Every [metric space](/page/Metric%20Space) carries a natural topology --- the collection of sets that are unions of open balls. This metric topology is extraordinarily well-behaved: sequences determine the topology (a set is closed if and only if it is sequentially closed), continuous functions can be characterised by the $\varepsilon$-$\delta$ criterion, and compact sets are precisely the sequentially compact ones. The question that launches the theory of metrizable spaces is: **given a topological space $(X, \tau)$, when does there exist a metric $d$ on $X$ whose open-ball topology recovers $\tau$?** This question is neither idle nor academic. Many of the spaces that arise in analysis --- $C^\infty(\mathbb{R}^n)$ with the topology of [uniform convergence](/page/Uniform%20Convergence) on compact sets, the weak* topology on bounded subsets of a dual Banach space, the product of countably many metric spaces --- are defined by topological data (families of seminorms, evaluation functionals, projection maps) rather than by a single metric. Whether these spaces are metrizable determines which analytical tools are available: sequential characterisations of [compactness](/page/Compact%20Space), the Baire Category Theorem, the existence of countable [dense subsets](/page/Dense%20Subset). [example: A Familiar Space That Is Not Metrizable] Consider the real line $\mathbb{R}$ equipped with the cofinite topology $\tau_{\text{cof}}$, in which a set $U \subset \mathbb{R}$ is declared open if and only if $\mathbb{R} \setminus U$ is finite or $U = \varnothing$. This is a legitimate topology, but it fails to be metrizable. To see why, suppose for contradiction that some metric $d: \mathbb{R} \times \mathbb{R} \to 0, \infty)$ generates $\tau_{\text{cof}}$. In any metric space, distinct points can be separated by disjoint [open sets](/page/Open%20Set): for $x \neq y$, the balls $B(x, d(x,y)/3)$ and $B(y, d(x,y)/3)$ are disjoint open sets containing $x$ and $y$ respectively. But in $(\mathbb{R}, \tau_{\text{cof}})$, any two nonempty open sets intersect --- if $U$ and $V$ are nonempty and open, then $\mathbb{R} \setminus U$ and $\mathbb{R} \setminus V$ are finite, so $\mathbb{R} \setminus (U \cap V) = (\mathbb{R} \setminus U) \cup (\mathbb{R} \setminus V)$ is finite, whence $U \cap V$ is nonempty (in fact cofinite). The space is not even Hausdorff, so it cannot be metrizable. This example reveals the first necessary condition for metrizability: the topology must be Hausdorff. But Hausdorffness alone is far from sufficient --- the real challenge is to identify exactly which separation and countability axioms characterise the metric topologies among all topologies. [/example] The cofinite topology fails metrizability at the most basic level --- it lacks the Hausdorff property. The definition below captures precisely what we are asking for when we say a topological space "behaves like a metric space." ## Definition [definition: Metrizable Space] A topological space $(X, \tau)$ is **metrizable** if there exists a metric $d: X \times X \to [0, \infty)$ such that $\tau$ equals the metric topology $\tau_d$ generated by the open balls \begin{align*} B_d(x, r) := \{y \in X : d(x, y) < r\}, \quad x \in X, \; r > 0. \end{align*} When such a metric exists, we call $d$ a **compatible metric** for $(X, \tau)$. [/definition] Several immediate observations sharpen this definition. First, the compatible metric is never unique: if $d$ is compatible, then so is $\min(d, 1)$, and so is $d/(1 + d)$. Both of these are bounded metrics generating the same topology. Thus metrizability is a property of the topology, not of any particular metric. Second, metrizability is a topological invariant: if $(X, \tau)$ is metrizable and $f: X \to Y$ is a homeomorphism, then $Y$ is metrizable (pull back the metric via $f$). In particular, metrizability is preserved under homeomorphism but is **not** preserved under [continuous](/page/Continuity) bijections that are not homeomorphisms. Third, while every metrizable space is Hausdorff (distinct points are separated by disjoint open balls), regular (a point and a disjoint closed set are separated by [open sets](/page/Open%20Set)), and normal (disjoint [closed sets](/page/Closed%20Set) are separated by open sets), these separation axioms alone do not suffice for metrizability. The additional ingredient is a countability condition, and the precise formulation of which countability condition is needed constitutes the content of the metrization theorems. [remark: Metrizability vs. Specific Metrics] It is essential to distinguish between a metrizable space and a [metric space](/page/Metric%20Space). A metric space is a pair $(X, d)$ --- it comes equipped with a specific metric. A metrizable space is a topological space $(X, \tau)$ for which some compatible metric exists, but no particular metric is preferred. This distinction matters because metric properties such as completeness, total boundedness, and uniform continuity depend on the choice of metric, not just on the topology. Two compatible metrics can disagree on whether a sequence is Cauchy, whether a set is bounded, or whether a function is uniformly continuous. We return to this subtlety in the section on completeness and Polish spaces. [/remark] ## Necessary Conditions for Metrizability Before stating the metrization theorems that give sufficient conditions, it is worth understanding which topological properties every metrizable space must possess. These necessary conditions serve as quick tests: if a space fails any one of them, it cannot be metrizable, and there is no need to search for a compatible metric. ### The Hausdorff Axiom Every metric space is Hausdorff, since for distinct points $x \neq y$ with $r = d(x, y) > 0$, the balls $B(x, r/2)$ and $B(y, r/2)$ are disjoint open neighbourhoods. In fact, [metric spaces](/page/Metric%20Space) satisfy a stronger separation property. [quotetheorem:1000] The function $f(x) = \min(d(x, C), 1)$ witnesses this, where $d(x, C) := \inf_{c \in C} d(x, c)$ is the distance function, which is $1$-Lipschitz and hence [continuous](/page/Continuity). ### First Countability A more subtle necessary condition involves the local structure of the topology. Recall that a **first-countable** space is one in which every point $x$ has a countable neighbourhood base $\{U_n\}_{n=1}^\infty$ --- a countable collection of [open sets](/page/Open%20Set) containing $x$ such that every open set containing $x$ includes some $U_n$. Every metrizable space is first-countable: the balls $\{B(x, 1/n)\}_{n=1}^\infty$ form a countable neighbourhood base at $x$. This is the topological property that makes sequences adequate for describing the topology in metrizable spaces. In a general topological space, sequences do not suffice --- one must use nets or filters. In a first-countable space, however, a set $C$ is closed if and only if it is sequentially closed, and a function $f: X \to Y$ is continuous at $x$ if and only if $f(x_n) \to f(x)$ whenever $x_n \to x$. First countability rules out many natural topologies in functional analysis. [example: Non-Metrizability via Failure of First Countability] Let $X$ be an infinite-dimensional [Banach space](/page/Banach%20Space). We show that $X$ equipped with the [weak topology](/page/Weak%20Topology) $\sigma(X, X^*)$ is not first-countable at the origin, and hence not metrizable. Suppose for contradiction that $\{U_n\}_{n=1}^\infty$ is a countable neighbourhood base at $0$ in the weak topology. Each $U_n$ contains a basic weak neighbourhood of $0$, which has the form \begin{align*} V(f_1, \ldots, f_{k_n}; \varepsilon) := \{x \in X : |f_j(x)| < \varepsilon \text{ for } j = 1, \ldots, k_n\} \end{align*} for some finite collection $f_1, \ldots, f_{k_n} \in X^*$. Let $S := \{f_j : j = 1, \ldots, k_n, \; n \in \mathbb{N}\}$ be the countable collection of all functionals appearing across all the basic neighbourhoods. The weak neighbourhood $W_f := \{x \in X : |f(x)| < 1\}$ is open for every $f \in X^*$, so there exists $n$ with $U_n \subset W_f$. Since $U_n$ contains a basic neighbourhood determined by finitely many functionals from $S$, say $f_{i_1}, \ldots, f_{i_m}$, the inclusion $\bigcap_{l=1}^{m} \ker f_{i_l} \subset \ker f$ holds. By a standard fact from linear algebra (a linear functional vanishing on the intersection of finitely many kernels is a linear combination of those functionals), we conclude $f \in \operatorname{span}\{f_{i_1}, \ldots, f_{i_m}\} \subset \operatorname{span}(S)$. Since $f$ was arbitrary, $X^* = \operatorname{span}(S)$, so $X^*$ is the algebraic span of countably many vectors. But this means $X^* = \bigcup_{n=1}^\infty \operatorname{span}\{f_1, \ldots, f_n\}$, a countable union of finite-dimensional --- hence closed, nowhere dense --- subspaces. Since $X^*$ is an infinite-dimensional Banach space, the [Baire Category Theorem](/theorems/630) forbids this (a complete metric space is not a countable union of nowhere dense sets). This contradiction shows that no countable neighbourhood base exists. Therefore the weak topology on an infinite-dimensional Banach space is not first-countable, and hence not metrizable. [/example] This example is the prototype for non-metrizability arguments in functional analysis. The same strategy applies to the [weak* topology](/page/Weak*%20Topology) on the full dual space $X^*$ of an infinite-dimensional normed space. ### Regularity A topological space $X$ is **regular** (or $T_3$) if it is Hausdorff and for every [closed set](/page/Closed%20Set) $C \subset X$ and every point $x \notin C$, there exist disjoint open sets $U$ and $V$ with $x \in U$ and $C \subset V$. Every metrizable space is regular: given $x \notin C$, set $r = d(x, C) > 0$ and take $U = B(x, r/2)$ and $V = \{y \in X : d(y, C) < r/2\}$. Regularity together with first countability does *not* suffice for metrizability --- the Sorgenfrey line (discussed below) is regular and first-countable but its square is not normal, so it cannot be metrizable by any of the standard metrization theorems. However, regularity together with a stronger countability axiom does suffice, as we now describe. ## The Urysohn Metrization Theorem The first major metrization theorem characterises metrizability among spaces with a countable topological base. It is the result one encounters first in a topology course, and it handles the majority of spaces arising in geometric applications. The underlying difficulty is this: we have a topological space $(X, \tau)$ specified by a collection of [open sets](/page/Open%20Set), and we must produce a single real-valued function $d: X \times X \to [0, \infty)$ satisfying the triangle inequality whose open balls recover all the sets in $\tau$. There is no natural candidate for $d$ --- the topology provides no notion of "distance." The idea, due to Urysohn, is to use the normality of the space (guaranteed by second countability plus regularity, via [Urysohn's Lemma](/theorems/887)) to construct sufficiently many continuous real-valued functions, and then to define $d$ via a weighted supremum over these functions. [quotetheorem:992] The power of this theorem lies in combining a separation axiom (regularity) with a countability axiom (second countability). Neither alone suffices: regularity without second countability admits non-metrizable spaces (the Sorgenfrey line is regular but not second-countable), and second countability without regularity admits non-Hausdorff spaces (any second-countable space with the indiscrete topology). The embedding into the Hilbert cube reveals the deeper structure: every second-countable regular space is a subspace of a compact metrizable space, and hence is [separable](/page/Separable%20Space)) and Lindel\"of. Since separability and second countability are equivalent in metrizable spaces (as recorded in the equivalence of separability conditions below), the Urysohn Metrization Theorem can equivalently be stated as: a regular space is second-countable if and only