Separation Axioms: the Hierarchy: Examples and Counterexamples

Let $S = \{a, b\}$ with $a \ne b$ carry the Sierpinski topology $\mathcal{T}_{\mathrm{Sier}} = \{\varnothing, \{b\}, S\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), so that $b$ is the open point and $a$ the closed one. Then:

1. $S$ is $T_0$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces): the open set $\{b\}$ contains $b$ and not $a$.
2. $S$ is not $T_1$: the singleton $\{b\}$ is not closed (A space is $T_1$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology). It is not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) either, and for an independent reason rather than as a consequence of the previous sentence: the only open set containing $a$ is $S$ itself, which also contains $b$.
3. $S$ is normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly): its only closed sets are $\varnothing$, $\{a\}$ and $S$, and no two disjoint closed sets are both nonempty.
4. $S$ is not regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly): the point $b$ and the closed set $\{a\}$ have no disjoint open neighbourhoods, because the only open set containing $a$ is $S$ itself.

Sierpinski space is therefore the sharpest small witness that normality without $T_1$ implies nothing: it is normal, satisfies the weakest axiom $T_0$, and fails the axiom immediately above it.

• Normality without $T_1$ does not imply regularity. This space is normal and not regular. Other combinations occur too: the indiscrete two-point space is regular and normal but not $T_0$, and the cofinite topology on an infinite set is $T_1$ and neither regular nor normal, both worked on this page. This page leaves the converse open; it is refuted later by Assuming countable choice, refuted: every regular space is normal ↗. Only with $T_1$ does normality imply regularity (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

• Why "the closed point" is the right name for $a$. $\{a\}$ is closed and $\{b\}$ is not, so the two points of $S$ are not interchangeable even though the set has only two elements; the labelling is fixed once and for all in The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies.

• The space is $T_0$ and nothing more. It sits at the very bottom of the chain, and it is the standard reminder that the bottom of the chain is not empty: $T_0$ is a real condition, satisfied here and failed by the indiscrete topology on the same underlying set.

Let $X$ be a set, fix $p \in X$, and give $X$ the particular-point topology $\mathcal{T}_p = \{\varnothing\} \cup \{\, U \subseteq X : p \in U \,\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), whose closed sets are $X$ together with the subsets of $X$ that do not contain $p$. Then:

1. $(X, \mathcal{T}_p)$ is $T_0$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces), for every $X$ and every $p$.
2. If $X$ has at least two points then $(X,\mathcal{T}_p)$ is not $T_1$, and hence not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).
3. If $X$ has at least two points then $(X,\mathcal{T}_p)$ is not regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly).
4. If $X$ has at least three points then $(X,\mathcal{T}_p)$ is not normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

Clause 4 needs one point more than clause 3, and the extra point is not slack: on a two-point set the particular-point topology is Sierpinski space, which is normal (Sierpinski space is $T_0$ and normal but neither $T_1$ nor regular: normality without $T_1$ implies nothing). So this family separates the two failures, and it shows that "not regular" and "not normal" begin at different sizes.

• Closures here are as large as they can be. For $A \ne \varnothing$ with $p \notin A$ one has $\overline{A} = A$, since $A$ is already closed; but $\overline{\{p\}} = X$, because the only closed set containing $p$ is $X$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$; the closure is the smallest closed superset and equals $A$ together with its derived set). So the particular point is dense, and a single point can be dense in a space with any number of points at all.

• Every nonempty open set contains $p$, which is the one fact behind clauses 3 and 4 alike. It makes the space as far from Hausdorff as possible while still distinguishing points: no two nonempty open sets are ever disjoint.

• Sierpinski space is the case of two points. With $X = \{p, x\}$ the open sets are $\varnothing$, $\{p\}$ and $X$, which is The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies's Sierpinski topology with the open point named $p$; so clause 4 genuinely needs its extra hypothesis.

