Metrisation and Separation Axioms

Introduction

A metric space carries more than its topology, but its topology is already special: metric spaces are normal, first countable, paracompact, and have a base that is a countable union of locally finite families. The metrisation problem asks the converse question. Given an abstract topological space, when is its topology the metric topology of some distance? A complete answer must supply conditions that are necessary, so that they are satisfied by every metric space, and sufficient, so that a space satisfying them can be given a distance whose balls recover the open sets.

The conditions separate into two layers. The first is the separation layer: a metric space distinguishes points from points, points from closed sets, and closed sets from closed sets, and there is a hierarchy of axioms — T$_0$, T$_1$, Hausdorff, regular, completely regular, normal — recording how much of this a general space retains. The second is a countability layer: the base of a metric space is generated by the balls of rational radius about the points of a dense subset, so it is σ-locally finite, and when the space is separable the base is countable. Urysohn's theorem proves that regularity together with a countable base suffices, and the Nagata–Smirnov theorem proves that regularity together with a σ-locally finite base suffices and that this condition is necessary. The countability condition cannot be dropped: the Sorgenfrey line is perfectly normal and separable and is not metrisable.

This article develops the separation axioms and their permanence, proves Urysohn's lemma and the Tietze extension theorem on which the metrisation theorems rest, and states and discusses the Urysohn, Nagata–Smirnov and Bing metrisation theorems. It closes with the spaces of the corpus that no distance induces. The distance itself and its completeness are the subject of Metric, Uniform and Complete Spaces; paracompactness, the shrinking lemma and the Smirnov theorem that completes the metrisation picture are not covered here; and the covering dimension is treated in Dimension Theory. No measure, integral or analytic limit is used. No physics is invoked.

The Separation Axioms

Points, Closed Sets and the Axioms

All the axioms below constrain a topological space $X$ by asking for open sets that separate specified objects. The objects can be a point and a point, a point and a closed set, or a closed set and a closed set, and the separating sets can be open sets or continuous functions into $[0,1]$.

Definition. A topological space $X$ is

$T_0$ if for all distinct $x, y$ there is an open set containing exactly one of them;
$T_1$ if for all distinct $x, y$ there is an open set containing $x$ and not $y$;
Hausdorff, or $T_2$, if distinct points have disjoint neighbourhoods;
regular if for every point $x$ and every closed set $F$ with $x \notin F$ there are disjoint open sets $U \ni x$ and $V \supseteq F$;
completely regular if for every $x$ and every closed $F \not\ni x$ there is a continuous $f : X \to [0,1]$ with $f(x) = 0$ and $f = 1$ on $F$;
normal if for all disjoint closed sets $F, G$ there are disjoint open sets $U \supseteq F$ and $V \supseteq G$.

The axioms $T_3$, $T_{3\frac12}$ and $T_4$ are the corresponding regularity, complete regularity and normality conditions together with $T_1$:

$$ T_3 = \text{regular} + T_1, \qquad T_{3\frac12} = \text{completely regular} + T_1, \qquad T_4 = \text{normal} + T_1 . $$

A space satisfying $T_{3\frac12}$ is called a Tychonoff space. A normal space in which every closed set is a countable intersection of open sets is perfectly normal.

The $T_1$ hypothesis in the second line is essential and not a technicality. A space with the trivial topology is regular, completely regular and normal by the definitions above, because its only closed sets are $\emptyset$ and $X$; but it is not Hausdorff when it has more than one point. Requiring $T_1$ excludes this, and it is the reason the chain of implications below is stated for the $T$-axioms.

Proposition. For a space $X$, $T_1$ holds if and only if every singleton is closed, and $T_0$ holds if and only if distinct points have distinct closures. Hence $T_1 \Rightarrow T_0$, and in a $T_1$ space the conditions "regular" and "$T_3$" are equivalent.

