A topology is a strange kind of ruler. It never tells you how far apart two points are — it only tells you which sets are open. So here is a natural worry: given a bare topological space, can the open sets even tell two distinct points apart? On the number line the answer is obviously yes: around any two different reals you can slip two little intervals that miss each other. But a general topology may be far coarser than that, and "coarser" can mean blind — unable to distinguish points that are, as sets, plainly different.

The separation axioms are a ladder of increasingly generous answers to the question "how well can this topology separate points?". The lowest rung asks only that some open set notices a difference; the famous middle rung, Hausdorff, asks that the two points can be quarantined into disjoint open neighbourhoods. This one page is about that ladder — T_0, T_1 and T_2 — and above all about why the Hausdorff rung is the one nearly every working mathematician silently assumes.

Everything below is phrased in terms of open sets and closed sets, closures and interiors; if the phrase "all singletons are closed" makes you pause, that page is the place to warm up first.

The three lowest rungs

Let (X, \tau) be a topological space and take two distinct points x \ne y. Read each axiom as a promise about what open sets exist.

• T_0 (Kolmogorov): some open set contains one of the points but not the other. Formally, there is an open U with x \in U,\ y \notin U, or an open V with y \in V,\ x \notin V. Just one of the two directions is required. This is the barest possible non-blindness: the topology can at least see that x and y are different.
• T_1 (Fréchet): each point has an open set missing the other. There is an open U \ni x with y \notin U and an open V \ni y with x \notin V. Now the requirement is symmetric — both directions at once.
• T_2 (Hausdorff): the two points have disjoint open neighbourhoods. There are open sets U \ni x and V \ni y with U \cap V = \varnothing. Not merely "each avoids the other point" — the two neighbourhoods avoid each other entirely.

The names climb for a reason: each rung strictly implies the one below it, and the picture that makes the whole hierarchy intuitive is the Hausdorff one — two points, each in its own bubble, the bubbles not touching.

The figure hides the essential trick. Given x and y a positive distance d apart, take the two open discs of radius r = d/2 centred at each point. If a point z lay in both, the triangle inequality would force d \le d(x,z) + d(z,y) < \tfrac{d}{2} + \tfrac{d}{2} = d — impossible. So the discs are disjoint. Remember this half-the-distance move; it is exactly why every metric space is Hausdorff.

An equivalent face of T_1: closed points

The T_1 axiom has a much more memorable disguise.

Why. Suppose X is T_1 and fix a point x. For each other point y \ne x the axiom hands us an open V_y \ni y with x \notin V_y. The union of all these V_y is exactly the complement X \setminus \{x\} — it is open, so \{x\} is closed. Conversely, if every singleton is closed then for x \ne y the set X \setminus \{y\} is an open neighbourhood of x missing y (and symmetrically), which is precisely T_1. Finite sets are closed because a finite union of closed points is closed.

This is the fact to keep in your pocket: T_1 means points are closed. It instantly explains why the cofinite topology (below) is the smallest T_1 topology on any set.

The strict hierarchy: T_2 \Rightarrow T_1 \Rightarrow T_0

Each implication is a one-line argument, and — crucially — each is strict: there are spaces sitting on exactly one rung. A hierarchy with no separating examples would be a distinction without a difference.

The implications. If X is Hausdorff, the disjoint neighbourhoods U \ni x and V \ni y already satisfy T_1: U misses y (it is disjoint from V, and y \in V) and V misses x. And T_1 plainly gives T_0: "each point has an open set missing the other" is stronger than "one of them does". So T_2 \Rightarrow T_1 \Rightarrow T_0.

And below T_0 lies the totally blind case: the indiscrete topology \{\varnothing, X\} on a set with two or more points. The only non-empty open set is X itself, so no open set separates anything — it fails even T_0. Keep it in view; it is about to wreck the uniqueness of limits.

Worked examples: classify the topology

Classifying a small space is a matter of running down the ladder. Take X = \{1, 2, 3\} and try three topologies.

