Strictness in the Separation Hierarchy
The separation axioms T0, T1, and Hausdorff impose progressively stronger requirements on pairs of distinct points. The implications between them do not reverse: a space can satisfy T0 without satisfying T1, and it can satisfy T1 without being Hausdorff. To see these failures, it is important to check the neighborhoods in the definition rather than rely on a diagram or on a familiar geometric picture. We will use the Sierpinski space for the first strict step and an infinite cofinite space for the second.
Throughout, a neighborhood means an open neighborhood. In the Hausdorff condition, the neighborhoods must be disjoint; merely finding neighborhoods that distinguish the points is not enough.
The difference between T0 and T1 is the direction of the requirement: T0 asks for a distinguishing open set in at least one direction, whereas T1 asks for one in each direction. Hausdorffness asks for something stronger still: two neighborhoods, one for each point, that do not intersect.
The Implications and Their Proofs
Proof. Suppose first that \(X\) is Hausdorff. Fix distinct \(x,y\in X\). Choose disjoint open neighborhoods \(U\) of \(x\) and \(V\) of \(y\). Since \(V\) contains \(y\) and is disjoint from \(U\), the open set \(U\) contains \(x\) but not \(y\). Interchanging the roles of \(x\) and \(y\) gives an open neighborhood of \(y\) that does not contain \(x\). This is the T1 condition.
Now suppose that \(X\) is T1. For each distinct pair \(x,y\), the T1 condition provides, in particular, an open set containing \(x\) but not \(y\). This open set distinguishes the pair, as required by T0. Therefore Hausdorff implies T1, and T1 implies T0. \(\square\)
These implications say that failure of T1 rules out Hausdorffness, and failure of T0 rules out T1. They do not show that either converse holds. The following examples verify, point by point, that both converses fail.
Worked Example: The Sierpinski Space Is T0 but Not T1
Let \(X=\{a,b\}\) and give it the topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). This is a topology: it contains \(\varnothing\) and \(X\); any union of members is one of \(\varnothing,\{a\},X\); and the intersection of any finite collection of members is also one of these three sets.
There is only one pair of distinct points to check. The open set \(\{a\}\) contains \(a\) and does not contain \(b\). Thus the pair is distinguished by an open set, so \(X\) is T0.
For T1, the requirement in the direction from \(b\) to \(a\) would need an open neighborhood of \(b\) that does not contain \(a\). But the only open set containing \(b\) is \(X\), and \(X\) contains \(a\). No such neighborhood exists, so \(X\) is not T1.
Check by closed singletons: the complement \(X\setminus\{a\}=\{b\}\) is not open, so \(\{a\}\) is not closed. In a T1 space every singleton is closed: for a fixed point \(a\), the T1 condition makes every \(x\ne a\) belong to an open set missing \(a\), and the union of those open sets is \(X\setminus\{a\}\). Here that closed-singleton condition fails, in agreement with the neighborhood test.
This example shows why checking only one direction would be insufficient for T1. The open set \(\{a\}\) distinguishes \(a\) from \(b\), but there is no open set that distinguishes \(b\) from \(a\). T0 permits this one-sided distinction; T1 does not.
An Infinite Cofinite Space
For an infinite set \(X\), the cofinite topology consists of \(\varnothing\) and every subset of \(X\) whose complement in \(X\) is finite. The size of the underlying set matters: two nonempty cofinite open sets in an infinite space must intersect, but the same assertion can fail when \(X\) is finite.
Proof. Let \(x,y\in X\) be distinct. The set \(X\setminus\{y\}\) has finite complement, so it is open; it contains \(x\) and excludes \(y\). Likewise \(X\setminus\{x\}\) is an open set containing \(y\) and excluding \(x\). This verifies T1.
To check Hausdorffness, take any two nonempty open sets \(U,V\). The complements \(X\setminus U\) and \(X\setminus V\) are finite. By the complement law, \(X\setminus(U\cap V)=(X\setminus U)\cup(X\setminus V)\), which is finite because it is a union of two finite sets. If \(U\cap V\) were empty, this equality would give \(X=(X\setminus U)\cup(X\setminus V)\), making \(X\) finite. That contradicts the assumption that \(X\) is infinite. Hence every two nonempty open sets intersect. In particular, no distinct pair can have disjoint open neighborhoods, so \(X\) is not Hausdorff. \(\square\)
Worked Example: The Cofinite Topology on the Natural Numbers
Take \(X=\mathbb{N}=\{1,2,3,\dots\}\) with the cofinite topology. We verify separation for the particular pair \(2,7\), then check why no pair can be separated by disjoint neighborhoods.
