Separation Axioms: the Hierarchy

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

• $X$ is $T_0$, or a Kolmogorov space, when any two distinct points are topologically distinguishable: for all $x,y\in X$ with $x\mathrm{\ne }y$ there is an open set containing exactly one of $x$ and $y$.
• $X$ is $T_1$, or a Frechet space, when each of any two distinct points has an open set containing it and missing the other: for all $x,y\in X$ with $x\mathrm{\ne }y$ there are $U,V\in \mathcal{T}$ with

$x\in U,y\mathrm{\notin }U,y\in V,x\mathrm{\notin }V\mathrm{.}$

Nothing is asserted about a pair of equal points, so a space with at most one point satisfies both conditions vacuously.

Since an open set containing a point is an open neighbourhood of it (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open), both conditions may be read with "open neighbourhood" in place of "open set"; and by the same equivalence recorded in Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open they may be read with arbitrary neighbourhoods, since a neighbourhood of $x$ contains an open one and an open neighbourhood is a neighbourhood.

Every $T_1$ space is $T_0$, and this is discharged here rather than left to the reader, because it is the bottom arrow of the whole hierarchy on this page. Let $x\mathrm{\ne }y$ and take $U,V$ as in the $T_1$ condition. Then $U$ is an open set containing $x$ and not $y$, so it contains exactly one of the two points, which is the $T_0$ condition. Only the first half of the $T_1$ condition is used, so the implication does not reverse formally, and it does not reverse in fact: Sierpinski space is a witness, recorded on the companion page.

The two conditions differ exactly in symmetry. $T_0$ asks for one open set that tells the pair apart, with no control over which of the two it contains; $T_1$ asks for both separations at once. In Sierpinski space $(\{a,b\},\{\mathrm{\varnothing },\{b\},\{a,b\}\})$ of The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies the open set $\{b\}$ contains $b$ and not $a$, so the space is $T_0$; but the only open set containing $a$ is the whole space, which also contains $b$, so it is not $T_1$.

Neither condition is a property of a set alone. Both are properties of the pair $(X,\mathcal{T})$, and both are inherited upwards along the comparison order of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison: if ${\mathcal{T}}_1\subseteq {\mathcal{T}}_2$ and $(X,{\mathcal{T}}_1)$ is $T_0$, respectively $T_1$, then so is $(X,{\mathcal{T}}_2)$, since the separating open sets of the coarser topology lie in the finer one. In particular the discrete topology satisfies both, and the indiscrete topology on a set with at least two points satisfies neither.

• The names. The numbering $T_0,T_1,T_2,\dots$ is Alexandroff and Hopf's Trennungsaxiome; the individual names honour Kolmogorov and Frechet. This page fixes each axiom by its condition and treats the numeral as an abbreviation, because the numerals above $T_3$ are used inconsistently in the literature (Conventions on this page, and the one implication of the classical chain that is not available at this point in the reading order).

• What $T_1$ says about closures. $T_1$ is equivalent to the closedness of every singleton, and hence to $\overline{\{x\}}=\{x\}$ for every point (Interior, closure, boundary, exterior, derived set and isolated point in a topological space); that equivalence is the next item and is a theorem, not a restatement. The corresponding characterisation of $T_0$, that distinct points have distinct closures, is not needed on this page and is not proved here.

• No separation is built into the word space. Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison assumes none; every separation property on this page is a hypothesis written out where it is used.

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}}$ be the cofinite topology on the set $X$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). The following four conditions are equivalent.

• (a) $X$ is $T_1$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).
• (b) $\{x\}$ is closed for every $x\in X$.
• (c) $F$ is closed for every finite $F\subseteq X$ (Finite, countably infinite, countable, uncountable).
• (d) ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}}\subseteq \mathcal{T}$, that is, the topology of $X$ is finer than the cofinite topology on the same set.

Condition (d) says that the cofinite topology is the coarsest $T_1$ topology on any set: it is $T_1$ by the equivalence, and every $T_1$ topology on that set contains it.

• The theorem is the reason $T_1$ is quoted as "points are closed". Every later use of $T_1$ on this page goes through clause (b): the $T_1$ hypothesis in $T_3$ and $T_4$ is used exactly to turn a point into a closed set so that regularity or normality applies to it.

