In any [metric space](/page/Metric%20Space), a convergent sequence has exactly one limit: if $x_k \to x$ and $x_k \to y$ with $x \neq y$, then for $r = d(x,y)/2 > 0$, the open balls $B(x,r)$ and $B(y,r)$ are disjoint, and the sequence cannot eventually belong to both. This argument uses nothing about the specific metric — only that distinct points can be separated by disjoint open sets. In a general [topological space](/page/Topology), this separation property can fail, and when it does, the consequences are severe: limits become non-unique, compact subsets need not be closed, and continuous bijections from compact spaces need not be homeomorphisms. The Hausdorff condition is the axiom that prevents these pathologies, and it is the minimal assumption under which the standard constructions of analysis remain valid.

[example: The Line with Two Origins] The following space demonstrates what goes wrong without the Hausdorff axiom. Let $X = (\mathbb{R} \setminus \{0\}) \cup \{p, q\}$, where $p$ and $q$ are two distinct points meant to serve as "competing origins." Define a topology on $X$ as follows: a subset $U \subset X$ is open if - the intersection $U \cap (\mathbb{R} \setminus \{0\})$ is open in the standard topology on $\mathbb{R} \setminus \{0\}$, and - if $p \in U$, then $(-\varepsilon, 0) \cup (0, \varepsilon) \cup \{p\} \subset U$ for some $\varepsilon > 0$, and - if $q \in U$, then $(-\delta, 0) \cup (0, \delta) \cup \{q\} \subset U$ for some $\delta > 0$. In other words, $p$ and $q$ each behave like the origin — they have the same collection of "punctured" neighbourhoods — but they are distinct points. Consider the sequence $x_k = 1/k$ for $k \in \mathbb{N}$. Every open set containing $p$ includes an interval $(-\varepsilon, 0) \cup (0, \varepsilon) \cup \{p\}$, so $x_k = 1/k \in U$ for all $k > 1/\varepsilon$. Thus $x_k \to p$. By the identical argument with $q$ in place of $p$, we also have $x_k \to q$. The sequence converges to two distinct limits simultaneously. The underlying problem is that $p$ and $q$ cannot be separated by disjoint open sets: any open set containing $p$ intersects every open set containing $q$ (both must contain points of $(0, \varepsilon)$ for small $\varepsilon$). The Hausdorff axiom, stated below, prohibits exactly this failure. [/example]

## Definition

The pathology above — the inability to separate distinct points by disjoint open sets — motivates the following axiom.

[definition: Hausdorff Space] A [topological space](/page/Topology) $(X, \tau)$ is a **Hausdorff space** (or **$T_2$ space**) if for every pair of distinct points $x, y \in X$, there exist open sets $U, V \in \tau$ such that \begin{align*} x \in U, \quad y \in V, \quad U \cap V = \varnothing. \end{align*} [/definition]

When we say that $x$ and $y$ can be "separated by open sets," we mean precisely that such disjoint open neighbourhoods exist. A topological space satisfying this condition is also called **separated**.

Every [metric space](/page/Metric%20Space) is Hausdorff: given distinct points $x \neq y$, set $r = d(x,y)/2 > 0$ and take $U = B(x,r)$, $V = B(y,r)$. The triangle inequality guarantees $U \cap V = \varnothing$, since any $z \in U \cap V$ would satisfy $d(x,y) \le d(x,z) + d(z,y) < r + r = d(x,y)$, a contradiction. In particular, $\mathbb{R}^n$ with the Euclidean topology, every [Banach space](/page/Banach%20Space) with the norm topology, and every [manifold](/page/Smooth%20Manifold) with its standard topology are all Hausdorff.

The Hausdorff condition sits in the middle of a hierarchy of separation axioms. To understand its role, we need to see what it gives beyond weaker axioms and what stronger axioms provide beyond it.

## The Hierarchy of Separation Axioms

A basic challenge in point-set [topology](/page/Topology) is deciding how much separation between points and closed sets a topological space should guarantee. The axioms of a topology are deliberately minimal — they ensure that unions and finite intersections of open sets are open, but say nothing about how effectively the topology distinguishes points from each other or from closed sets. Different levels of separation are captured by the $T_i$ axioms, and the choice of which axiom to impose determines how much of classical analysis can be carried out.

### The $T_0$ and $T_1$ Axioms

The weakest useful separation axiom asks only that the topology can tell points apart *somehow*.

[definition: $T_0$ Space] A topological space $(X, \tau)$ is **$T_0$** (or **Kolmogorov**) if for every pair of distinct points $x, y \in X$, there exists an open set $U \in \tau$ that contains one of the two points but not the other. [/definition]

The $T_0$ axiom says the topology is fine enough to distinguish points, but it does not say which point gets the separating open set. This asymmetry is a real limitation: in a $T_0$ space, it is possible for $x$ to have no open neighbourhood avoiding $y$, even though $y$ has an open neighbourhood avoiding $x$.

The $T_1$ axiom eliminates this asymmetry.

[definition: $T_1$ Space] A topological space $(X, \tau)$ is **$T_1$** (or **Fr\'{e}chet**) if for every pair of distinct points $x, y \in X$, there exists an open set $U \in \tau$ with $x \in U$ and $y \notin U$. [/definition]

The $T_1$ condition is equivalent to requiring that every singleton $\{x\}$ is a [closed set](/page/Closed%20Set). To see this, note that if $(X, \tau)$ is $T_1$, then for every $y \neq x$, there is an open set $U_y$ containing $y$ but not $x$. The union $\bigcup_{y \neq x} U_y = X \setminus \{x\}$ is open, so $\{x\}$ is closed. Conversely, if every singleton is closed, then for $x \neq y$, the set $U = X \setminus \{y\}$ is open, contains $x$, and excludes $y$.

However, the $T_1$ axiom is still insufficient for analysis. In a $T_1$ space, limits of sequences can fail to be unique, because the axiom provides an open set containing $x$ but not $y$, and a separate open set containing $y$ but not $x$ — but these sets need not be disjoint. Without disjointness, a sequence can belong eventually to both.

[example: The Cofinite Topology on an Infinite Set] Let $X$ be any infinite set and define the **cofinite topology** $\tau = \{U \subset X : X \setminus U \text{ is finite}\} \cup \{\varnothing\}$. This topology is $T_1$: for any distinct $x, y \in X$, the set $U = X \setminus \{y\}$ is open (its complement is the finite set $\{y\}$), contains $x$, and excludes $y$. However, $(X, \tau)$ is not Hausdorff. Suppose $U$ and $V$ are nonempty open sets. Then $X \setminus U$ and $X \setminus V$ are both finite, so $X \setminus (U \cap V) = (X \setminus U) \cup (X \setminus V)$ is a finite union of finite sets, hence finite. Since $X$ is infinite, $U \cap V \neq \varnothing$. Therefore no two nonempty open sets are disjoint, and no pair of distinct points can be separated. As a consequence, every sequence in $X$ whose range is infinite converges to every point of $X$. Indeed, if $\{x_k\}_{k=1}^\infty$ takes infinitely many distinct values and $p \in X$ is any point, then for any open set $U$ containing $p$, the complement $X \setminus U$ is finite, so at most finitely many terms $x_k$ lie outside $U$. Thus $x_k$ is eventually in $U$, and $x_k \to p$. Limits are completely meaningless in this topology. [/example]

The Hausdorff axiom ($T_2$) resolves this by requiring that the separating open sets be disjoint, which — as demonstrated in the opening — is exactly the condition needed to guarantee uniqueness of limits.