Proof. If every singleton is closed, then given $x \neq y$ the set $X \setminus \{y\}$ is an open set containing $x$ and not $y$, which is $T_1$; conversely, if $X$ is $T_1$ and $y \neq x$, the open set witnessing the axiom for the pair $(x, y)$ has complement containing $x$ but excluding $y$, so $y \notin \overline{\{x\}}$ and $\{x\} = \overline{\{x\}}$ is closed. For the closure statement, $x$ and $y$ have the same closure exactly when every open set containing one contains the other, which is the negation of $T_0$ applied to both orders.

The Implication Diagram

Theorem. The axioms are ordered as follows:

$$ T_4 \Longrightarrow T_{3\frac12} \Longrightarrow T_3 \Longrightarrow T_2 \Longrightarrow T_1 \Longrightarrow T_0 . $$

Proof. $T_4 \Rightarrow T_{3\frac12}$: in a normal $T_1$ space a point $x$ outside a closed set $F$ has $\{x\}$ and $F$ disjoint closed sets, and Urysohn's lemma, proved from normality alone in the next section, supplies a continuous $f : X \to [0,1]$ with $f(x) = 0$ and $f = 1$ on $F$, which is complete regularity. $T_{3\frac12} \Rightarrow T_3$: if $f$ separates $x$ from the closed set $F$ as above, then $f^{-1}[0,1/2)$ and $f^{-1}(1/2,1]$ are disjoint open sets containing $x$ and $F$ respectively, which is regularity. $T_3 \Rightarrow T_2$: given $x \neq y$, the singleton $\{y\}$ is closed by $T_1$, and regularity separates $x$ from it. $T_2 \Rightarrow T_1$: a Hausdorff separation of $x$ and $y$ gives an open neighbourhood of $x$ missing $y$. $T_1 \Rightarrow T_0$ was proved above.

None of the implications reverses, and the two middle ones reverse the order of the labels $T_3$ and $T_{3\frac12}$ often seen in the literature: complete regularity is the stronger condition, and it is the one Urysohn's lemma supplies. The Sorgenfrey plane is Tychonoff and not normal, so $T_{3\frac12}$ does not imply $T_4$; Mysior's example is a regular $T_1$ space that is not completely regular, so $T_3$ does not imply $T_{3\frac12}$; the Niemytzki plane is a Tychonoff space that is not normal; the space $[0,1]^{I}$ with the product topology for uncountable $I$ is compact Hausdorff and not first countable; and the trivial topology on two points is regular, completely regular and normal but not $T_1$.

Comparing the Axioms: Counterexamples

Each of the standard examples isolates one implication, and they are collected here because the metrisation theorems are read against them.

Example (the Sorgenfrey line). Let $\mathbb{R}_S$ have the topology generated by the half-open intervals $[a, b)$ with $a < b$. It is first countable, since $\{[x, x + 1/n)\}$ is a countable neighbourhood base at $x$; it is separable, since $\mathbb{Q}$ is dense; it is perfectly normal and Lindelöf; but it is not second countable, because no countable family of half-open intervals can generate the topology at every point, and a separable metric space is second countable. Hence $\mathbb{R}_S$ is not metrisable, and it shows that separable plus first countable plus perfectly normal does not suffice.

Example (the Sorgenfrey plane). The square $\mathbb{R}_S \times \mathbb{R}_S$ is separable and first countable, and it is not normal. The antidiagonal $D = \{(x, -x) : x \in \mathbb{R}_S\}$ is closed and discrete, and the two sets $$ A = \{(x, -x) : x \in \mathbb{Q}\}, \qquad B = D \setminus A $$ are disjoint closed sets that cannot be separated by disjoint open sets. So separability is not productive for normality, and the product of separable spaces need not be normal even when both factors are perfectly normal.

Example (the Niemytzki plane). On the closed upper half-plane give the open upper half-plane its usual topology, and give a point $x$ of the boundary the neighbourhoods formed by $\{x\}$ together with an open disc in the upper half-plane tangent to the boundary at $x$. The resulting space is separable and first countable and a Moore space; it is Tychonoff and not normal. It is the standard example of a Moore space that is not metrisable.

