Question regarding definition of limit point and uniqueness I found the following definition in a paper

A sequence in a space $X$ is an ordered family $(x_j)_{j\in \mathbb{N}}$ where $x_j \in X$ but not necessarily $x_i \ne x_j$ for $i \ne j$, that is, such a sequence is an element of $X^{\mathbb{N}}$. A point $x$ in a topological space $(X, O)$ is called a limit point of the sequence $(x_j)_{j\in \mathbb{N}}$ if, for every open set $M \in O$ containing $x$, there is $j_0 \in \mathbb{N}$ such that $x_j \in M$ for all $j, j \ge j_0$. The set of all limit points of a sequence $(x_j)_{j\in \mathbb{N}}$ is denoted by lim $x_j$. Observe that a sequence may have more than one limit point or no limit point at all.

I guess with this definition of limit point, it is not possible that there exists more then one limit point, because it is demanded that $x_j \in M$ for every $j \ge j_0$. I know sequence which have more than one limit point, like $(-1)^n$, but then the limit of a sequence is defined such that $x$ is a limit point iff for every open set $M$ with $x\in M$ there exists some point $x_j$ in the sequence with $x_j \in M$. This is a much weaker requirement and allows for multiple limits. But with the definition above, assume a sequence $(x_j)$ has two limit distinct points $x, y$ and consider an open set $M$ such that $x \in M$ but $y \notin M$, then all $x_j$ for an $j_0$ with $j \ge j_0$ must lie within $M$, but this contradicts that $y$ is a limit point???
 A: You are confused about two different notions. The sequence $(-1)^n$ in $\mathbb R$ has no limit point, but it has two adherent points.
Consider the space $\{-1, 1\}$ with the topology $\mathcal T = \{\emptyset, \{-1,1\}\}$, then the sequence $(-1)^n$ has two limit points.

In an Hausdorff topological space, any sequence can have only one limit point.

A: Consider the "double zero space" $X = \mathbb{R} \cup \{ 0^* \}$ with the topology generated by the sets of the form:


*

*$( a , b )$ where $a < b$ are real numbers; and

*$( a , 0 ) \cup \{ 0^* \} \cup ( 0 , b )$ where $a < 0 < b$ are real numbers.


(Essentially, the basic open neighbourhoods of $0^*$ are the basic open neighbourhoods of $0$ in $\mathbb{R}$ with $0$ switched out for $0^*$.)
Now consider the sequence $\langle \frac 1n \rangle_{n \in \mathbb{N}}$ in $X$.  It is easy to see that this sequence converges to $0$.  But it is equally easy to see that it converges to $0^*$!
We could also consider a triple zero space, or a quadruple zero space, or even an infinite zero space.

In your proof of uniqueness of limit points, you have implicitly assumed that the underlying space is T$_1$, which is not in itself enough to show that sequences have at most one limit point (the double zero space mentioned above is T$_1$).  Indeed, even if $y \notin M$ and $x_j \in M$ for all $j \geq j_0$, it might be that every open neighbourhood of $y$ includes $M$ as a (proper) subset, and so we don't get the contradiction you are thinking of.
As Damien L. states in his answer, Hausdorff spaces have the property that sequences have at most one limit point.  The proof is almost immediate from the definition:

If $x , y$ are distinct limit points of $\langle x_n \rangle_{n \in \mathbb{N}}$ then there are disjoint open neighbourhoods $U, V$ of $x,y$, respectively.  As $x$ is a limit point of $\langle x_n \rangle_{n \in \mathbb{N}}$ there is a $N_x$ such that $x_n \in U$ for all $n \geq N_x$.  As $y$ is a limit point of $\langle x_n \rangle_{n \in \mathbb{N}}$ there is an $N_y$ such that $x_n \in V$ for all $n \geq N_y$.  Consider $n = \max \{ N_x , N_y \}$: then $x_n$ must be in both $U$ and $V$; but these sets are disjoint!

Finally, to add total confusion to this topic, there are examples of non-Hausdorff topological spaces in which all convergent sequences have unique limits.  Consider an uncountable set $X$, and give it the co-countable topology.  Given a sequence $\langle x_n \rangle_{n \in \mathbb{N}}$ in $X$ and some $x \in X$ it is quite easy to show that $x = \lim_{n \rightarrow \infty} x_n$ iff there is an $N \in \mathbb{N}$ such that $x_n = x$ for all $n \geq N$.
A: Consider $\mathbb R$ with the topology $\mathcal T=\{ U\cup\{0\}\mid U\text{ is non-empty and open in the standard topology}\}\cup\{\varnothing\}$. Now consider the sequence $x_n=0$ for all $n\in\mathbb N$.
Given any point $t\in\mathbb R$ and $U\in\cal T$ such that $t\in U$ we have that for every $n\in\mathbb N$, $x_n\in U$. So by definition of a limit point, $t$ is a limit point of $(x_n)$. Not only there is more than one limit point, every point is a limit point of this sequence!
The reason you find this counterintuitive is that we often think of the real numbers as the first example of a topological space, but this is a really nice space. It's normal and completely metrizable, it's second countable and locally compact... but most topological spaces are actually bizarre and full of strange properties which may completely contradict the naive and initial approach when we think about the real numbers.
We do know, however, that if topological space is Hausdorff then limits are unique (when they exist, of course). This is because given two points we can find two disjoint open neighborhoods which separate them and then no sequence can have both points as limit points. 
The other implication is not true, that is if every sequence has a unique limit the space need not be Hausdorff, this is because there can be spaces which are "too big" to be described only by sequences. For them we need to utilize a machinery called nets, which is a generalization of sequences. It is true that if every convergent net has a unique limit point then the space is Hausdorff.