The set \(U=\mathbb{N}\setminus\{7\}\) is open because its complement \(\{7\}\) is finite. It contains \(2\), since \(2\ne7\), and it excludes \(7\). In the other direction, \(V=\mathbb{N}\setminus\{2\}\) is open, contains \(7\), and excludes \(2\). This pair therefore satisfies the T1 test. The same construction works for any distinct \(m,n\in\mathbb{N}\), using \(\mathbb{N}\setminus\{n\}\) and \(\mathbb{N}\setminus\{m\}\); thus the whole space is T1.
For this chosen pair, \(U\cap V=\mathbb{N}\setminus\{2,7\}\), which contains \(1,3,4,5,6,8,\dots\), so these neighborhoods are not disjoint. More generally, let \(U,V\) be any two nonempty open sets. Their union of excluded points, \((\mathbb{N}\setminus U)\cup(\mathbb{N}\setminus V)\), is finite. Since \(\mathbb{N}\) is infinite, there is a natural number outside that finite set, and it belongs to both \(U\) and \(V\). Therefore no two nonempty open sets are disjoint, and \(\mathbb{N}\) is not Hausdorff.
Check against the definitions: the sets \(U,V\) each exclude the other chosen point, which is exactly what T1 requires. Their intersection is nonempty, so they do not demonstrate Hausdorff separation. In fact, the argument applies to every possible choice of neighborhoods.
Why the Infinite Hypothesis Matters
The cofinite-space argument depends on the fact that deleting finitely many points from an infinite set cannot delete every point. On a finite set, every subset has finite complement, so the cofinite topology is instead the discrete topology: every subset is open. Distinct points in a discrete space have disjoint singleton neighborhoods. The next calculation makes this contrast explicit.
Worked Example: A Finite Cofinite Topology Is Hausdorff
Let \(X=\{1,2,3\}\) have the cofinite topology. Every subset \(A\subseteq X\) has a finite complement \(X\setminus A\), since \(X\setminus A\subseteq X\) and \(X\) has only three elements. Thus every subset of \(X\) is open, and the topology is \(\mathcal{P}(X)\).
For any distinct \(i,j\in X\), the singleton sets \(\{i\}\) and \(\{j\}\) are open. They contain their respective points and have intersection \(\varnothing\), because \(i\ne j\). Hence every distinct pair has disjoint open neighborhoods, so this space is Hausdorff. It is also T1 and T0 by the implication theorem.
Check for the pair \(1,3\): the neighborhoods \(\{1\}\) and \(\{3\}\) are both open, and \(\{1\}\cap\{3\}=\varnothing\). This direct calculation shows why the infinite cofinite example cannot be transferred to a finite underlying set.
A Common Pitfall: Forgetting What Is Being Separated
A common incorrect argument says that any two cofinite open sets intersect because their complements are finite. The missing hypothesis is that the underlying set must be infinite. For a concrete countercalculation, take \(X=\{1,2,3\}\) with the cofinite topology. The open sets \(U=\{1\}\) and \(V=\{2\}\) are cofinite, since their complements \(\{2,3\}\) and \(\{1,3\}\) are finite, but \(U\cap V=\varnothing\). The complements are finite, yet their union is all of \(X\); there is no point left outside the excluded points.
The correction is to use infinitude explicitly: in an infinite cofinite space, the union of two finite complements is finite and therefore cannot equal \(X\). That fact forces a point into the intersection of the two open sets. Without infinitude, the conclusion does not follow.
It is also useful to keep the three tests distinct. T0 asks for one distinguishing open set; T1 asks for a distinguishing open set in both directions; Hausdorffness asks for disjoint neighborhoods. The examples establish that each strengthening has content:
| Space | T0? | T1? | Hausdorff? | Reason for the strict step |
|---|---|---|---|---|
| Sierpinski space \(\{a,b\}\) | Yes | No | No | \(\{a\}\) distinguishes \(a\) from \(b\), but no open neighborhood of \(b\) excludes \(a\). |
| Infinite cofinite space | Yes | Yes | No | Each point can be excluded individually, but any two nonempty open sets intersect. |
| Finite cofinite space \(\{1,2,3\}\) | Yes | Yes | Yes | Every subset is open, so distinct points have disjoint singleton neighborhoods. |
Thus the hierarchy is strict at both steps: T0 does not imply T1, and T1 does not imply Hausdorff. The examples also show what each stronger condition adds: directional separation, followed by separation by disjoint neighborhoods.
Check Your Understanding
For each question, identify the relevant open sets and say which separation condition they verify or fail.
- In the Sierpinski space, which open set verifies T0 for the pair \(a,b\), and which direction of the T1 test fails?
- In an infinite cofinite space, explain why \(X\setminus\{y\}\) and \(X\setminus\{x\}\) verify T1 when \(x\ne y\).
- Why does the proof that two nonempty cofinite open sets intersect require the underlying set to be infinite?
- Give disjoint open neighborhoods of \(1\) and \(2\) in the cofinite topology on \(\{1,2,3\}\).
- State the two implications in the separation hierarchy and identify which examples show that their converses fail.