if it embeds into the Hilbert cube. The construction of the embedding proceeds as follows. Let $\{B_n\}_{n=1}^\infty$ be a countable base for $\tau$. By regularity, for each pair $(n, m)$ with $\overline{B_n} \subset B_m$, [Urysohn's Lemma](/theorems/887) provides a continuous function $f_{nm}: X \to [0, 1]$ with $f_{nm}|_{\overline{B_n}} \equiv 0$ and $f_{nm}|_{X \setminus B_m} \equiv 1$. Enumerating these functions as $\{g_k\}_{k=1}^\infty$, the map \begin{align*} \Phi: X &\to [0,1]^{\mathbb{N}} \\ x &\mapsto (g_1(x), g_2(x), g_3(x), \ldots) \end{align*} is a [continuous](/page/Continuity) injection. That it is an embedding (a homeomorphism onto its image) follows from the fact that the functions $\{g_k\}$ separate points from [closed sets](/page/Closed%20Set). [example: Metrizability of Smooth Manifolds] Every smooth manifold $M$ (assumed second-countable and Hausdorff, as is standard) is metrizable by the Urysohn Metrization Theorem. More concretely, if $M$ is a smooth manifold of dimension $n$, choose a countable atlas $\{(U_k, \varphi_k)\}_{k=1}^\infty$ and a subordinate [partition of unity](/page/Partition%20of%20Unity) $\{\psi_k\}_{k=1}^\infty$. Define \begin{align*} d(p, q) := \sum_{k=1}^\infty 2^{-k} \min\!\big(|\psi_k(p) \varphi_k(p) - \psi_k(q) \varphi_k(q)|, 1\big), \end{align*} where we set $\psi_k(p) \varphi_k(p) := 0$ when $p \notin U_k$. This is a well-defined metric on $M$: the sum converges uniformly since each term is bounded by $2^{-k}$, and the triangle inequality holds because $\min(|\cdot|, 1)$ is a metric on $\mathbb{R}^n$ and the triangle inequality is preserved termwise under non-negative summation. The metric $d$ generates the manifold topology because the maps $p \mapsto \psi_k(p)\varphi_k(p)$ are continuous and separate points from closed sets. The Urysohn theorem is ideally suited to manifolds because second countability is built into the standard definition. By contrast, a "manifold" without the second countability axiom (such as the long line) may fail to be metrizable. [/example] ## The Nagata--Smirnov Metrization Theorem The Urysohn Metrization Theorem requires second countability, which is a strong global condition --- it implies separability, the Lindel\"of property, and that the space has at most continuum-many [open sets](/page/Open%20Set). Many important spaces are metrizable without being second-countable. The discrete topology on an uncountable set, for instance, is metrizable (via the discrete metric $d(x, y) = 1$ for $x \neq y$) but not second-countable. A complete characterisation of metrizability requires replacing the countable base with a weaker condition. The obstacle is to formulate a condition that captures exactly the right amount of "countable control" over the topology. The key concept is that of a $\sigma$-locally finite base. [definition: Sigma-Locally Finite Base] Let $(X, \tau)$ be a topological space. A collection $\mathcal{B}$ of open sets is a **$\sigma$-locally finite base** if: 1. $\mathcal{B}$ is a base for $\tau$: every open set is a union of members of $\mathcal{B}$. 2. $\mathcal{B} = \bigcup_{n=1}^\infty \mathcal{B}_n$, where each $\mathcal{B}_n$ is **locally finite**: every point $x \in X$ has an open neighbourhood $U$ that intersects only finitely many members of $\mathcal{B}_n$. [/definition] The difference between a countable base and a $\sigma$-locally finite base is significant. A countable base is literally a countable collection. A $\sigma$-locally finite base is a countable union of families, each of which may be uncountable but is "spread out" so that no point sees more than finitely many members of each family. For second-countable spaces, the base itself is a $\sigma$-locally finite base (set $\mathcal{B}_n = \{B_n\}$ for each $n$). For the discrete topology on an uncountable set $X$, take $\mathcal{B}_1 = \{\{x\} : x \in X\}$, which is locally finite because every point has a neighbourhood (namely $\{x\}$ itself) meeting exactly one member. [quotetheorem:995] This is a complete characterisation: metrizability is equivalent to regularity plus the existence of a $\sigma$-locally finite base. The Urysohn Metrization Theorem is an immediate corollary, since every countable base is $\sigma$-locally finite. The "only if" direction is the easier half: if $(X, d)$ is a [metric space](/page/Metric%20Space), then for each $n \in \mathbb{N}$, the collection \begin{align*} \mathcal{B}_n := \{B(x, 1/n) : x \in X\} \end{align*} is an open cover, and refining by a maximal $1/(3n)$-separated subset produces a locally finite subcollection that is still a base. (Alternatively, in a metrizable space, every open cover has a $\sigma$-locally finite open refinement --- this is Stone's theorem on paracompactness of metrizable spaces.) The "if" direction is the deep content of the theorem: given a $\sigma$-locally finite base, one constructs a compatible metric. The construction again uses Urysohn-type functions, but the argument is substantially more involved than in the second-countable case because one cannot simply enumerate the base elements. [remark: Relationship to Paracompactness] Every metrizable space is paracompact (Stone's theorem, 1948): every open cover has a locally finite open refinement. The Nagata--Smirnov theorem can be viewed as a refinement of this fact: not only does every open cover have a locally finite refinement, but one can find a *base* that decomposes into countably many locally finite families. The relationship between metrizability, paracompactness, and normality is a recurring theme in general topology. [/remark] ## Metrizability of Product Spaces One of the most common sources of metrizable spaces in analysis is the product construction. If each factor is metrizable, is the product metrizable? The answer depends critically on whether the index set is countable. ### Countable Products When the index set is countable, the product of metrizable spaces is again metrizable. The difficulty is purely technical: the [product