Example (the one-point compactification of an uncountable discrete space). Let $X = D \cup \{\infty\}$ with $D$ uncountable and discrete and the neighbourhoods of $\infty$ the cocountable sets containing $\infty$. Then $X$ is compact Hausdorff and not first countable at $\infty$, so no metric induces its topology. It is the simplest compact Hausdorff non-metrisable space.

Example (the Zariski topology). On $\mathbb{R}$ or $\mathbb{C}$ the Zariski topology has for closed sets the zero sets of polynomials, which are finite; the space is $T_1$ but not Hausdorff, and it is not metrisable. It is the topology of the spectrum of a ring, treated in the companion categories of algebraic geometry, and it is one of the standard non-metrisable topologies of the corpus.

Permanence

Whether an axiom passes to subspaces, to products and to images is what makes it usable, and the answers differ sharply between the separation axioms and the countability axioms that metrisation also needs.

Theorem. The following are hereditary (inherited by every subspace): $T_0$, $T_1$, $T_2$, regular, completely regular, normal, perfectly normal, metrisability, first countability, second countability. The closed subspaces of a normal space are normal, and every subspace of a metrisable space is metrisable (with the restricted metric).

Proof. The point-separating axioms are immediate, since a subspace inherits the traces of the open sets. For regularity, given a point $x$ of a subspace $A$ and a closed subset $F$ of $A$ with $x \notin F$, write $F = A \cap G$ with $G$ closed in $X$ and separate $x$ from $G$ in $X$; intersecting with $A$ gives the separation in $A$. For normality of a closed subspace $F$ of a normal space, separate two disjoint closed subsets of $F$, which are closed in $X$, by normality of $X$. Metrisability and the countability axioms are inherited by restriction of the metric and of the base.

Theorem. The following are productive: $T_0$, $T_1$, $T_2$, regular, completely regular, first countability, and metrisability over countable products. Normality and perfect normality are productive over finite products but not over arbitrary products; second countability is productive over countable products but not over arbitrary products; and the arbitrary product $\prod_i X_i$ is metrisable only when all but countably many factors are singletons and the remaining countable product is metrisable.

Proof. For a product of regular spaces, separate $(x_i)$ from a basic closed set by finitely many coordinates and intersect the corresponding separated open sets, each of which contains all but finitely many coordinates. For the arbitrary product of metrisable spaces, a countable base of the product is determined by countably many coordinates, so any factor contributing a nonsingleton must occur among countably many coordinates; and $\{0,1\}^{I}$ is metrisable only for countable $I$, since it is then a countable product of discrete two-point spaces. For the failure of normality, the Sorgenfrey plane is a finite product of normal spaces and is not normal, and the product of uncountably many copies of the unit interval is compact Hausdorff and not normal.

Theorem. Continuous images of compact spaces are compact, but none of $T_2$, regular, normal or metrisable is preserved by arbitrary continuous images; they are preserved by closed continuous images under additional hypotheses.

Proof. The quotient of $\mathbb{R}$ obtained by identifying all integers to one point is a Hausdorff space that is not first countable at the identified point, so it is not metrisable; this is the standard witness. On the positive side, a quotient of a metrisable space by a closed relation is metrisable when the decomposition is upper semicontinuous with compact fibres, and a quotient of a compact metrisable space by a closed equivalence relation is metrisable.

Urysohn's Lemma and the Tietze Theorem

Urysohn's lemma is the tool that converts the separation of closed sets by open sets into the separation of closed sets by continuous functions. Every metrisation theorem in the next section uses it, and it is stated in the written article Topological Spaces without proof.

Theorem (Urysohn's lemma). Let $X$ be a normal space and let $A$ and $B$ be disjoint closed subsets of $X$. Then there is a continuous function $f : X \to [0,1]$ with $f = 0$ on $A$ and $f = 1$ on $B$.

