Existence of Limit points in Hausdorff Spaces I'm having trouble understanding how limit points can exist in Hausdorff Spaces.
From Munkres,
"If A is a subset of the topological space X and if x is a point of X, we say that x is a limit point (or "cluster point," or "point of accumulation") of A if every neighborhood of x intersects A in some point other than x itself. Said differently, x is a limit point
of A if it belongs to the closure of      A—{x}"
"A topological space X is called a Hausdorff space if for each pair x1, x2
of distinct points of X, there exist neighborhoods U1, and U2 of x1 and x2, respectively,that are disjoint"
From the definition of a Hausdorff space it is my understanding that it implies that there exists at least one neighborhood in which x only includes itself, the singleton set {x}. If this is true for every point x, then doesn't it contradict with the definition of a limit point considering that every neighborhood of x must include a point other than itself?
I'd like to know in what way I am misunderstanding the definitions and what is the right way to think about this. I think most of my confusion comes from the phrase "every neighborhood" in the limit point definition.
 A: You are misunderstanding the definition of Hausdorff space.And I´m gonna try to clarify it.
Definition:
A topological space $(X,\tau)$ is Hausdorff or $\bf{T_2}$ if $\forall x,y\in X$ and $x\neq y$ there are non empty open sets $U,V\subset X$ such that $x\in U$ ,$y\in V$ and $U\cap V =\emptyset$.
It means that if $X$ is a topological space that is Hausdorff given two different arbitrary points you can always find two open sets that contains the points and  are disjoint or similary given two distinct  points you can always separate it by open sets .
The set $\lbrace x \rbrace$ not is necessary open in any topological space, and hence if $X$ is Hausdorff it not implies that always $\lbrace x \rbrace$ is the open neighbourhood that contains $x$.
If $X$ is Hausdorff $\lbrace x\rbrace$ is closed (it is more general true for a finitely many points $\lbrace x_1,x_2,\dots ,x_n\rbrace$)it is easy to prove using the definition of Hausdorff space.
Definition:
Let $(X,\tau)$ a topological space and $A\subset X$, a  point $x\in X$ is a limit point of the set $A$ if for every open set $U$ that contains $x$ occur that $(U\setminus \lbrace x \rbrace) \cap A\neq \emptyset$.
Intuitively this definition say that if you have a topological space, $x\in X$ and you chose a subset $A\subset X$ then if $x$ is limit point of $A$ then you can always chose a point $x^{\prime}\in A$ such that $x^{\prime}$ is very very near of $x$ if your $X$ were a metric space you can said that $x$ limit point of $A$ implies that you can find $x^{\prime }\in A$ such that $d(x,x^{\prime})<\varepsilon$.
Finally  one definition not  exclude the other.
