a real convergent sequence has a unique limit point How to show that a real convergent sequence has a unique limit point viz. the limit of the sequene?
I've used the result several times but I don't know how to prove it!
Please help me!
 A: Hint: Suppose that a sequence $\{x_n\}$ has two different limits $l_1$ and $l_2$. There exists a number $\varepsilon>0$ such that $2\varepsilon<|l_1-l_2|$. Then $\varepsilon$-neighborhoods of the points $l_1$ and $l_2$ are disjoint. Therefore, for sufficiently large $n$ we have that ...
A: Assume to the contrary that the sequence $\{X_n\}$ has two limit points, say $a$ and $b$. Then we have $|a-X_n| < ε$  for $n\ge N_1$ and $|b-X_n| < ε$ for $n\ge N_2$. Let $ε*= (a-b)/4$. 
Then we have 
$$\begin{aligned}
               |a-b| &= |a-b+Xn-Xn|\\
                     &= |(a-Xn)+(-b+Xn)|\\
                     &\le|(a-Xn)|+|(b-Xn)|\\
                     &\le 2ε
                      = 2(a-b)/4\\
                      &= (a-b)/2,
\end{aligned}$$
which is not correct for our $ε$. Thus, we showed one $ε$, which is already enough, that contradicts the assertion.  
A: I am assuming that limit points are defined as in Section $6.4$ of the book Analysis $1$ by the author Terence Tao. We assume that the sequence of real numbers $(a_{n})_{n=m}^{\infty}$ converges to the real number $c$. Then we have to show that $c$ is the unique limit point of the sequence. First, we shall show that $c$ is indeed a limit point. By the definition of convergence we have
$$\forall\epsilon > 0\:\exists N_{\epsilon}\geq m\:\text{s.t.}\:\forall n\geq N_{\epsilon}\:|a_{n}-c|\leq\epsilon$$
We need to prove that $\forall\epsilon >0\:\forall N\geq m\:\exists n\geq N\:\text{s.t.}\:|a_{n}-c|\leq\epsilon$. Fix $\epsilon$ and $N$. If $N\leq N_{\epsilon}$ then we could pick and $n\geq N_{\epsilon}$. If $N>N_{\epsilon}$ then picking any $n\geq N > N_{\epsilon}$ would suffice.
Now, we shall prove that $c$ is the unique limit point. Let us assume for the sake of contradiction that $\exists$ a limit point $c'\in\mathbb{R}$ and $c'\neq c$. Since the sequence converges to $c$, $\forall\epsilon > 0\:\exists M_{\epsilon}\geq m$ such that $\forall n\geq M_{\epsilon}$ we have $|a_{n}-c|\leq \epsilon/2$. Further, since $c'$ is a limit point, $\forall\epsilon > 0$ and $\forall N\geq m\:\exists k_{N}\geq N$ such that $|a_{k_{N}}-c'|\leq \epsilon/2$. In particular, $|a_{k_{M_{\epsilon}}}-c'|\leq \epsilon/2$ and $|a_{k_{M_{\epsilon}}}-c|\leq \epsilon/2$. Using the triangle inequality we get $\forall\epsilon > 0\: |c-c'|\leq\epsilon\implies |c-c'| = 0\implies c = c'$ which contradicts the assumption.