Proof. Let $D$ be the set of dyadic rationals in $[0,1]$. One constructs open sets $U_q$ for $q \in D$ with $$ A \subseteq U_0, \qquad U_1 = X \setminus B, \qquad \overline{U_q} \subseteq U_r \quad (q < r). $$ Start with $U_1 = X \setminus B$ and use normality to find $U_0$ open with $A \subseteq U_0 \subseteq \overline{U_0} \subseteq U_1$. Having defined $U_q$ for the dyadic rationals with denominator $2^n$, let $q < r$ be consecutive ones and insert the new midpoint: normality applied to the closed set $\overline{U_q}$ and the closed set $X \setminus U_r$ produces an open $U_{(q+r)/2}$ with $\overline{U_q} \subseteq U_{(q+r)/2} \subseteq \overline{U_{(q+r)/2}} \subseteq U_r$. Define $$ f(x) = \inf\{\, q \in D : x \in U_q \,\}, \qquad f(x) = 1 \text{ if } x \notin U_q \text{ for every } q \in D . $$ Then $f = 0$ on $A$, and $f = 1$ on $B$ because $U_1 = X \setminus B$. For continuity, note that $f(x) < a$ if and only if $x \in U_q$ for some $q < a$, an open condition, and $f(x) > a$ if and only if $x \notin \overline{U_q}$ for some $q > a$, also open; these conditions generate the topology of $[0,1]$.

Corollary. A normal space is completely regular, and a $T_4$ space is Tychonoff. The function $f$ of the lemma separates $A$ and $B$ in the unit interval, and rescaling gives a function with values in $[a,b]$ for any $a < b$.

Theorem (the Tietze extension theorem). Let $X$ be a normal space, $A \subseteq X$ closed, and $f : A \to [a,b]$ continuous. Then $f$ extends to a continuous $F : X \to [a,b]$ with the same range; if $f$ is real-valued and bounded, it extends to a bounded real-valued function with the same supremum.

Proof sketch. It suffices to treat $[a,b] = [-1,1]$ and $f$ bounded by $1$. Apply Urysohn's lemma successively: with $C_0 = f^{-1}[-1,-1/3]$ and $D_0 = f^{-1}[1/3,1]$, choose $g_0 : X \to [-1/3, 1/3]$ vanishing on $C_0$ and equal to $1/3$ on $D_0$; then $|f - g_0| \leq 2/3$ on $A$. Repeat with the residual function on $A$ and the bounds scaled by $2/3$ to produce $g_1$ with $|g_1| \leq \tfrac13 \cdot \tfrac23$, and so on. The series $\sum_n g_n$ converges uniformly on $X$, by the Weierstrass $M$-test with the geometric bound, and its sum $F$ extends $f$; the partial sums approximate $f$ on $A$ by construction, so the limit agrees with $f$ there.

Remark. The Tietze theorem is the extension statement dual to Urysohn's lemma: the lemma is the case in which $f$ is the two-valued function on the closed set $A \cup B$ taking the values $0$ and $1$, and the extension theorem is the general case. The theorem also holds for functions into $\mathbb{R}^n$, by applying it coordinatewise, and for a function into a separable metric space.

Theorem (Urysohn embedding lemma). Let $X$ be a $T_1$ space and let $(f_n)_{n \in \mathbb{N}}$ be a countable family of continuous functions $f_n : X \to [0,1]$ such that for every $x \in X$ and every neighbourhood $U$ of $x$ there is $n$ with $f_n(x) > 0$ and $f_n = 0$ outside $U$. Then $$ F : X \longrightarrow [0,1]^{\mathbb{N}}, \qquad F(x) = (f_n(x))_{n \in \mathbb{N}}, $$ is an embedding onto a subspace of the Hilbert cube.