topology](/page/Product%20Topology) is defined by finitely many coordinate conditions at a time (basic open sets restrict only finitely many coordinates), so any compatible metric must simultaneously encode convergence in all coordinates while respecting this finiteness. [quotetheorem:959] The explicit metric in this theorem deserves attention. Let $\{(X_k, d_k)\}_{k=1}^\infty$ be the factor spaces, where we may assume $d_k \le 1$ by replacing $d_k$ with $\min(d_k, 1)$. The standard construction defines \begin{align*} d(x, y) := \sum_{k=1}^\infty 2^{-k} d_k(x_k, y_k) \end{align*} for $x = (x_k)_{k=1}^\infty$ and $y = (y_k)_{k=1}^\infty$ in $\prod_{k=1}^\infty X_k$. That this series converges uniformly (each term is at most $2^{-k}$) is immediate. That it defines a metric follows from the triangle inequality applied termwise. The key verification is that $d$ generates the product topology: a sequence $(x_m)_{m=1}^\infty$ converges to $x$ in the metric $d$ if and only if $(x_m)_k \to x_k$ in $(X_k, d_k)$ for every $k$, which is precisely convergence in the product topology. The weighting $2^{-k}$ ensures that later coordinates contribute less to the distance, reflecting the fact that the product topology "cares" about all coordinates but does not privilege any finite subset over the others. Any summable sequence of positive weights would work equally well; the specific choice affects the metric but not the topology. [example: The Space of Sequences as a Countable Product] The space $\mathbb{R}^{\mathbb{N}} = \prod_{k=1}^\infty \mathbb{R}$, equipped with the product topology (also called the topology of pointwise convergence), is metrizable. A compatible metric is \begin{align*} d(x, y) := \sum_{k=1}^\infty 2^{-k} \frac{|x_k - y_k|}{1 + |x_k - y_k|}, \end{align*} where the bounded metric $\rho(s, t) = |s - t|/(1 + |s - t|)$ on $\mathbb{R}$ replaces the standard (unbounded) metric. This metrizable space is fundamental: it is homeomorphic to the irrational numbers (a classical result of Alexandroff and Urysohn), and it serves as a universal space for [separable](/page/Separable%20Space)) metrizable spaces (every separable metrizable space embeds into $\mathbb{R}^{\mathbb{N}}$). [/example] This result has a direct application to locally convex spaces defined by countably many seminorms. If a locally convex space%20Vector%20Space) has its topology generated by a countable separating family of seminorms, then the same weighted-sum construction produces a translation-invariant compatible metric --- this is the content of the metrizability theorem for locally convex spaces stated in the next section. The Fr\'echet spaces --- complete, metrizable, locally convex spaces --- are precisely those whose topology is generated by a countable separating family of seminorms. The space $C^\infty(U)$ of smooth functions on an [open set](/page/Open%20Set) $U \subset \mathbb{R}^n$, topologised by [uniform convergence](/page/Uniform%20Convergence) of all derivatives on compact subsets, is the prototypical example: the countably many seminorms \begin{align*} p_{K,m}(f) := \max_{|\alpha| \le m} \sup_{x \in K} |D^\alpha f(x)|, \end{align*} indexed over a countable exhaustion $K_1 \subset K_2 \subset \cdots$ of $U$ by compact sets and $m \in \mathbb{N}$, generate the topology. ### Uncountable Products When the index set is uncountable, the situation changes dramatically. [quotetheorem:993] The reason is a failure of first countability. A basic open set in the product topology restricts only finitely many coordinates. If a countable neighbourhood base $\{U_n\}_{n=1}^\infty$ existed at some point $x$, each $U_n$ would restrict only finitely many coordinates, so the union of all restricted coordinates would be countable. But then the neighbourhood base cannot detect the coordinate $\alpha$ for any $\alpha$ outside this [countable set](/page/Countable%20Set), exactly as in the weak-topology argument above. This result has a striking consequence: the Hilbert cube $[0,1]^{\mathbb{N}}$ is compact and metrizable (by [Tychonoff's Theorem](/theorems/953) and the countable product metrization), but $[0,1]^{\mathbb{R}}$ is compact (again by Tychonoff) yet not metrizable. ## Metrizability in Functional Analysis The interplay between metrizability and the various weak topologies on infinite-dimensional spaces is one of the most important applications of metrization theory. The guiding principle is: **metrizability of a [weak topology](/page/Weak%20Topology) on a subset is governed by the "size" (in the sense of separability) of the predual or the ambient space.** ### The Weak* Topology on Bounded Sets The full [weak* topology](/page/Weak*%20Topology) $\sigma(X^*, X)$ on the dual of an infinite-dimensional Banach space $X$ is never metrizable (by the first-countability argument from the example above, applied to $X^*$ with $X$ playing the role of the predual). However, when restricted to bounded subsets, the picture changes. [quotetheorem:495] The hypothesis that $X$ is [separable](/page/Separable%20Space))%20Space) is essential. The dense sequence $\{x_k\}_{k=1}^\infty$ is the entire mechanism: on the bounded set $B_R$, functionals are equicontinuous (by the norm bound), so their values on a [dense subset](/page/Dense%20Subset) determine them completely. The denominator $1 + |f(x_k) - g(x_k)|$ ensures each term is at most $2^{-k}$, guaranteeing convergence. Without separability --- when no countable dense set exists --- no countable family of evaluation functionals can separate points of $B_R$, and the argument collapses (as the example below demonstrates). This result is indispensable in PDE theory: combining it with the Banach-Alaoglu theorem (which says $B_R$ is weak*-compact) gives that $B_R$ is a **compact metrizable** space when $X$ is separable. [Compactness](/page/Compact%20Space) plus metrizability implies sequential compactness, so every bounded sequence in $X^*$ has a weak*-convergent subsequence. This is the workhorse extraction principle behind