Hausdorff and Separation Axioms

The Hausdorff property is a standing assumption in essentially all of differential geometry: manifolds are defined to be Hausdorff (plus second countable) precisely so that limits, tangent spaces, and coordinate charts behave predictably — without it, "the line with two origins" shows that even locally Euclidean spaces can have non-unique limits. In functional analysis, Hausdorff topological vector spaces are needed for norms and seminorms to actually separate points, which underlies the Hahn–Banach separation theorems. Normality (T4) is the exact hypothesis behind Urysohn's Lemma, which produces continuous real-valued "bump functions" separating closed sets — this is the engine behind partitions of unity, used constantly in constructing Riemannian metrics, in sheaf theory, and in proving the existence of smooth cutoff functions in PDE theory. In theoretical computer science, the Kolmogorov T0 quotient of a space appears in domain theory, where non-Hausdorff spaces (Scott topologies) model computation and partial information, and the T0 axiom exactly captures when two "computational states" are genuinely distinguishable by observable open conditions. In algebraic geometry, the Zariski topology on an algebraic variety is T1 (closed points are... points) but famously not Hausdorff, which is one of the first surprises students encounter and which motivates entirely separate technology (separatedness of morphisms, valuative criteria) to recover a working substitute for the Hausdorff condition.

Ask the tutor: Can you show me step by step why the Sorgenfrey line is normal but its square is not?

Think of each axiom as answering a sharper version of the same question: can open sets tell apart the following data?

• T0 asks only that open sets distinguish points at all — one point might be in the closure of the other's every neighborhood, but not both ways.
• T1 asks for symmetric distinguishability of points: each point has a neighborhood avoiding the other. This turns out to be equivalent to points being closed sets — a purely topological finiteness condition.
• T2 asks for the neighborhoods to be disjoint, which is the geometric picture we usually associate with "different points don't overlap."
• T3 upgrades this to separating a point from an entire closed set — a whole "wall" rather than a single other point.
• T4 upgrades further to separating two disjoint walls from each other.

A useful mental model is a ladder of increasingly fine "resolving power." A space failing T1 has points that are topologically blurred together (like the indiscrete topology, where no two points can be told apart at all). A space satisfying T2 but not T3 can separate points but cannot always pry a point away from a closed set that "wraps around" it topologically. Normality (T4) is the strongest of these four and, remarkably, is not generally inherited by subspaces — a fact that surprises many students meeting it for the first time.

Example 1: The cofinite topology on an infinite set is T1 but not T2. Let $X$ be an infinite set (say $X = \mathbb{Z}$), and let

First check $\tau_{cf}$ is a topology: $\emptyset \in \tau_{cf}$ by definition and $X \in \tau_{cf}$ since $X \setminus X = \emptyset$ is finite. If $U_1, \dots, U_n \in \tau_{cf}\setminus\{\emptyset\}$, then $X \setminus (U_1 \cap \cdots \cap U_n) = (X\setminus U_1)\cup \cdots \cup (X \setminus U_n)$ is a finite union of finite sets, hence finite, so finite intersections of nonempty basic opens stay in $\tau_{cf}$ (and intersections involving $\emptyset$ trivially remain in $\tau_{cf}$). If $\{U_i\}_{i\in I}$ is any family in $\tau_{cf}\setminus\{\emptyset\}$, then $X\setminus \bigcup_i U_i = \bigcap_i (X\setminus U_i)$ is a subset of the finite set $X \setminus U_{i_0}$ for any fixed $i_0$, hence finite, so the union lies in $\tau_{cf}$. Thus arbitrary unions are open too. T1 check. Fix $x \in X$. Is $\{x\}$ closed, i.e. is $X\setminus\{x\}$ open? We need $X \setminus (X\setminus\{x\}) = \{x\}$ to be finite — yes, it has one element. So $X \setminus \{x\} \in \tau_{cf}$, hence $\{x\}$ is closed for every $x$. By part (3) of the theorem, $X$ is T1. Failure of T2. Suppose toward contradiction that $U, V$ are disjoint nonempty open sets in $\tau_{cf}$. Then $X \setminus U$ and $X \setminus V$ are finite, so