Let $X$ be an infinite set — that is, a set that is not finite (Finite, countably infinite, countable, uncountable) — and give it the cofinite topology $\mathcal{T}_{\mathrm{cof}} = \{\varnothing\} \cup \{\, U \subseteq X : X \setminus U \text{ is finite} \,\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), whose closed sets are $X$ together with the finite subsets of $X$. Then:

1. $(X, \mathcal{T}_{\mathrm{cof}})$ is $T_1$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).
2. No two nonempty open sets are disjoint. Consequently the space is not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not), not regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly) and not normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

So the cofinite topology on an infinite set satisfies $T_1$ and $T_0$ and fails every axiom above them. It is the standard witness that $T_1$ is strictly weaker than the Hausdorff condition, and that is how it is used on the main page (FALSE: every $T_1$ space is Hausdorff); here it is pushed further, to show that $T_1$ implies neither of the two axioms that sit above $T_2$ either.

• The three failures have one cause. In the cofinite topology on an infinite set the open sets are so large that no two nonempty ones are disjoint, so every separation axiom whose conclusion is a pair of disjoint nonempty open sets fails at once. What survives is $T_1$, whose conclusion asks for open sets that are allowed to overlap.

• The hypothesis that $X$ is infinite is necessary. On a finite set the cofinite topology is the discrete one (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), which satisfies every axiom on the main page.

• This is the coarsest $T_1$ topology on $X$ (A space is $T_1$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology, clause (d)), which is why it is the natural place to look for a $T_1$ space that fails everything else: any $T_1$ topology on $X$ contains it, so any counterexample to a $T_1$ implication should be sought here first.

Give $\mathbb{R}$ the cocountable topology $\mathcal{T}_{\mathrm{coc}} = \{\varnothing\} \cup \{\, U \subseteq \mathbb{R} : \mathbb{R} \setminus U \text{ is at most countable} \,\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), whose closed sets are $\mathbb{R}$ together with the at most countable subsets of $\mathbb{R}$ (Finite, countably infinite, countable, uncountable). Then:

1. $(\mathbb{R}, \mathcal{T}_{\mathrm{coc}})$ is $T_1$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).
2. Every convergent sequence is eventually constant, so every sequence has at most one limit (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).
3. No two nonempty open sets are disjoint. Consequently the space is not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not), not regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly) and not normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

Clauses 1, 2 and the Hausdorff half of clause 3 are what refute the claim that unique sequential limits force the Hausdorff condition (FALSE: a space in which every sequence has at most one limit is Hausdorff); clause 3's other two halves place the space in the hierarchy exactly where the cofinite topology sits, at $T_1$ and no higher.

• Sequences cannot see this topology. A sequence reaches at most countably many points, and every at most countable set is closed, so the complement of the values other than the limit is an open set that forces the sequence to be eventually constant. Uniqueness of limits is therefore free, and it carries no separation information at all — which is the point of FALSE: a space in which every sequence has at most one limit is Hausdorff.

• The failure of $T_2$, $T_3$ and $T_4$ has the same one-line cause as in the cofinite case: two at most countable sets cannot cover an uncountable one, so two nonempty open sets always meet. What changes between the two examples is only how large a set has to be for the topology to be interesting: infinite for cofinite, uncountable for cocountable.

• On an at most countable set the cocountable topology is discrete (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), so the uncountability of $\mathbb{R}$ is doing real work here and not merely supplying a familiar underlying set.

Let $X$ be a set with the discrete topology $\mathcal{T}_{\mathrm{disc}} = \mathcal{P}(X)$, and let $Y = \{a,b\}$ with $a \ne b$ carry the indiscrete topology $\mathcal{T}_{\mathrm{ind}} = \{\varnothing, Y\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). Then:

1. $(X, \mathcal{T}_{\mathrm{disc}})$ satisfies every axiom named on the main page: it is $T_1$, Hausdorff, Urysohn, regular, completely regular, normal, completely normal and perfectly normal, hence $T_0$, $T_1$, $T_2$, $T_{2\frac12}$, $T_3$, $T_{3\frac12}$, $T_4$, $T_5$ and $T_6$.
2. $(Y, \mathcal{T}_{\mathrm{ind}})$ is regular, completely regular, normal, completely normal and perfectly normal, and it is not $T_0$, hence not $T_1$, not Hausdorff and not Urysohn; and it is not metrizable (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

Clause 2 is the sharpest form of the observation that the unnumbered adjectives carry no information about points: a space may satisfy all five of them and still fail to distinguish any pair of its points. That is exactly what the numerals $T_3$ to $T_6$ are for.

