Much of analysis works by approximation: a function is the [limit](/page/Limit) of polynomials, a [distribution](/page/Distribution) is the limit of [test functions](/page/Test%20Function), an $L^p$ function is the limit of simple [functions](/page/Function). In each case, the approximating objects are drawn from a "nice" subcollection, and the target is recovered as a limit. The question of *how large* this subcollection needs to be — whether a countable collection suffices, or whether uncountably many approximants are required — is the question of **separability**. The distinction matters because countability is the gateway to sequential arguments. When a space is separable, bounded [sequences](/page/Sequence) have convergent subsequences (in appropriate weak topologies), diagonal extraction arguments work, and the topology on compact [sets](/page/Set) can be described by a metric. When separability fails, these tools break down: compactness must be formulated in terms of nets or filters, extracting convergent subsequences may be impossible, and the gap between sequential and [topological](/page/Topology) statements becomes a genuine obstacle. The theory of PDEs, the [calculus of variations](/page/Calculus%20of%20Variations), and much of modern analysis operate almost exclusively in separable spaces — not by accident, but because separability is the structural condition that makes the standard machinery run. ## Motivation [motivation] ### The Finite-Dimensional Baseline In $\mathbb{R}^n$, countability questions never arise. The space is spanned by $n$ vectors, every bounded sequence has a convergent subsequence (Bolzano-Weierstrass), and the topology is completely described by the Euclidean metric. The rational points $\mathbb{Q}^n$ form a countable dense subset, but this is an afterthought — it plays no essential role in the theory because all the tools work regardless. ### What Goes Wrong in Infinite Dimensions In infinite-dimensional spaces, the Bolzano-Weierstrass theorem fails catastrophically: the closed unit ball is never compact in the norm topology. The standard rescue is to pass to a weaker topology — the [weak topology](/page/Weak*%20Topology) on a Banach space, or the weak* topology on its dual — where the Banach-Alaoglu theorem restores compactness of the unit ball. But this compactness is in the topological sense (every open cover has a finite subcover), not the sequential sense (every sequence has a convergent subsequence). In general, topological compactness does not imply sequential compactness. The problem is concrete: in PDE theory, one routinely constructs a bounded sequence $\{u_n\}$ in some function space and needs to extract a convergent subsequence. If the relevant compact set is not metrizable — so that sequences do not determine the topology — this extraction is impossible by sequential methods. ### The Resolution: Separability Provides Metrizability The key insight is that when the underlying Banach space $X$ is separable, the weak* topology on bounded subsets of $X^*$ becomes metrizable. Metrizability restores the equivalence between compactness and sequential compactness, and the standard diagonal extraction arguments work. Similarly, separability of $X^*$ makes the weak topology on bounded subsets of $X$ metrizable. In both cases, separability is the structural hypothesis that converts abstract topological compactness into the sequential compactness that analysis actually uses. This is why the spaces of PDE theory — $L^p$ for $1 \le p < \infty$, [Sobolev spaces](/page/Sobolev%20Space) $W^{k,p}$, and their duals — are all separable, while $L^\infty$ (which is not separable) requires special treatment and is generally avoided as an ambient space. [/motivation] ## The General Definition The notion of separability is purely topological: it asks only for the existence of a countable dense subset, with no reference to metrics, norms, or algebraic structure. [definition:Separable Topological Space] A topological space $(X, \tau)$ is **separable** if it contains a countable dense subset: there exists a [countable set](/page/Countable%20Set) $D \subseteq X$ such that $\overline{D} = X$, where the closure is taken in the topology $\tau$. Equivalently, $X$ is separable if every nonempty [open set](/page/Open%20Set) in $X$ contains an element of $D$. [/definition] The two formulations are equivalent by definition of closure: $\overline{D} = X$ means every open set meets $D$. The second formulation is often more practical for verifying separability — one checks that the countable set $D$ "touches" every open set. A countable dense subset can be thought of as a countable "skeleton" that is topologically indistinguishable from the full space: any property that depends only on the closure operator ([continuity](/page/Continuity), convergence, density) can be tested on $D$ alone. This is the source of separability's power — it reduces questions about an uncountable space to questions about a countable approximation. [example:Separable Spaces] The following are separable: 1. **$\mathbb{R}^n$** with the Euclidean topology: $\mathbb{Q}^n$ is a countable dense subset. 