using $U \cap V = \emptyset$. But $(X\setminus U)\cup(X\setminus V)$ is a union of two finite sets, hence finite — contradicting that $X$ is infinite. So no two nonempty open sets are disjoint, and in particular, for any two distinct points $x,y \in X$, there do not exist disjoint open sets separating them (any open set containing $x$ and any open set containing $y$ must intersect, since both are nonempty and cofinite topology forces nonempty opens to overlap). Hence $(X, \tau_{cf})$ is T1 but not Hausdorff. This single example simultaneously shows: T1 does not imply T2, and (taking $C = \{y\}$, which is closed by T1) shows the space is not even T3 in a nontrivial sense once we probe further — indeed no two disjoint nonempty closed sets with nonempty complement structure can be separated here either, since any two nonempty open sets meet.

Example 2: Regular but not normal — the "bowtie" tangent-disc-type construction via subspace of a normal space fails to preserve normality. To see concretely that normality is not hereditary (T4 can fail to pass to subspaces, unlike T2 and T3 which are hereditary), consider the Moore plane (Niemytzki plane), a standard asymmetric counterexample. Let $X = \{(x,y) \in \mathbb{R}^2 : y \geq 0\}$, the closed upper half-plane. Define a topology as follows.

• For a point $p = (x,y)$ with $y > 0$: basic neighborhoods are ordinary open Euclidean disks around $p$ contained in the open upper half-plane.
• For a point $p = (x, 0)$ on the boundary line $L = \{y=0\}$: basic neighborhoods are sets of the form $\{p\} \cup D$, where $D$ is an open Euclidean disk of some radius $r>0$ tangent to $L$ at $p$ from above (i.e. $D$ is the open disk with center $(x, r)$ and radius $r$, so it touches $L$ exactly at $p$).

This is a valid topology (one checks the basic sets satisfy the base axioms: any two basic neighborhoods' intersection around a common point contains a smaller basic neighborhood of that same type — this is routine and standard). Hausdorff / regular. This space $X$ is Hausdorff: two points in the open half-plane are separated as in the usual plane topology; a boundary point and an interior point, or two boundary points, are separated using sufficiently small tangent disks and Euclidean disks that can be made disjoint by shrinking radii. One can further check $X$ is regular (T3): given a point $p$ and a closed set $C$ not containing it, the basic neighborhoods of $p$ are already "small" tangent disks or Euclidean balls, and their closures can be taken inside the complement of $C$ by shrinking radius, using the same tangency mechanism. Non-normality. The boundary line $L = \{y = 0\}$, as a subspace of $X$, is discrete: each singleton $\{(x,0)\}$ is open in $L$'s subspace topology because a tangent disk neighborhood $\{(x,0)\}\cup D$ meets $L$ only at $(x,0)$ (a tangent disk touches the line at exactly one point). A discrete space on the cardinality of $\mathbb{R}$ (continuum many points) has $2^{\mathfrak c}$ subsets, all of which are simultaneously open and closed in $L$; one shows (via a counting/separation argument comparing the number of open subsets of $X$, which is bounded by considering countable data attached to each disk, against the $2^{2^{\aleph_0}}$ subsets of $L$ that would all need distinct normality-separating open sets in $X$) that not every pair of disjoint closed subsets of $L$ extends to disjoint open sets in $X$. Concretely, splitting $L$ into two disjoint subsets $A, B$ that are both closed in $X$ (using that every subset of the discrete space $L$ is closed in $L$, and $L$ itself is closed in $X$, so any subset of $L$ is closed in $X$ as an intersection of closed sets $L \cap (\text{closed in } L)$... one must choose $A,B$ so that no pair of open sets in $X$ separates them, which is the content of the classical Jones' lemma-style argument), one produces disjoint closed sets in $X$ with no disjoint open neighborhoods. The moral of Example 2, stated precisely as a theorem-level fact even though its full proof (via Jones' Lemma) is a standard but longer argument typically given beyond an introductory pass: regularity is hereditary to all subspaces, but normality is not — the Moore plane is a regular (indeed Tychonoff) space that is not normal, and its non-normality is witnessed exactly by the discrete subspace $L$ failing to inherit separating power from the ambient space.