• Clause (c) is not a strengthening of clause (b). It follows from it by a finite union, and the finite union is genuinely finite: an arbitrary union of closed sets need not be closed, and in the cofinite topology on an infinite set no infinite proper subset is closed at all, although every singleton is.

• Clause (d) locates the cofinite topology. It is the smallest $T_1$ topology on a given set, in the sense of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison's comparison order, and this is why it is the standard witness for a $T_1$ space that fails every stronger separation axiom; the witness is worked on the companion page.

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $A,B\subseteq X$, with closures taken in $X$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space). Then $A$ and $B$ are separated when

$\bar{A}\cap B=\mathrm{\varnothing }\text{and}A\cap \bar{B}=\mathrm{\varnothing }\mathrm{.}$

Equivalently, neither set meets the closure of the other. The condition is symmetric in $A$ and $B$ by construction, and it is inherited downwards: if $A$ and $B$ are separated and $A^{\mathrm{\prime }}\subseteq A$, $B^{\mathrm{\prime }}\subseteq B$, then $A^{\mathrm{\prime }}$ and $B^{\mathrm{\prime }}$ are separated, because $A^{\mathrm{\prime }}\subseteq A$ forces $\overline{A^{\mathrm{\prime }}}\subseteq \bar{A}$, the closure $\bar{A}$ being a closed superset of $A^{\mathrm{\prime }}$ and $\overline{A^{\mathrm{\prime }}}$ the smallest such (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, claim 2).

Separated sets are disjoint, and being disjoint is not enough. From $A\subseteq \bar{A}$ one gets $A\cap B\subseteq \bar{A}\cap B=\mathrm{\varnothing }$. The converse fails: in $\mathbb{R}$ with its usual topology the sets $A=(0,1)$ and $B=[1,2)$ are disjoint, yet $1\in \bar{A}\cap B$, so they are not separated.

Two sufficient conditions, both used constantly below.

1. Disjoint closed sets are separated. If $A$ and $B$ are closed and disjoint then $\bar{A}=A$ and $\bar{B}=B$ (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, claim 2), so both displayed intersections are $A\cap B=\mathrm{\varnothing }$.
2. Disjoint open sets are separated. Let $U,V$ be open and disjoint. If $y\in V$ then $V$ is an open set containing $y$ and missing $U$, so $y\mathrm{\notin }\bar{U}$ by clause (c) of 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; hence $\bar{U}\cap V=\mathrm{\varnothing }$, and symmetrically $U\cap \bar{V}=\mathrm{\varnothing }$.

Separation is absolute rather than relative to a subspace. Let $A,B\subseteq S\subseteq X$ with $S$ carrying the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then $A$ and $B$ are separated in the space $S$ if and only if they are separated in $X$. Indeed ${\mathrm{cl}}_S(A)=\bar{A}\cap S$ (For $A\subseteq S\subseteq X$ the closure of $A$ in $S$ is ${\bar{A}}^X\cap S$, while the interior only contains ${\mathrm{int}}^X(A)\cap S$, with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$, claim 1), so

${\mathrm{cl}}_S(A)\cap B=\bar{A}\cap S\cap B=\bar{A}\cap B$

because $B\subseteq S$, and symmetrically for the other intersection. So the phrase "$A$ and $B$ are separated" needs no ambient space named once both sets are fixed, and this is exactly what makes the notion the right hypothesis for complete normality later on this page.

• Why the notion is not "disjoint closures". Requiring $\bar{A}\cap \bar{B}=\mathrm{\varnothing }$ is strictly stronger, and it is too strong to be useful: in $\mathbb{R}$ the sets $(0,1)$ and $(1,2)$ are separated in the sense above, while their closures $[0,1]$ and $[1,2]$ meet. The definition asks only that each set avoid the other's closure.

• The vocabulary collides with two others and neither is meant here. "$A$ and $B$ are separated by disjoint open sets" is a different, stronger condition, and it is the conclusion of the normality and complete-normality axioms below, not the hypothesis. "Separable", meaning "has an at most countable dense subset", is unrelated and is defined later in Separability: the existence of an at most countable dense subset ↗.

