1. ## Metric Space, limit

Let $(X, d)$ be a metric space. If $(x_n)$ is a sequence in $X$ which converges to both $a \in X$ and $b \in X$, prove that $a=b.$

2. 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 $\epsilon= d(x,y)/2$ to get a contradiction:
Since $x_n$ converges to a, there exist $N_1$ such that if n> $N_1$, then $|x_n- a|< \epsilon$. Since $x_n$ converges to b, there exist $N_2$ such that if n> $N_2$ then $|x_n- b|< \epsilon$. Use the triangle inequality to show that those are impossible.

3. Suppose to the contrary that $(x_{n}) , n=1,2,.... \infty$ converges to two distinct limit points a and b in the metric space (X, d).

Then $d(a,b) \leq d(a, x_{n}) + d(x_{n}, b)$ 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 $d(x_{n}, a) < \epsilon$ and if n >= N2, then $d(x_{n}, b) < \epsilon$. Let N= max (N1, N2).

Then, if n >= N,
$d(a,b) \leq d(a, x_{n}) + d(x_{n}, b) < \epsilon + \epsilon = 2\epsilon$.

Let $\epsilon = (1/2) \times d(a,b)$.
Then,
$d(a,b) \leq d(a, x_{n}) + d(x_{n}, b) < \epsilon + \epsilon = 2\epsilon = d(a,b)$, 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.

4. There is an intuitive way to consider this problem.
If $\left( {x_n } \right) \to a$ then almost all of the terms of $\left( {x_n } \right)$ are in any ball that contains $\color{blue}a$.
So if $a \not= b$ then there are two disjoint balls, one containing a the other containing b.
BUT THAT IS IMPOSSIBLE. Almost all of the terms of $\left( {x_n } \right)$ are in any ball that contains $a$.

5. 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 $N_1$ such that if $n\leqslant N_1$ then $d(x_n,a)<\frac{\varepsilon}{2}$

Also, choose $N_2$ such that if $n\leqslant N_2$ then $d(x_n,b)<\frac{\varepsilon}{2}$

Now choose $N=\max\left\{N_1,N_2\right\}$ then for $n\leqslant N$ we have that $d(a,b)\leqslant d(x_n,a)+d(x_n,b)\leqslant\frac{\varepsilon}{2}+\f rac{\varepsilon}{2}=\varepsilon$.

Since $\varepsilon$ was arbitrary this implies that $d(a,b)=0$ which by the axioms of metric spaces is only true if $a=b\quad\blacksquare$