Proof. The map $F$ is continuous because each coordinate is. If $x \neq y$, choose a neighbourhood $U$ of $x$ with $y \notin U$; the function $f_n$ of the hypothesis vanishes outside $U$, so $f_n(y) = 0$ while $f_n(x) > 0$, and the images differ. Thus $F$ is injective. To see that $F$ is an embedding, let $U$ be a neighbourhood of $x$ and choose $n$ with $f_n(x) > 0$ and $f_n = 0$ off $U$; then $F^{-1}(V_n) \subseteq U$, where $V_n = \{z : z_n > 0\}$ is an open subset of the cube: if $F(y) \in V_n$ then $f_n(y) > 0$, so $y \in U$. Hence $F$ carries neighbourhoods of $X$ to traces of neighbourhoods of the cube, and the inverse is continuous on the image.

Corollary (local compactness version). If $X$ is locally compact Hausdorff, $K \subseteq X$ compact and $U$ open with $K \subseteq U$, then there is a continuous $f : X \to [0,1]$ with $f = 1$ on $K$ and $f = 0$ outside a compact subset of $U$. Such an $f$ is a bump function for the pair $(K, U)$.

Proof. In a locally compact Hausdorff space there is, by Urysohn's lemma applied to the compact Hausdorff (hence normal) space $X^+$ or directly, an open set $V$ with $K \subseteq V \subseteq \overline{V} \subseteq U$ and $\overline V$ compact; the lemma for the normal space $\overline V$ separates $K$ from its boundary and the result extends by zero.

Metrisation

Necessary Conditions

Before asking for sufficient conditions, it is worth recording what a metrisable space must satisfy. Regularity is the first item, and the base condition is the second.

Theorem. Every metrisable space is $T_4$ (normal Hausdorff), perfectly normal, first countable, and paracompact; it has a σ-discrete base and hence a σ-locally finite base. It is second countable if and only if it is separable, if and only if it is Lindelöf.

Proof. Hausdorffness is the metric separation axiom, and normality was proved in Metric, Uniform and Complete Spaces. For perfect normality, let $F$ be closed and $f(x) = d(x, F)$; then $F = \bigcap_n \{x : f(x) < 1/n\}$, a countable intersection of open sets. The balls of rational radius about the points of a dense subset form a countable base when the space is separable, which gives the second countability; conversely a second-countable metric space is separable by choosing a point from each member of a countable base. The σ-locally-finite base comes from refining the cover by balls of radius $2^{-n}$ about the points of a maximal $2^{-n}$-separated set; the standard construction (Stone) yields a σ-discrete refinement, and paracompactness follows .

Theorem. Let $X$ be metrisable by $d$. Then $d$ induces a bounded metric, $\min(d,1)$ say, with the same topology; the metric $d$ is complete if and only if the topologically complete condition below holds; and a topological space admits a metric if and only if it is metrisable.

Proof. $\min(d,1)$ is a metric by the triangle inequality for the truncated function, and its balls of radius below $1$ coincide with those of $d$. The last statement is the definition of metrisability.

Urysohn's Metrisation Theorem

Theorem (Urysohn). Every regular second-countable $T_1$ space is metrisable.

Proof. Let $\mathcal{B} = (B_n)_{n \in \mathbb{N}}$ be a countable base. For every pair $(m,n)$ with $\overline{B_m} \subseteq B_n$, apply Urysohn's lemma — available because a regular Lindelöf space is normal, and a second-countable space is Lindelöf — to the disjoint closed sets $\overline{B_m}$ and $X \setminus B_n$, obtaining a continuous $f_{m,n} : X \to [0,1]$ that is $1$ on $\overline{B_m}$ and $0$ outside $B_n$. The countable family of the $f_{m,n}$ satisfies the hypothesis of the embedding lemma: given $x$ and a neighbourhood $U$ of $x$, choose a base element $B_n$ with $x \in B_n \subseteq U$ and, by regularity, a base element $B_m$ with $x \in B_m \subseteq \overline{B_m} \subseteq B_n$; then $f_{m,n}(x) = 1 > 0$ and $f_{m,n} = 0$ outside $B_n \subseteq U$. The embedding lemma embeds $X$ into the Hilbert cube $[0,1]^{\mathbb{N}}$, which is metrisable as a countable product of metrisable spaces, and a subspace of a metrisable space is metrisable.

