# Hausdorff space

The Hausdorff condition is one of several additional conditions one can impose on a topological space. By adding additional constraints, the theorems that may be proven are stronger, but they apply to fewer classes of topological spaces. Hausdorff spaces are also called $T_2$ spaces.[footnotes 1]

A topological space $S$ is said to be a Hausdorff space if any two points $x$ and $y$ can be separated by disjoint neighborhoods $N_x$ and $N_y$. Specifically, let $(X,\tau)$ be a topological space. We say that $(X,\tau)$ is Hausdorff if for any pair $x, y \in X$, there is a $U \in \tau$ with $x \in U$ and a $V \in \tau$ with $y \in V$ such that $U \cap V = \emptyset$.

Intuitively, this can be explained simply using the topological space comprised of only the real number line $\mathbb{R}$. In this space, if the limit of a function $f(x)$ exists, it must be unique. There are topological spaces which do not meet this requirement, however, and are thus said to not be closed.[footnotes 2] A Hausdorff space can thus be thought of as a topological space where the "limit" at a point must be unique. In Topology, however, the conventions of limit points and neighborhoods are used, as they make more intuitive sense in dimensions greater than $\mathbb{R}$.

## Proof

THEOREM: If $X$ is a Hausdorff space, then a sequence of points of $X$ converges to at most one point of $X$.
Proof:
Suppose that $x_n$ is a sequence of points of $X$ that converges to $x$. If $y \neq x$, let $U$ and $V$ be disjoint neighborhoods of $x$ and $y$, respectively. Since $U$ contains $x_n$ for all but finitely many values of $n$, the set $V$ cannot. Therefore, $x_n$ cannot converge to $y$.

## Footnotes

1. Note as well that every compact Hausdorff space is normal
2. It must be noted that the Law of the Excluded Middle does not automatically apply in this case; the fact that a set is not closed does not automatically imply it is open.