virtually every existence theorem in the [calculus of variations](/page/Calculus%20of%20Variations). [example: Failure of Weak* Metrizability Without Separability] Let $X = \ell^\infty(\mathbb{N})$, the space of bounded real sequences with the supremum norm. This space is not separable (the uncountable collection $\{\mathbb{1}_S : S \subset \mathbb{N}\}$ of indicator functions of subsets of $\mathbb{N}$ forms an uncountable $1$-separated set, so no countable subset can be dense). The [dual space](/page/Dual%20Space) $X^* = (\ell^\infty)^*$ is enormous (it contains finitely additive measures on $\mathbb{N}$, not just the elements of $\ell^1$). We claim that the weak* topology on the unit ball $B_{X^*}$ is not metrizable. The argument uses a generalisation of the first-countability test. Suppose $d$ is a compatible metric for the weak* topology on $B_{X^*}$. For each $n \in \mathbb{N}$, the ball $B_d(0, 1/n) \cap B_{X^*}$ is a weak* neighbourhood of $0$ in $B_{X^*}$. Each such neighbourhood contains a basic weak* neighbourhood determined by finitely many elements $x_1, \ldots, x_{k_n} \in X$. The collection of all such elements across all $n$ is countable. Since $X = \ell^\infty$ is not separable, the norm-closed span of this countable collection is a proper separable subspace $Y \subsetneq X$. By the Hahn-Banach theorem, there exists $f \in X^*$ with $\|f\| = 1$ and $f|_Y = 0$. Then $t f \in B_d(0, 1/n)$ for every $n$ and every $|t| \le 1$, which is impossible if $d$ is a metric (since $d(f, 0) > 0$ but $f \in B_d(0, 1/n)$ for all $n$). This failure of metrizability is not merely a curiosity --- it means that in non-separable settings, one cannot extract weak*-convergent subsequences from bounded sequences. One must instead use nets or ultrafilter arguments, which are substantially more delicate. [/example] ### Metrizability of Locally Convex Spaces The connection between countable families of seminorms and metrizability extends beyond the special case of product metrics. [quotetheorem:708] This result completely characterises metrizable locally convex spaces: they are precisely the spaces whose topology is generated by a countable separating family of seminorms. It also shows that the topology of every metrizable locally convex space is generated by a translation-invariant metric --- a property that is stronger than bare metrizability. The translation-invariance is crucial for applications to [topological vector spaces](/page/Topological%20Vector%20Space)%20Vector%20Space), because it ensures that the metric is compatible with the vector space structure: convergence of $x_k \to x$ is equivalent to convergence of $x_k - x \to 0$. When the family of seminorms is not countable, metrizability can fail dramatically. [quotetheorem:713] The space $\mathcal{D}(\Omega) = C_c^\infty(\Omega)$ of compactly supported smooth functions, equipped with the standard [strict inductive limit topology](/page/Strict%20Inductive%20Limit%20Topology), is the most important example. Write $\Omega$ as a union of compact exhaustions $K_1 \subset K_2 \subset \cdots$ with $K_n \subset \operatorname{int}(K_{n+1})$, and let $V_n = C^\infty_{K_n}(\Omega)$ be the Fr\'echet space of smooth functions supported in $K_n$. The strict inductive limit $\mathcal{D}(\Omega) = \varinjlim_n V_n$ is a locally convex space that is not metrizable, yet it is complete in a generalised sense (it is a complete locally convex space, sometimes called an LF-space). The non-metrizability means that sequences do not determine the topology of $\mathcal{D}(\Omega)$: there exist [linear maps](/page/Linear%20Map) $T: \mathcal{D}(\Omega) \to \mathbb{R}$ that are sequentially continuous but not continuous. This subtlety underlies the need for the full theory of [distributions](/page/Distribution)). ## Non-Metrizable Spaces: A Catalogue of Obstructions Understanding why specific spaces fail to be metrizable sharpens one's intuition for the metrization theorems. Each example below illustrates a distinct mechanism by which metrizability can fail. ### The Sorgenfrey Line and Its Square The **Sorgenfrey line** $\mathbb{R}_\ell$ is the set $\mathbb{R}$ equipped with the topology generated by the half-open intervals $[a, b) = \{x \in \mathbb{R} : a \le x < b\}$. This topology is strictly finer than the standard topology on $\mathbb{R}$. [example: The Sorgenfrey Line Is Not Metrizable] The Sorgenfrey line $\mathbb{R}_\ell$ is first-countable (the intervals $\{x, x + 1/n)\}_{n=1}^\infty$ form a countable neighbourhood base at $x$), Hausdorff, regular, normal, and even perfectly normal. It is also [separable](/page/Separable%20Space) ($\mathbb{Q}$ is dense) and Lindel\"of. Yet it is not second-countable. To see this, suppose $\mathcal{B}$ is a countable base. For each $x \in \mathbb{R}$, the [open set](/page/Open%20Set) $[x, x+1)$ contains some $B_x \in \mathcal{B}$ with $x \in B_x$. Since $B_x$ is open in the Sorgenfrey topology and contains $x$, we have $B_x \subset [x, x + \delta)$ for some $\delta > 0$, which gives $\inf B_x = x$. If $x \neq y$, then $\inf B_x \neq \inf B_y$, so $B_x \neq B_y$. This produces uncountably many distinct members of $\mathcal{B}$, contradicting countability. Since $\mathbb{R}_\ell$ is separable but not second-countable, and these two properties are equivalent in metrizable spaces (as shown below in the section on [separability and metrizability), the space cannot be metrizable. [/example] The Sorgenfrey line shows that the Urysohn Metrization Theorem's hypothesis of second countability cannot be replaced by separability alone. The failure is even more dramatic in the product. [example: The Sorgenfrey Plane Is Not Normal] The **Sorgenfrey plane** $\mathbb{R}_\ell \times \mathbb{R}_\ell$ --- the product of two copies of the Sorgenfrey line with the [product topology](/page/Product%20Topology) --- is not normal. Since every metrizable space is normal, this gives another proof that the Sorgenfrey plane (and hence the Sorgenfrey line itself, whose square