• Nothing here needs a separation axiom. The definition and all four observations above hold in an arbitrary topological space, points closed or not.

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $A\subseteq X$.

• $A$ is a $G_{\delta }$ set of $X$ when there is a sequence $(V_n{)}_{n\in \mathbb{N}}$ of open subsets of $X$ with $A={\bigcap }_{n\in \mathbb{N}}{V}_n\mathrm{.}$
• $A$ is an $F_{\sigma }$ set of $X$ when there is a sequence $(F_n{)}_{n\in \mathbb{N}}$ of closed subsets of $X$ with $A={\bigcup }_{n\in \mathbb{N}}{F}_n\mathrm{.}$

As everywhere in this library $\mathbb{N}$ contains $0$, so both indexings start at $0$. An at most countable family may always be presented as a sequence (Finite, countably infinite, countable, uncountable): a finite list $V_0,\dots ,V_m$ is extended by $V_n:=V_m$ for $n>m$, which changes neither the intersection nor the union, so nothing is lost by indexing over $\mathbb{N}$.

The two classes are exchanged by complementation. $A$ is $F_{\sigma }$ in $X$ if and only if $X\setminus A$ is $G_{\delta }$ in $X$. If $A={\bigcup }_n{F}_n$ with each $F_n$ closed then $X\setminus A={\bigcap }_n(X\setminus F_n)$ by De Morgan and each $X\setminus F_n$ is open (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison); the converse is the same computation read backwards.

Every open set is $G_{\delta }$ and every closed set is $F_{\sigma }$, by the constant sequence $V_n:=A$, respectively $F_n:=A$. Neither converse holds, and $\mathbb{R}$ with its usual topology already refutes both. The singleton $\{0\}$ is a $G_{\delta }$ that is not open: it is ${\bigcap }_{n\in \mathbb{N}}(-1\mathrm{/}(n+1),1\mathrm{/}(n+1))$, since $0$ lies in every one of those intervals while a real $t\mathrm{\ne }0$ is excluded at some index, the Archimedean property giving a natural $k\ge 1$ with $1\mathrm{/}k<\mathrm{\mid }t\mathrm{\mid }$ and $k$ being a successor $n+1$ (For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$, Every nonzero natural number is a successor, The canonical natural $\iota (n)=n\cdot 1_F$ of a field); and $\{0\}$ is not open because every bounded open interval $(a,b)$ with $a<0<b$ contains the point $b\mathrm{/}2\mathrm{\ne }0$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, claim 3). Complementing, $\mathbb{R}\setminus \{0\}$ is an $F_{\sigma }$ that is not closed, its complement $\{0\}$ not being open.

The condition that is a real restriction is the other pairing, namely that every closed set be a $G_{\delta }$, equivalently that every open set be an $F_{\sigma }$. That is not automatic in an arbitrary space, and it is exactly the second conjunct of perfect normality later on this page. It must not be confused with the two automatic inclusions above: they hold everywhere and say nothing about a space.

Agreement with the real-line notion, stated because a second notion of the same name would be a defect. $F_{\sigma }$ and $G_{\delta }$ subsets of $\mathbb{R}$ defines $F_{\sigma }$ and $G_{\delta }$ subsets of $\mathbb{R}$ by the same two displayed conditions, with "open" and "closed" read in the sense of Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen. Those two words name the same two collections of subsets of $\mathbb{R}$ as the usual topology of $\mathbb{R}$ does, and the verification is one line of unfolding. Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen calls $U$ open when every $x\in U$ admits $\epsilon >0$ with $N_{\epsilon }(x)\subseteq U$, where $N_{\epsilon }(x)=(x-\epsilon ,x+\epsilon )$ (The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$); The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement calls $U$ open in $(\mathbb{R},d_{\mathbb{R}})$ when every $x\in U$ admits $r>0$ with $B(x,r)\subseteq U$, and $B(x,r)=(x-r,x+r)$ by claim 2 of The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded. The two conditions are therefore the same condition word for word, so the two collections of open subsets of $\mathbb{R}$ are one collection, and hence so are the two collections of closed subsets, each being the complements of the other collection. The usual topology of $\mathbb{R}$ is the metric topology of $d_{\mathbb{R}}$ (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). Since the two definitions quantify over one collection of open sets and one collection of closed sets, a subset of $\mathbb{R}$ is $G_{\delta }$ in the sense above, for $\mathbb{R}$ with its usual topology, if and only if it is $G_{\delta }$ in the sense of $F_{\sigma }$ and $G_{\delta }$ subsets of $\mathbb{R}$; and likewise for $F_{\sigma }$. There is one notion here, not two, and every statement proved about $F_{\sigma }$ or $G_{\delta }$ subsets of $\mathbb{R}$ elsewhere in this library may be quoted verbatim as a statement about the topological space $\mathbb{R}$.