Example A — \tau_A = \{\varnothing,\ \{1\},\ \{1,2\},\ \{1,2,3\}\} (a "chain"). Is it T_0? For 1 vs 2, \{1\} catches 1 alone; for 2 vs 3, \{1,2\} catches 2 but not 3; for 1 vs 3, \{1\} works. So yes, T_0. Is it T_1? We would need an open set containing 3 but not 1 — but the only open set containing 3 is X. So not T_1. This is just the Sierpiński pattern stretched over three points.

Example B — the discrete topology \tau_B = \mathcal{P}(X) (every subset open). Every singleton is open, so for any x \ne y the sets \{x\} and \{y\} are disjoint open neighbourhoods. Hausdorff — the discrete topology sits at the top of the ladder (and satisfies every higher axiom too).

Example C — the indiscrete topology \tau_C = \{\varnothing, X\}. The only non-empty open set is X, so no open set isolates any point: it fails T_0, the very bottom.

Example D — a genuinely middle case. On the infinite set \mathbb{N}, the cofinite topology is T_1 (points are closed) yet not T_2 (any two non-empty opens meet). It is the canonical inhabitant of the gap between the two most-used rungs — and a reminder that the T_1/T_2 distinction only bites on infinite spaces (every finite T_1 space is discrete, hence Hausdorff).

Why Hausdorff is the one everyone assumes

The hierarchy would be idle taxonomy if Hausdorff did not buy you something. It buys a great deal. Here are the three payoffs to remember.

Proof. Suppose for contradiction that p \ne q. By Hausdorff there are disjoint open sets U \ni p and V \ni q, so U \cap V = \varnothing. Because x_n \to p, the tail of the sequence eventually lies in every neighbourhood of p: there is an N_1 with x_n \in U for all n \ge N_1. Likewise x_n \to q gives an N_2 with x_n \in V for all n \ge N_2. Then for any n \ge \max(N_1, N_2) the term x_n lies in both U and V — but those are disjoint. Contradiction. Hence p = q. \blacksquare

Look at what the proof used: only that the two limit candidates could be walled off into disjoint neighbourhoods. That is the entire content of the "\lim" notation making sense — writing \lim x_n = p, with a definite article, quietly presumes Hausdorff.

Sketch. Fix a point p \notin K. For each k \in K, Hausdorff gives disjoint opens U_k \ni k and V_k \ni p. The U_k cover K; compactness extracts a finite subcover U_{k_1}, \dots, U_{k_m}. The finite intersection V = V_{k_1} \cap \dots \cap V_{k_m} is an open neighbourhood of p disjoint from every U_{k_i}, hence from K. So each point outside K has a neighbourhood outside K: the complement is open, and K is closed. (Drop Hausdorff and this fails — in the cofinite topology on \mathbb{N} the whole space is compact but non-closed subsets abound.)

Proof. Take x \ne y and let d = d(x, y) > 0. Set r = d/2 and take the open balls U = B(x, r) and V = B(y, r). If some z lay in both, the triangle inequality would give

a strict inequality d(x,y) < d(x,y) — absurd. So U \cap V = \varnothing, and the metric topology is Hausdorff. \blacksquare This is the deepest reason the pathologies above feel exotic: anything you can put a distance on is automatically Hausdorff, and most spaces you meet before topology carry a metric. Hausdorff is the price of admission for the "one limit" world you grew up in.

Climbing higher: T_3, T_4 and the road to metrization

The ladder does not stop at Hausdorff. The next rungs separate a point from a whole closed set, or two closed sets from each other.

• Regular (T_3): Hausdorff, and every point can be separated from every closed set not containing it by disjoint open sets.
• Normal (T_4): Hausdorff, and every two disjoint closed sets have disjoint open neighbourhoods.

Why care? Because these stronger separations are exactly what you need to answer "which topological spaces secretly come from a metric?" — the metrization problem. The Urysohn metrization theorem says a space that is regular, T_1, and second-countable (has a countable base) is metrizable — it can be given a distance function inducing its topology. Normality is the setting for Urysohn's lemma (disjoint closed sets can be separated by a continuous function) and the Tietze extension theorem. Hausdorff, in other words, is not the summit — it is base camp. But it is the rung where limits become unique and analysis becomes possible, which is why it, and not its neighbours, is the axiom quietly baked into almost every definition you will meet downstream.