Corollary. Every second-countable regular space is normal, hence perfectly normal; every second-countable normal space is metrisable and separable. The Hilbert cube itself is compact, metrisable and separable, and it contains a homeomorph of every separable metrisable space.

Remark. Urysohn's theorem is the sharpest possible result using only a countable base; the theorem of the next subsection replaces countability by σ-local finiteness, which is the countability condition that a metric space actually satisfies.

The Nagata–Smirnov and Bing Theorems

The countability hypothesis of Urysohn's theorem is that the whole topology has a countable base. A metric space need not have one — an uncountable discrete space is metrisable and has no countable base — so the hypothesis must be weakened to a condition that is still strong enough to build distances and weak enough to hold in every metric space. The correct weakening is a σ-locally finite base, a base that is a countable union of families each of which is locally finite.

Definition. A family $\mathcal{A}$ of subsets of $X$ is locally finite if every point has a neighbourhood meeting only finitely many members of $\mathcal{A}$; it is σ-locally finite if $\mathcal{A} = \bigcup_{n} \mathcal{A}_n$ with each $\mathcal{A}_n$ locally finite; it is discrete if every point has a neighbourhood meeting at most one member, and σ-discrete if it is a countable union of discrete families. A base of the space that is σ-locally finite is a σ-locally finite base.

Theorem (Nagata–Smirnov metrisation theorem). A $T_1$ space $X$ is metrisable if and only if it is regular and has a σ-locally finite base.

Theorem (Bing metrisation theorem). A $T_1$ space $X$ is metrisable if and only if it is regular and has a σ-discrete base.

Proof sketch. The necessity is the construction of the preceding subsection: a metric space has the σ-discrete base obtained from the covers by balls of radius $2^{-n}$, refined locally finitely and made discrete by a standard shrinking argument. For sufficiency one builds a metric directly from the base. Let $\mathcal{B} = \bigcup_n \mathcal{B}_n$ be a σ-locally finite base and for each $n$ and each $B \in \mathcal{B}_n$ define, using local finiteness and normality, a continuous $f_{n,B} : X \to [0,1]$ with $f_{n,B} = 1$ on $B$ and $f_{n,B} = 0$ outside a slightly larger locally assigned open set; the local finiteness makes the sums $\sum_{B \in \mathcal{B}_n} f_{n,B}$ locally finite. Put $$ d(x, y) = \sum_{n=1}^{\infty} \frac{1}{2^n} \sum_{B \in \mathcal{B}_n} \bigl| f_{n,B}(x) - f_{n,B}(y) \bigr| . $$ The series converges, $d$ is symmetric and satisfies the triangle inequality, and $d(x,y) = 0$ implies $x = y$ because the functions distinguish the points of a $T_1$ space. The metric induces the original topology: an open set $U$ and $x \in U$ contain a base element $B \in \mathcal{B}_n$ with $x \in B$, and the function $f_{n,B}$ is $1$ at $x$ and $0$ off a set contained in $U$, so a small $d$-ball about $x$ lies in $U$; conversely the functions are continuous for the original topology, so the $d$-balls are open in it. This is the Nagata–Smirnov metric, and Bing's argument replaces the σ-locally finite base by a σ-discrete one and the sums by an equivalent metric built on the same pattern.

Corollary. A regular space with a σ-locally finite base is paracompact and normal. Every metrisable space has such a base, so the two conditions of the theorem are independent of the particular metric: the theorem characterises the topology, not the distance.

Remark. Neither the Urysohn nor the Nagata–Smirnov condition is sufficient without regularity, and neither the σ-local finiteness nor the Lindelöf property may be weakened. The Sorgenfrey line is regular, perfectly normal and Lindelöf, and it is not metrisable, so by the Nagata–Smirnov theorem it has no σ-locally finite base; the natural base of half-open intervals fails local finiteness at every point, since each point lies in infinitely many members $[q, q + 1/n)$ with $q$ rational. The base condition of Nagata–Smirnov is therefore strictly stronger than Lindelöfness, and the example is the sharpest caution against reading the theorem as a countability statement.