• The letters. $F$ for ferme with $\sigma$ for somme, $G$ for Gebiet with $\delta$ for Durchschnitt, as $F_{\sigma }$ and $G_{\delta }$ subsets of $\mathbb{R}$ records.

• Neither class is closed under complementation, which is why both names are needed; and neither is a topology, an arbitrary union of $G_{\delta }$ sets being no longer $G_{\delta }$ in general. What is true, and all that is used on this page, is the complementation duality above together with the fact that a finite intersection of $G_{\delta }$ sets and a finite union of $F_{\sigma }$ sets stay in their class, by rearranging a finite array of sequences.

• In a metric space every closed set is $G_{\delta }$. That is proved later on this page from the distance function, and it is the reason every metrizable space is perfectly normal. In a general space it can fail, so it is a genuine hypothesis and not a convenience.

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

• $X$ is regular when a point can be separated from a closed set not containing it: for every closed $C\subseteq X$ and every $x\in X\setminus C$ there are $U,V\in \mathcal{T}$ with $x\in U,C\subseteq V,U\cap V=\mathrm{\varnothing }\mathrm{.}$
• $X$ is $T_3$ when it is regular and $T_1$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).

Since an open set containing a point is an open neighbourhood of it (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open), regularity reads: $x$ and $C$ have disjoint open neighbourhoods. The case $C=\mathrm{\varnothing }$ is allowed and is satisfied by $U=X$, $V=\mathrm{\varnothing }$, so no nonemptiness is hidden in the condition.

The convention fork, and this library's side of it. Textbooks disagree about whether the word regular carries a $T_1$ hypothesis. Munkres builds it in, defining a regular space to be one in which points are closed and the separation condition above holds; Kelley, Willard and Engelking do not, and reserve $T_3$ for the conjunction. This library takes the second side: regular names the separation condition alone, $T_3$ names regular plus $T_1$, and every statement that needs points to be closed writes the $T_1$ hypothesis out. The reason is that the two halves are genuinely independent and each is used alone below: the indiscrete topology on a two-point set is regular and not $T_0$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), and the cofinite topology on an infinite set is $T_1$ and not regular, both witnessed on the companion page.

Regularity alone implies no other separation axiom. It does not imply $T_0$, $T_1$ or Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not): in the indiscrete topology on a set $X$ the only closed sets are $\mathrm{\varnothing }$ and $X$, so the only pair $(C,x)$ to be separated has $C=\mathrm{\varnothing }$, and $U=X$, $V=\mathrm{\varnothing }$ separates it; yet no two distinct points are distinguished by any open set. Conversely $T_1$ does not imply regularity. It is the conjunction $T_3$ that sits above Hausdorff in the hierarchy, and the proof of that is three items below.

• A regular space is not required to separate two closed sets, which is the stronger condition of normality defined later on this page; and a normal space is not required to separate a point from a closed set, since a point need not be closed. Normality does not imply regularity, and the witness is Sierpinski space on the companion page. Whether regularity implies normality is a question this page leaves open, and no statement here asserts an answer (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

• What regularity is really about. The reformulation proved next — every point has a neighbourhood base of closed neighbourhoods — is the form in which regularity is used in practice, and the form in which it is verified for the ordinal spaces later on this page, whose basis consists of clopen sets.

• The numeral. Because of the fork above, "$T_3$" in the literature may mean either what is defined here or the bare separation condition. This library always writes the numeral for the conjunction and never uses it to abbreviate the separation condition alone (Conventions on this page, and the one implication of the classical chain that is not available at this point in the reading order).