is not normal) is not metrizable. The argument uses **Jones' Lemma**: if $X$ is a normal space with a [dense subset](/page/Dense%20Subset) $A$ and a closed discrete subset $D$, then $|D| \le 2^{|A|}$; in particular, if $X$ is separable) and normal, then every closed discrete subspace is countable. The proof of Jones' Lemma is a counting argument. In a normal space, [Urysohn's Lemma](/theorems/887) guarantees that for any two disjoint closed sets $C_1, C_2 \subset X$, there exists a continuous function $f: X \to [0,1]$ with $f|_{C_1} \equiv 0$ and $f|_{C_2} \equiv 1$. If $D$ is closed and discrete, then every subset $E \subset D$ is closed (any subset of a discrete closed set is closed). For each $E \subset D$, choose a [continuous](/page/Continuity) function $f_E: X \to [0,1]$ with $f_E|_E \equiv 0$ and $f_E|_{D \setminus E} \equiv 1$. If $E \neq F$, then $f_E \neq f_F$ as functions on $X$ (they differ on $D$). This gives $2^{|D|}$ distinct continuous functions $X \to [0,1]$. But if $A \subset X$ is dense, then a continuous function $X \to [0,1]$ is determined by its values on $A$, so the number of continuous functions is at most $|[0,1]^A| = \mathfrak{c}^{|A|}$. When $A$ is countable, this gives at most $\mathfrak{c}^{\aleph_0} = \mathfrak{c} = 2^{\aleph_0}$ continuous functions, so $2^{|D|} \le 2^{\aleph_0}$, whence $|D| \le \aleph_0$. Now apply this to the Sorgenfrey plane. Consider the anti-diagonal $D = \{(x, -x) : x \in \mathbb{R}\}$. In the Sorgenfrey plane topology, $D$ is a closed discrete subspace: the basic open set $[x, x + \varepsilon) \times [-x, -x + \varepsilon)$ meets $D$ only at the point $(x, -x)$, so $\{(x, -x)\}$ is open in the subspace topology on $D$. Thus $D$ is discrete, and $D$ is closed because its complement is open (for any $(a,b)$ with $a + b \neq 0$, a small enough basic rectangle around $(a,b)$ avoids $D$). The set $D$ is uncountable ($|D| = |\mathbb{R}|$). On the other hand, $\mathbb{R}_\ell \times \mathbb{R}_\ell$ is separable: $\mathbb{Q} \times \mathbb{Q}$ is countable and dense (every basic open set $[a, a + \varepsilon) \times [b, b + \varepsilon)$ contains a point of $\mathbb{Q} \times \mathbb{Q}$). By Jones' Lemma, $\mathbb{R}_\ell \times \mathbb{R}_\ell$ cannot be normal. [/example] ### The Long Line The **long line** $L$ is a connected, locally Euclidean, Hausdorff space that is not second-countable. It is constructed by taking $\omega_1$ (the first uncountable ordinal) copies of $0, 1)$ and gluing them end to end in ordinal order. Each point has a neighbourhood homeomorphic to $\mathbb{R}$, so $L$ is a $1$-dimensional topological manifold --- except that it violates the second countability axiom that is conventionally included in the definition of a manifold. The long line is not metrizable. If it were, it would be paracompact (since metrizable spaces are paracompact), and a connected, locally compact, paracompact space is $\sigma$-compact (a countable union of compact sets). But the long line is not $\sigma$-compact: any compact subset is contained in the union of countably many copies of $0, 1)$, and the entire space requires uncountably many. This example motivates the convention of including second countability in the definition of a smooth manifold. ### The Weak Topology on Infinite-Dimensional Spaces We have already shown that the [weak topology](/page/Weak%20Topology) $\sigma(X, X^*)$ on an infinite-dimensional [Banach space $X$ is not metrizable (failure of first countability). It is worth recording the precise statement. [quotetheorem:984] Both statements follow from the failure of first countability established in the earlier example. The key step is the Baire category argument: if a countable neighbourhood base existed at the origin, the functionals determining it would algebraically span all of $X^*$, presenting $X^*$ as a countable union of finite-dimensional subspaces --- contradicting the Baire Category Theorem for the [Banach space](/page/Banach%20Space) $X^*$. The second statement uses the same argument with $X$ playing the role of the "test functionals." Despite the non-metrizability of the full weak and weak* topologies, the restrictions to bounded sets often are metrizable (as we saw in the weak* case when $X$ is separable). For the weak topology, the situation is analogous: $\sigma(X, X^*)$ restricted to bounded subsets of $X$ is metrizable if and only if $X^*$ is separable. ## Metrizability and Completeness: Polish Spaces The interplay between metrizability and completeness gives rise to one of the most important classes of spaces in descriptive set theory and probability. A metrizable space $(X, \tau)$ may admit both complete and incomplete compatible metrics. For example, the open interval $(0, 1)$ is homeomorphic to $\mathbb{R}$, and the standard metric $d(x, y) = |x - y|$ restricted to $(0, 1)$ is incomplete (the sequence $x_n = 1/n$ is Cauchy but has no limit in $(0, 1)$), while the pullback of the standard metric on $\mathbb{R}$ via any homeomorphism $(0, 1) \to \mathbb{R}$ is a complete compatible metric. This shows that completeness is a property of the metric, not of the topology --- a given topology may be compatible with both complete and incomplete metrics. [definition: Polish Space] A topological space $(X, \tau)$ is called a **Polish space** if it is [separable](/page/Separable%20Space)) and there exists a complete metric $d$ on $X$ such that $\tau = \tau_d$. Equivalently, $X$ is homeomorphic to a complete separable [metric space](/page/Metric%20Space). [/definition] The definition requires the existence of *some* complete compatible metric, not that every compatible metric be complete. The open interval $(0, 1)$ is Polish (it is homeomorphic to $\mathbb{R}$), even though its most natural metric is incomplete. Polish spaces form the natural setting for measure theory and probability: the regular conditional probability theorem, the Prokhorov theorem on tightness, and the theory of standard Borel spaces all require the Polish