Complete Metrisability

A metrisable space may admit some metrics that are complete and some that are not, and Metric, Uniform and Complete Spaces observes that completeness is not a topological property. The topological spaces that admit a complete metric can nevertheless be characterised.

Theorem. A metrisable space $X$ admits a complete metric if and only if it is a $G_\delta$ subset of its completion; equivalently, if and only if it is completely metrisable, that is, $X$ is a $G_\delta$ in some (and hence in every) compactification. A closed subset of a completely metrisable space is completely metrisable, and a countable intersection of open subsets of a completely metrisable space is completely metrisable.

Proof sketch. If $d$ is a complete metric inducing the topology and $(\hat X, \hat d)$ is the completion of a metric space homeomorphic to $X$, then $X$ is the intersection of the open sets $\{y : \text{some } x_n \to y \text{ with } x_n \in X\}$ arising from the embedding, a countable intersection obtained from a countable dense set of $X$; conversely a $G_\delta$ subset of a complete space carries a complete metric by a standard reparametrisation of the restricted metric along the defining open sets. The last two statements are the previous subsection of Metric, Uniform and Complete Spaces.

Remark. Every completely metrisable space is a Baire space, by the Baire category theorem for complete metric spaces, and the converse fails: the irrationals are completely metrisable and a Baire space, while the rationals are neither. The Baire property and its consequences are not covered here but one.

Non-metrisable Spaces in the Corpus

The corpus uses several topologies that carry other structure and are not metrisable, and it is worth recording which of them are, so that no later argument silently assumes a distance.

Example (the Stone space of $\mathbb{N}$). By the dictionary of nets and filters, the set $\beta\mathbb{N}$ of ultrafilters on $\mathbb{N}$ carries the topology with base $\{U : A \in U\}$ for $A \subseteq \mathbb{N}$. It is compact Hausdorff, and it is not first countable at any free ultrafilter: given countably many sets $A_n$ in an ultrafilter $\mathcal{U}$, partition $\mathbb{N}$ into finite or infinite cells refining them and choose a set $B$ meeting every cell in a way incompatible with all but finitely many of the $A_n$; the details give an open neighbourhood of $\mathcal{U}$ containing no member of any proposed countable base. Hence $\beta\mathbb{N}$ is not metrisable. It is the Stone–Čech compactification of the discrete space $\mathbb{N}$, and it is the presentation of the compact Hausdorff spaces that is available once ultrafilters are, as in Nets, Filters and Convergence.

Example (arbitrary products). The space $\{0,1\}^{I}$ is compact Hausdorff and metrisable exactly when $I$ is countable. For uncountable $I$ it is not first countable, so it carries no metric; it is the standard compact Hausdorff non-metrisable space, and it is the ambient space of the inverse limits that are treated in Topological Groups.

Example (profinite groups). A profinite group $G = \varprojlim_i G_i$ with the inverse limit topology is a closed subgroup of $\prod_i G_i$ and is compact Hausdorff and totally disconnected. It is metrisable exactly when it has a countable neighbourhood base at the identity, equivalently when the open normal subgroups form a countable family, equivalently when the inverse system may be indexed by a countable directed set. Thus the Galois group of a countable field with the Krull topology is metrisable, while an uncountable product of nontrivial finite groups is not. The point for this article is that the Krull topology is a legitimate topology of the corpus which is metrisable in exactly this countable case, and no more.

Example (the Gelfand spectrum). Gelfand duality attaches to a commutative C$^*$-algebra a compact Hausdorff space, its spectrum, whose algebra of continuous functions is the given algebra; the construction is treated, and the present article records only the outcome: the spectrum is metrisable exactly when the algebra is separable. The spectrum of $\ell^\infty(\mathbb{N})$ is the space $\beta\mathbb{N}$ of the previous example, so spectra supply compact Hausdorff spaces that are not second countable.