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), with closures as in Interior, closure, boundary, exterior, derived set and isolated point in a topological space and neighbourhoods as in Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, so that a neighbourhood need not be open. The following three conditions are equivalent.

• (a) $X$ is regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly).
• (b) For every $x\in X$ and every open $U$ with $x\in U$ there is an open $V$ with $x\in V\subseteq \bar{V}\subseteq U\mathrm{.}$
• (c) Every point of $X$ has a neighbourhood base consisting of closed neighbourhoods: for every $x\in X$ and every neighbourhood $N$ of $x$ there is a closed neighbourhood $K$ of $x$ with $K\subseteq N$.

• Clause (b) is the working form. Every application of regularity below uses it in the shape "shrink an open set around a point so that even its closure stays inside", which is what makes regularity behave like a one-sided version of the normality shrinking lemma proved later on this page.

• Clause (c) is what makes a clopen basis decisive. If a space has a basis of clopen sets then the basic sets containing a point are closed neighbourhoods of it and form a neighbourhood base (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open), so (c) holds and the space is regular with no further work. That is exactly the route by which the ordinal spaces later on this page are shown to be regular.

• No separation hypothesis is used anywhere above. Points need not be closed, and the lemma is a statement about regularity alone; combining it with $T_1$ is the separate step that produces $T_3$.

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), with closures as in Interior, closure, boundary, exterior, derived set and isolated point in a topological space. Then $X$ is an Urysohn space, also written $T_{2\frac{1}{2}}$, when any two distinct points have open neighbourhoods whose closures are disjoint: for all $x,y\in X$ with $x\mathrm{\ne }y$ there are $U,V\in \mathcal{T}$ with

$x\in U,y\in V,\bar{U}\cap \bar{V}=\mathrm{\varnothing }\mathrm{.}$

Equivalently, by Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, distinct points have disjoint closed neighbourhoods: if $U$ and $V$ are as displayed then $\bar{U}$ and $\bar{V}$ are disjoint closed neighbourhoods of $x$ and $y$; conversely disjoint closed neighbourhoods $K\ni x$ and $L\ni y$ contain open $U\ni x$ and $V\ni y$ with $\bar{U}\subseteq K$ and $\bar{V}\subseteq L$, since $K$ and $L$ are closed, so the closures are disjoint.

The condition is vacuous for a space with at most one point, and nothing is asserted about equal points.

The condition strictly strengthens the Hausdorff condition on its face: $U\subseteq \bar{U}$ and $V\subseteq \bar{V}$, so disjointness of the closures forces disjointness of $U$ and $V$ and hence the Hausdorff property (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not). That implication is proved as the next item, together with the implication that puts Urysohn spaces below the regular $T_1$ spaces. This page does not exhibit a Hausdorff space that is not Urysohn, and it does not assert that one exists: every witness reachable from the material developed here would need machinery this page does not have, so the question whether the implication reverses is left open here.

A live naming collision, flagged here and settled in this page's conventions. Two different conditions travel under Urysohn's name:

• the one defined above, separation of points by disjoint closed neighbourhoods, which is what this library calls Urysohn and $T_{2\frac{1}{2}}$;
• separation of points by a continuous real-valued function, which is usually called completely Hausdorff and which this library does not define.

Some texts exchange the two names. Neither is Urysohn's lemma, a theorem about normal spaces that is not proved on this page at all (Conventions on this page, and the one implication of the classical chain that is not available at this point in the reading order).

• The fractional numeral is an interpolation, not an arithmetic fact. It records that the condition sits between $T_2$ and $T_3$ in the standard ordering and carries no other meaning; the same is true of $T_{3\frac{1}{2}}$ later on this page.

• Closures, not interiors. Replacing "$\bar{U}\cap \bar{V}=\mathrm{\varnothing }$" by "$U\cap V=\mathrm{\varnothing }$" gives back the Hausdorff condition exactly, so the whole content of the axiom is the passage to closures.

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). Then:

1. If $X$ is Urysohn (Urysohn ($T_{2\frac{1}{2}}$) space: distinct points have neighbourhoods with disjoint closures) then $X$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).
2. If $X$ is Hausdorff then $X$ is $T_1$, and hence $T_0$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).
3. If $X$ is regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly) and $T_1$ — that is, if $X$ is $T_3$ — then $X$ is Urysohn.