• The two extremes bracket the whole page. Every topology on a set lies between the indiscrete and the discrete one in the comparison order (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), and the two ends of that order sit at opposite ends of the separation hierarchy: the discrete topology satisfies everything, the indiscrete topology on two points satisfies every unnumbered adjective and no numbered axiom at all.

• The indiscrete space is the reason the numerals exist. It refutes at a stroke any reading of "normal", "completely normal", "perfectly normal", "regular" or "completely regular" as implying a separation of points; the main page records the normal case as a false statement, and the argument here shows the same for the other four adjectives.

• The discrete space is metrizable and the indiscrete one is not. Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not records the second, the failure of the Hausdorff condition being an obstruction to metrizability; the first is not needed here, since every axiom was verified directly rather than quoted from the metric theorems of the main page.

Let $\mathbb{R}$ carry its usual metric $d_{\mathbb{R}}(s,t) = |s-t|$ and its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and write $1/(n+1)$ for the inverse of the canonical natural $\iota(n+1)$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field). Let $A \subseteq \mathbb{R}$ be nonempty and closed, and put $d(x,A) := \inf\{\, |x - a| : a \in A \,\}$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). Then, as the general metric theorem (In a metric space every closed set is a zero set and a $G_\delta$, and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal) specialises:

1. $x \mapsto d(x,A)$ is continuous and $A = Z(d(\cdot,A))$ (Zero sets and cozero sets of continuous real-valued functions), so $A$ is a zero set.
2. $A = \bigcap_{n \in \mathbb{N}} \{\, x \in \mathbb{R} : d(x,A) < 1/(n+1) \,\}$, an intersection of open sets, so $A$ is a $G_\delta$ of the topological space $\mathbb{R}$ ($G_\delta$ and $F_\sigma$ subsets of a topological space, agreeing with the real-line notion) and hence a $G_\delta$ subset of $\mathbb{R}$ in the sense of $F_\sigma$ and $G_\delta$ subsets of $\mathbb{R}$, the two notions being the same one.

Two worked instances:

• $A = [0,1]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). Here $d(x,[0,1]) = \begin{cases} -x & x < 0 \\ 0 & 0 \le x \le 1 \\ x - 1 & x > 1 \end{cases}$ so $Z(d(\cdot,[0,1])) = [0,1]$ and, for every real $\varepsilon > 0$, $\{\, x : d(x,[0,1]) < \varepsilon \,\} = (-\varepsilon,\ 1 + \varepsilon)$. Taking $\varepsilon = 1/(n+1)$ gives $[0,1] \;=\; \bigcap_{n \in \mathbb{N}} \big(-1/(n+1),\ 1 + 1/(n+1)\big).$
• $A = \{0\}$. Here $d(x,\{0\}) = |x|$, so $\{0\} \;=\; \bigcap_{n \in \mathbb{N}} \big(-1/(n+1),\ 1/(n+1)\big),$ the standard presentation of a point of $\mathbb{R}$ as a $G_\delta$.

The converse fails. A $G_\delta$ subset of $\mathbb{R}$ need not be closed: $(0,1)$ is open, hence a $G_\delta$ by the constant sequence, and it is not closed.

• The index starts at $n = 0$, where the radius is $1/(0+1) = 1$, so the first set in each intersection is $(-1, 2)$ for $[0,1]$ and $(-1,1)$ for $\{0\}$. Writing $1/n$ instead would divide by zero (The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

• The two presentations are not independent. A zero set is always a $G_\delta$ (Zero sets and cozero sets of continuous real-valued functions), so claim 2 follows from claim 1; the explicit intersection is written out because it is the presentation an argument actually uses, and because it makes the radii visible.

• Why this is the metric case of perfect normality. That every closed set is a $G_\delta$ is one of the two conjuncts of perfect normality (In a metric space every closed set is a zero set and a $G_\delta$, and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal); the other, that $\mathbb{R}$ is normal, comes from the same theorem's completely normal clause. So $\mathbb{R}$ is $T_6$, and everything below it in the chain.