hypothesis. The class of Polish spaces is also the correct domain for descriptive set theory, where one studies the complexity of subsets defined by countable logical operations. [example: The Irrationals Are Polish] The space $\mathbb{R} \setminus \mathbb{Q}$ of irrational numbers, with the subspace topology inherited from $\mathbb{R}$, is Polish. This is not evident from the standard metric, which is incomplete on $\mathbb{R} \setminus \mathbb{Q}$ (for instance, choose a sequence of irrationals converging to a rational). To see that $\mathbb{R} \setminus \mathbb{Q}$ is Polish, we exhibit a homeomorphism to a complete separable metric space. Consider the space $\mathbb{N}^{\mathbb{N}} = \prod_{k=1}^\infty \mathbb{N}$ of infinite sequences of natural numbers, equipped with the [product topology](/page/Product%20Topology) (where each factor $\mathbb{N}$ has the discrete topology). This is metrizable by the Metrizability of Countable Products theorem, with compatible metric \begin{align*} d(x, y) := \sum_{k=1}^\infty 2^{-k} \frac{|x_k - y_k|}{1 + |x_k - y_k|} \quad \text{for } x = (x_k), \; y = (y_k) \in \mathbb{N}^{\mathbb{N}}. \end{align*} In fact, an even simpler complete metric is available: define $\rho(x, y) := 1/(1 + \min\{k : x_k \neq y_k\})$ for $x \neq y$ and $\rho(x, x) = 0$. This is an ultrametric, and $(\mathbb{N}^{\mathbb{N}}, \rho)$ is complete (a [Cauchy sequence](/page/Cauchy%20Sequence) is eventually constant in each coordinate). The space $\mathbb{N}^{\mathbb{N}}$ is also separable (the sequences that are eventually zero form a countable dense subset). The classical result of Alexandroff and Urysohn states that $\mathbb{R} \setminus \mathbb{Q}$ is homeomorphic to $\mathbb{N}^{\mathbb{N}}$ via continued fraction expansion. Since $\mathbb{N}^{\mathbb{N}}$ is a complete separable metric space, $\mathbb{R} \setminus \mathbb{Q}$ is Polish. [/example] The following theorem characterises exactly which subsets of a Polish space are again Polish. [quotetheorem:966] This explains why open subsets of $\mathbb{R}^n$ are Polish ([open sets](/page/Open%20Set) are $G_\delta$), why $\mathbb{R} \setminus \mathbb{Q}$ is Polish ($\mathbb{R} \setminus \mathbb{Q} = \bigcap_{q \in \mathbb{Q}} \mathbb{R} \setminus \{q\}$ is a countable intersection of open sets), and why $\mathbb{Q}$ is not Polish ($\mathbb{Q}$ is an $F_\sigma$ subset of $\mathbb{R}$ but not a $G_\delta$, by the Baire Category Theorem). [example: The Rationals Are Not Polish] The space $\mathbb{Q}$ with the subspace topology from $\mathbb{R}$ is metrizable and separable but not Polish. Suppose for contradiction that $d$ is a complete compatible metric on $\mathbb{Q}$. Enumerate $\mathbb{Q} = \{q_1, q_2, q_3, \ldots\}$. Each singleton $\{q_n\}$ is closed in $\mathbb{Q}$ (since $\mathbb{Q}$ is Hausdorff) and has empty interior (since every open set in $\mathbb{Q}$ is infinite). Thus $\{q_n\}$ is nowhere dense, and $\mathbb{Q} = \bigcup_{n=1}^\infty \{q_n\}$ is a countable union of nowhere dense sets. By the [Baire Category Theorem](/theorems/630), a complete metric space cannot be written as a countable union of nowhere dense closed sets. Since we assumed $(\mathbb{Q}, d)$ is complete, this is a contradiction. Therefore no complete compatible metric on $\mathbb{Q}$ exists. This is a topological invariant: since $\mathbb{Q}$ is not Polish, any topological space homeomorphic to $\mathbb{Q}$ also fails to be Polish. The distinction between $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ --- both countable dense (resp. cocountable dense) subsets of $\mathbb{R}$, both metrizable and separable --- is entirely a question of whether a complete compatible metric exists. [/example] ## Separability and Metrizability In metrizable spaces, several notions of "smallness" for the topology collapse into a single condition. This equivalence is one of the most useful features of metrizable spaces, and it fails in general topological spaces. [quotetheorem:545] This theorem is a workhorse: it allows one to move freely among separability (easy to verify by exhibiting a dense sequence), second countability (the hypothesis for the Urysohn Metrization Theorem), and the Lindel\"of property (needed for covering arguments) whenever the ambient space is known to be metrizable. The equivalence fails outside the metrizable setting. The Sorgenfrey line $\mathbb{R}_\ell$ is [separable](/page/Separable%20Space)) (with $\mathbb{Q}$ dense) and Lindel\"of but not second-countable, witnessing the failure of (1) $\Rightarrow$ (2) and (3) $\Rightarrow$ (2) without metrizability. Conversely, the ordinal space $[0, \omega_1)$ with the order topology is first-countable but neither separable, second-countable, nor Lindel\"of --- the open cover $\{[0, \alpha) : \alpha < \omega_1\}$ has no countable subcover, since any countable union of countable ordinals is bounded by a countable ordinal. A related result records the hereditary nature of separability in metrizable spaces. [quotetheorem:942] This hereditary separability of metrizable spaces is a direct consequence of the equivalence with second countability: every subspace of a second-countable space is second-countable, and second countability implies separability. In non-metrizable spaces, separability is not hereditary --- for instance, the Sorgenfrey plane $\mathbb{R}_\ell \times \mathbb{R}_\ell$ is separable (the [countable set](/page/Countable%20Set) $\mathbb{Q} \times \mathbb{Q}$ is dense), but its anti-diagonal $\{(x, -x) : x \in \mathbb{R}\}$ is an uncountable discrete subspace, hence not separable. ## Working With Metrizable Spaces: Standard Techniques The power of metrizability lies in the concrete tools it makes available. This section collects the standard methods one uses when working with metrizable spaces, each illustrated with a representative application. ### Sequential Characterisations In a metrizable space, topology reduces to sequences. This is the single most important consequence of metrizability for analysis. **Closure:** A