Consequently $T_3$ implies $T_{2\frac{1}{2}}$, which implies $T_2$, which implies $T_1$, which implies $T_0$. Nothing here asserts that any of the four implications reverses; two of the failures are recorded among this page's false statements.

• Claim 3 is where the $T_1$ hypothesis earns its place. Regularity separates a point from a closed set, and the closed set used in the proof is the singleton $\{y\}$; without $T_1$ that singleton need not be closed and the argument has nothing to start from. The indiscrete topology on two points is regular and not Urysohn, which shows the hypothesis cannot simply be dropped.

• Regularity is applied twice, and the second application is the whole point. The first shrink puts $y$ outside $\bar{U}$; the second separates $y$ from the closed set $\bar{U}$, which is what upgrades disjointness of the sets to disjointness of their closures.

• Claim 2 explains why the Hausdorff condition alone is often quoted as "points are closed". By 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 the two are not the same condition; $T_1$ is strictly weaker, and the cofinite topology on an infinite set separates them.

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $\mathbb{R}$ carry its usual topology, the metric topology of $d_{\mathbb{R}}(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ 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). For a continuous $f:X\to \mathbb{R}$ (Continuity of a map of topological spaces at a point and globally) put

$Z(f):=f^{-1}[\{0\}]=\{x\in X:f(x)=0\},\mathrm{coz}(f):=X\setminus Z(f)=\{x\in X:f(x)\mathrm{\ne }0\}\mathrm{.}$

$Z(f)$ is the zero set of $f$ and $\mathrm{coz}(f)$ its cozero set. A subset of $X$ is a zero set of $X$ when it is $Z(f)$ for some continuous $f:X\to \mathbb{R}$, and a cozero set of $X$ when it is the complement of one. Where the target is written $[0,1]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with its subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace), a continuous map $X\to [0,1]$ is the same thing as a continuous map $X\to \mathbb{R}$ with all values in $[0,1]$, by the characteristic property of a map into a subspace recorded in Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace; so nothing below depends on which of the two targets is written.

Every zero set is closed and every cozero set is open. $\{0\}$ is closed in $\mathbb{R}$: its complement $\mathbb{R}\setminus \{0\}$ is open, since a point $t\mathrm{\ne }0$ has the bounded open interval $(t-\mathrm{\mid }t\mathrm{\mid },t+\mathrm{\mid }t\mathrm{\mid })$ around it inside $\mathbb{R}\setminus \{0\}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, claim 3). The preimage of a closed set under a continuous map is closed (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\bar{A})\subseteq \overline{f(A)}$, clause (c)).

Every zero set is a $G_{\delta }$ and every cozero set an $F_{\sigma }$ ($G_{\delta }$ and $F_{\sigma }$ subsets of a topological space, agreeing with the real-line notion). Writing $\iota$ for the canonical natural of $\mathbb{R}$ (The canonical natural $\iota (n)=n\cdot 1_F$ of a field), so that $1\mathrm{/}(n+1)$ abbreviates the inverse of $\iota (n+1)$, put

$V_n:=f^{-1}[(-1\mathrm{/}(n+1),1\mathrm{/}(n+1))](n\in \mathbb{N})\mathrm{.}$

Each $V_n$ is open, being the preimage of an open interval (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\bar{A})\subseteq \overline{f(A)}$, clause (b)). Clearly $Z(f)\subseteq {\bigcap }_n{V}_n$. Conversely, if $f(x)\mathrm{\ne }0$ then $\epsilon :=\mathrm{\mid }f(x)\mathrm{\mid }>0$, and For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$ gives a natural $k\ge 1$ with $1\mathrm{/}k<\epsilon$; since $k\mathrm{\ne }0$ it is a successor, $k=n+1$ with $n\in \mathbb{N}$ (Every nonzero natural number is a successor), so $\mathrm{\mid }f(x)\mathrm{\mid }>1\mathrm{/}(n+1)$ and $x\mathrm{\notin }{V}_n$. Hence $Z(f)={\bigcap }_n{V}_n$ is a $G_{\delta }$, and $\mathrm{coz}(f)$ is an $F_{\sigma }$ by complementation.