# 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.

• The definition of Hausdorff space does not imply that $x$ has a neighborhood containning only $x$. It implies that, for each $y\neq x$, there is a neighborhood $U$ of $x$ that doesn't contain $y$, but different choices of $y$ may (and usually do) require different choices of $U$. Only if a single $U$ worked for all $y$ simultaneously could you say $U=\{x\}$. Commented Jun 12, 2021 at 21:18
• I believe I understand, so there are points in the hausdorff space that don't have neighborhoods U that exclude all distinct points y not equal to x. Commented Jun 12, 2021 at 21:35
• If I may, here's a general hint for studying abstract topics like topology: I suggest to always back up your intuition / ideas about an abstract notion with several simple concrete examples. For instance, when discussing Hausdorff spaces one should be aware that the real numbers with the Euclidean topology are an easy standard example of a Hausdorff space. Such examples are useful since they immediatley show you why, for instance, the claim in your penultimate paragraph is not correct. Commented Jun 12, 2021 at 22:02
• On a side note, the only topology in which all the singletons are open is the discrete topology.
– Alan
Commented Jun 13, 2021 at 1:49

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.