point $x$ lies in the closure $\overline{A}$ of $A \subset X$ if and only if there exists a sequence $(a_n)_{n=1}^\infty$ in $A$ with $a_n \to x$. **Continuity:** A function $f: X \to Y$ (with $X$ metrizable) is [continuous](/page/Continuity) if and only if $f(x_n) \to f(x)$ whenever $x_n \to x$ in $X$. **[Compactness](/page/Compact%20Space):** A subset $K$ of a metrizable space is compact if and only if it is sequentially compact (every sequence in $K$ has a convergent subsequence with limit in $K$). In non-metrizable spaces, all three of these equivalences fail. The typical substitute is the theory of nets (or filters), which is substantially more technical. This is why establishing metrizability is often the first step in any analysis on a topological space: once metrizability is known, one can work with sequences exclusively. [example: Extracting Weak*-Convergent Subsequences] Let $X$ be a separable [Banach space](/page/Banach%20Space) and let $(f_n)_{n=1}^\infty$ be a bounded sequence in $X^*$ with $\sup_n \|f_n\|_{X^*} \le R$. We extract a weak*-convergent subsequence. By the Banach-Alaoglu theorem, the ball $B_R = \{f \in X^* : \|f\| \le R\}$ is weak*-compact. By the Weak* Metrizability on Bounded Sets theorem, $B_R$ is metrizable in the [weak* topology](/page/Weak*%20Topology) (using the separability of $X$). Since $(f_n)$ is a sequence in the compact metrizable space $B_R$, it has a convergent subsequence: there exist $f \in B_R$ and a subsequence $(f_{n_j})$ with $f_{n_j} \overset{*}{\rightharpoonup} f$. Without separability of $X$, the ball $B_R$ is still weak*-compact (Banach-Alaoglu holds in general), but it is not metrizable, so one cannot extract subsequences. One must instead use nets or invoke Eberlein-\v{S}mulian-type results. [/example] ### The Baire Category Theorem The Baire Category Theorem applies to complete [metric spaces](/page/Metric%20Space) (and more generally to locally compact Hausdorff spaces): a countable intersection of dense open sets is dense. Equivalently, a complete metric space cannot be written as a countable union of nowhere dense [closed sets](/page/Closed%20Set). This theorem has no analogue for general metrizable spaces that lack a complete compatible metric. The standard applications --- open mapping theorems, uniform boundedness principles, existence of continuous-nowhere-differentiable functions --- all require completeness. When a space is known to be Polish (separable + completely metrizable), the Baire Category Theorem applies even though the given metric might not be complete. The point is that *some* complete compatible metric exists, and the Baire property depends only on the topology, not on the specific metric. This is why Polish spaces, rather than merely metrizable spaces, are the correct setting for Baire-category arguments on [separable spaces](/page/Separable%20Space). ### Metrisation as a Tool for Compactness Arguments A powerful pattern in functional analysis combines the Banach-Alaoglu theorem with metrizability to extract convergent subsequences from bounded sequences. The template is: 1. **Identify the bounded set.** Show that a sequence $(u_n)$ is bounded in a reflexive Banach space $X$, or that a sequence $(f_n)$ is bounded in a dual space $X^*$. 2. **Invoke weak or weak* compactness.** Banach-Alaoglu (for [dual spaces](/page/Dual%20Space)) or the Eberlein-\v{S}mulian theorem (for [reflexive spaces](/page/Reflexive%20Space)) guarantees that the sequence lies in a compact set in the appropriate weak topology. 3. **Establish metrizability.** If $X$ (or $X^*$) is separable, then the restriction of the weak (or weak*) topology to bounded sets is metrizable. 4. **Extract a subsequence.** Compactness plus metrizability implies sequential compactness, yielding a convergent subsequence. This pattern underlies the "direct method" in the [calculus of variations](/page/Calculus%20of%20Variations), the extraction of weak solutions to elliptic PDE, and the construction of Young measures in the theory of oscillations. ### Constructing Compatible Metrics When one needs an explicit compatible metric (not just the assertion of metrizability), the standard constructions are: **Countable products.** Use the weighted sum with bounded metrics $\bar{d}_k = \min(d_k, 1)$: \begin{align*} d(x, y) = \sum_{k=1}^\infty 2^{-k} \bar{d}_k(x_k, y_k). \end{align*} **Countable seminorms.** For a locally convex space with separating seminorms $\{p_n\}_{n=1}^\infty$: \begin{align*} d(x, y) = \sum_{n=1}^\infty 2^{-n} \frac{p_n(x - y)}{1 + p_n(x - y)}. \end{align*} **Embedding.** If $X$ embeds into a metrizable space $Y$ (e.g., the Hilbert cube), pull back the metric from $Y$. In each case, the key properties to verify are: (i) the series converges, (ii) the function $d$ satisfies the metric axioms (especially the triangle inequality), and (iii) $d$ generates the original topology (convergence in $d$ is equivalent to convergence in $\tau$). ## References - Munkres, J. R., *Topology* (2nd edition, 2000). Chapters 4 and 6 cover the Urysohn Metrization Theorem and the Nagata--Smirnov theorem with full proofs. - Kelley, J. L., *General Topology* (1955). The classical reference for metrization theory, including the Bing and Nagata--Smirnov theorems. - Willard, S., *General Topology* (1970). Comprehensive treatment of metrization with extensive examples of non-metrizable spaces. - Kechris, A. S., *Classical Descriptive Set Theory* (1995). The standard reference for Polish spaces and their role in descriptive set theory. - Rudin, W., *Functional Analysis* (2nd edition, 1991). Metrizability of locally convex spaces and weak* topologies on bounded sets. - Engelking, R., *General Topology* (revised edition, 1989). The encyclopedic reference, containing all metrization theorems and their interrelationships. - Brezis, H., *Functional Analysis, [Sobolev Spaces](/page/Sobolev%20Space) and Partial Differential Equations* (2011). Applications of weak* metrizability to PDE existence theory.