# 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 $\mathbb{R}$.^{[1]}

## 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 $y$.^{[2]} █