Example (the Zariski topology). The prime spectrum $\operatorname{Spec} R$ of a commutative ring, with the topology whose closed sets are the sets of primes containing a fixed ideal, is $T_0$ but not $T_1$ whenever a nonzero prime is contained in another; it is therefore not metrisable. This is the topology of the spectra of the later categories of this Part, and its failure of metrisability is a failure of the separation axioms and not merely of countability.

Summary

The separation axioms record how far a space can separate points, closed sets and pairs of closed sets: $T_0$, $T_1$, Hausdorff, regular, completely regular and normal, with $T_3$, $T_{3\frac12}$ and $T_4$ denoting the corresponding conditions together with $T_1$; they are ordered $T_4 \Rightarrow T_3 \Rightarrow T_{3\frac12} \Rightarrow T_2 \Rightarrow T_1 \Rightarrow T_0$, and no implication reverses. The Sorgenfrey line is perfectly normal and separable and not metrisable, the Sorgenfrey plane and the Niemytzki plane show that separability and first countability do not give normality, and the one-point compactification of an uncountable discrete space is compact Hausdorff and not metrisable.

The axioms are hereditary, and $T_0$ through complete regularity are preserved by arbitrary products; normality and perfect normality fail for infinite products, second countability fails for uncountable products, and metrisability of a product needs all but countably many factors to be singletons. Continuous images preserve none of them in general.

Urysohn's lemma produces a continuous function separating two disjoint closed sets in a normal space, and the Tietze theorem extends a continuous function from a closed subset to the whole space with the same range. The Urysohn embedding lemma turns a countable separating family of functions into an embedding into the Hilbert cube, and Urysohn's metrisation theorem concludes that a regular second-countable $T_1$ space is metrisable. The Nagata–Smirnov theorem characterises metrisability by regularity together with a σ-locally finite base, Bing's theorem by a σ-discrete base, and the condition is necessary as well as sufficient, so it is a property of the topology and not of any particular distance. A metrisable space admits a complete metric exactly when it is completely metrisable, that is, a $G_\delta$ in its completion. Among the non-metrisable spaces of the corpus are the Stone space $\beta\mathbb{N}$, the uncountable products, and the compact and spectral spaces of the later categories of this Part, among them the profinite groups with uncountably many open subgroups.

Summary of Notation

Symbol	Meaning
$X, Y$	Topological spaces
$T_0, T_1, T_2, T_3, T_{3\frac12}, T_4$	Separation axioms; Hausdorff, regular, completely regular, normal with $T_1$
regular, normal, completely regular	Point-versus-closed-set, closed-versus-closed-set, and function separation
Tychonoff	A $T_{3\frac12}$ space
perfectly normal	Normal and every closed set a countable intersection of open sets
$F, G$	Closed sets; $A, B$ disjoint closed sets in Urysohn's lemma
$f : X \to [0,1]$	Urysohn function; $F$ its Tietze extension
$[0,1]^{\mathbb{N}}$	Hilbert cube; universal separable metrisable space
$\mathcal{B}$, $(B_n)$	A base; a countable base
locally finite, σ-locally finite, discrete, σ-discrete	The refinement conditions of Nagata–Smirnov and Bing
$d(x,y) = \sum_n 2^{-n} \sum_{B} \|f_{n,B}(x) - f_{n,B}(y)\|$	The Nagata–Smirnov metric built from a σ-locally finite base
$G_\delta$	Countable intersection of open sets; complete metrisability
$\mathbb{R}_S$	Sorgenfrey line; half-open interval topology
$\beta\mathbb{N}$	Ultrafilters on $\mathbb{N}$ with the Stone topology
$\operatorname{Spec} R$	Prime spectrum with the Zariski topology
$\overline{A}$, $A^\circ$	Closure and interior, as in Topological Spaces