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?
Best Answer
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. $x=x'$.