption
Consequently $x = x'$ and we are done.
Could someone please verify if I am reasoning correctly?
 A: Offering another proof which doesn't use contradiction.
Fix $\epsilon > 0$. Choose a natural number so large that
$$d(x^{(n)}, x) \leq \epsilon, \quad d(x^{(n)}, x') \leq \epsilon$$
Then from the triangle inequality we get $$d(x,x') \leq 2 \epsilon$$
We thus have proven that 
$$\forall \epsilon > 0: d(x,x') \leq 2\epsilon$$
which implies (why?) that $d(x,x') = 0$, i.e. $x=x'$.
A: Suppose ${x_n}$ is a convergent sequence which converges to
two different limits $x\neq y$. Then $ε =\frac{1}{2}
d(x, y)$ is positive, so there
exist integers $N_1$, $N_2$ such that
$d(x_n, x) < ε$ for all $n ≥ N_1$,
$d(x_n, y) < ε$ for all $n ≥ N_2$.
Setting $N= max\{N_1, N_2\}$ for convenience, we conclude that
$2ε = d(x, y) ≤ d(x, x_n) + d(x_n, y) < 2ε$
for all $n ≥ N$. This is a contradiction, so the limit is unique.
A: You shouldn't use something like
$$
\begin{cases}
n\geq N_{1}\\\\
n\geq N_{2}
\end{cases}\Longrightarrow
\begin{cases}
d(x^{(n)},x) \leq \varepsilon\\\\
d(x^{(n)},x') \leq \varepsilon 
\end{cases}
$$
It's better to state the two statements separately. Finally you take $N\ge\max\{N_1,N_2\}$, but then use an unspecified $n$. Did you mean $n$ in both places?
Otherwise your proof is good.
Your definition of limit seems to use $\le$ instead of the more common $<$. Not a big deal, the two definitions are completely equivalent. With $<$, you can choose $\varepsilon=d(x,x')/2$, with $\le$ you need to choose something smaller. It's even easier with $d(x,x')/4$, but dividing by $3$ is good as well.

Suppose the sequence converges at $x$ and $x'$, with $x\ne x'$. Set $\varepsilon=d(x,x')/4$. Then, by definition of limit, there exist $N_1\ge m$ and $N_2\ge m$ such that


*

*for $n\ge N_1$, $d(x^{(n)},x)\le\varepsilon$

*for $n\ge N_2$, $d(x^{(n)},x')\le\varepsilon$
If we take $n\ge\max\{N_1,N_2\}$, we have, by the triangle inequality,
$$
d(x,x')\le d(x^{(n)},x)+d(x^{(n)},x')\le\varepsilon+\varepsilon=\dfrac{d(x,x')}{2}
$$
which implies $d(x,x')=0$, a contradiction.
