Let $\displaystyle (X, d)$ be a metric space. If $\displaystyle (x_n)$ is a sequence in $\displaystyle X$ which converges to both $\displaystyle a \in X$ and $\displaystyle b \in X$, prove that $\displaystyle a=b.$
Suppose x were not equal to y. Then d(x,y) is not 0. Apply the definition of convergence to both x and y with $\displaystyle \epsilon= d(x,y)/2$ to get a contradiction:
Since $\displaystyle x_n$ converges to a, there exist $\displaystyle N_1$ such that if n> $\displaystyle N_1$, then $\displaystyle |x_n- a|< \epsilon$. Since $\displaystyle x_n$ converges to b, there exist $\displaystyle N_2$ such that if n> $\displaystyle N_2$ then $\displaystyle |x_n- b|< \epsilon$. Use the triangle inequality to show that those are impossible.
Suppose to the contrary that $\displaystyle (x_{n}) , n=1,2,.... \infty$ converges to two distinct limit points a and b in the metric space (X, d).
Then $\displaystyle d(a,b) \leq d(a, x_{n}) + d(x_{n}, b)$ by the the axiom of triangle inequality in the metric space.
By definition of a convergent sequence, there exists integers N1 and N2 such that if n >= N1, then $\displaystyle d(x_{n}, a) < \epsilon $ and if n >= N2, then $\displaystyle d(x_{n}, b) < \epsilon $. Let N= max (N1, N2).
Then, if n >= N,
$\displaystyle d(a,b) \leq d(a, x_{n}) + d(x_{n}, b) < \epsilon + \epsilon = 2\epsilon $.
Let $\displaystyle \epsilon = (1/2) \times d(a,b)$.
Then,
$\displaystyle d(a,b) \leq d(a, x_{n}) + d(x_{n}, b) < \epsilon + \epsilon = 2\epsilon = d(a,b)$, which leads a contradiction since d(a,b) < d(a,b) is not true.
Thus, a sequence in a metric space cannot converge to more than one limit.
There is an intuitive way to consider this problem.
If $\displaystyle \left( {x_n } \right) \to a$ then almost all of the terms of $\displaystyle \left( {x_n } \right)$ are in any ball that contains $\displaystyle \color{blue}a$.
So if $\displaystyle a \not= b$ then there are two disjoint balls, one containing a the other containing b.
BUT THAT IS IMPOSSIBLE. Almost all of the terms of $\displaystyle \left( {x_n } \right)$ are in any ball that contains $\displaystyle a$.
This may be similar to HallsofIvy, but it is slightly different and he/she seems to have LaTeX mistakes.. All you need is the triangle inequality
Choose $\displaystyle N_1$ such that if $\displaystyle n\leqslant N_1$ then $\displaystyle d(x_n,a)<\frac{\varepsilon}{2}$
Also, choose $\displaystyle N_2$ such that if $\displaystyle n\leqslant N_2$ then $\displaystyle d(x_n,b)<\frac{\varepsilon}{2}$
Now choose $\displaystyle N=\max\left\{N_1,N_2\right\}$ then for $\displaystyle n\leqslant N$ we have that $\displaystyle d(a,b)\leqslant d(x_n,a)+d(x_n,b)\leqslant\frac{\varepsilon}{2}+\f rac{\varepsilon}{2}=\varepsilon$.
Since $\displaystyle \varepsilon$ was arbitrary this implies that $\displaystyle d(a,b)=0$ which by the axioms of metric spaces is only true if $\displaystyle a=b\quad\blacksquare$