2. **$C([0,1])$** with the supremum norm: by the [Weierstrass approximation theorem](/theorems/480), polynomials with rational coefficients are dense. This set is countable (finite rational combinations of the countable set $\{1, t, t^2, \dots\}$). 3. **$\ell^p(\mathbb{N})$** for $1 \le p < \infty$: the set of eventually-zero sequences with rational entries is countable and dense. Given $x = (x_1, x_2, \dots) \in \ell^p$, the truncation $(x_1, \dots, x_N, 0, 0, \dots)$ converges to $x$ as $N \to \infty$ (since $\sum_{k > N} |x_k|^p \to 0$), and each truncation can be approximated by rational entries. [/example] [example:Non-Separable Spaces] The following are *not* separable: 1. **$\ell^\infty(\mathbb{N})$** with the supremum norm: for each subset $S \subseteq \mathbb{N}$, the indicator sequence $\mathbb{1}_S := (\mathbb{1}_S(1), \mathbb{1}_S(2), \dots) \in \ell^\infty$. For $S \ne T$, $\|\mathbb{1}_S - \mathbb{1}_T\|_\infty = 1$. The family $\{\mathbb{1}_S\}_{S \subseteq \mathbb{N}}$ is uncountable (it has the cardinality of the power set $\mathcal{P}(\mathbb{N})$), and the balls $B(\mathbb{1}_S, 1/3)$ are pairwise disjoint. Any dense subset must place at least one element in each ball, so no countable set can be dense. 2. **Uncountable discrete spaces:** if $X$ is uncountable with the discrete topology (every subset is open), then the only dense subset of $X$ is $X$ itself, which is uncountable. 3. **$L^\infty(0, 1)$**: the indicator functions $\{\mathbb{1}_{(0,t)}\}_{t \in (0,1)}$ are at mutual distance $1$ in the $L^\infty$ norm, and the same packing argument applies. [/example] ### Basic Permanence Properties Separability is preserved by several standard operations. The proofs are direct from the definition and we state them without proof. **Continuous images.** If $f: X \to Y$ is a continuous surjection and $X$ is separable, then $Y$ is separable. (If $D$ is countable and dense in $X$, then $f(D)$ is countable and dense in $Y$, since for any open $V \subseteq Y$, $f^{-1}(V)$ is open and nonempty, hence meets $D$.) **Subspaces of metrizable spaces.** A subspace of a separable metrizable space is separable. This uses second countability: a subspace of a second-countable space is second-countable, and for metrizable spaces, second countability is equivalent to separability. In general topological spaces, subspaces of separable spaces need not be separable (the Sorgenfrey plane provides a counterexample). **Countable products.** A countable product of separable spaces is separable (in the product topology). This is because the product topology on $\prod_{n=1}^\infty X_n$ has a countable base when each $X_n$ is second-countable, and for metrizable spaces the result follows from the equivalence theorem below. For uncountable products, separability can fail. ## Separability in Metrizable Spaces For general topological spaces, separability, second countability, and the Lindelöf property are three distinct conditions with no implications between them (beyond the trivial "second countable implies separable and Lindelöf"). The situation simplifies dramatically for metrizable spaces, where all three conditions coincide. This equivalence is important because each formulation has different uses: separability is the easiest to verify (exhibit a countable dense set), second countability gives the strongest structural consequences (subspaces inherit it, and it implies paracompactness), and the Lindelöf property provides covering arguments. [quotetheorem:545] The proof strategy reveals why metrizability is essential. The implication "separable $\Rightarrow$ second countable" uses the metric to build a countable base from rational-radius balls centred at the dense set — this construction has no analogue in a general topological space, where there is no notion of "radius." The implication "Lindelöf $\Rightarrow$ separable" uses the metric to produce the covering $\{B(x, 1/n)\}_{x \in X}$ and extract a countable subcover — again, a construction that requires the metric. ### Subspace Separability in [Metric Spaces](/page/Metric%20Space) A useful consequence of the equivalence is that subspaces of separable metric spaces are separable. In a general topological space, a subspace of a separable space can be non-separable (the Sorgenfrey plane is a separable space containing a non-separable subspace). For metrizable spaces, the equivalence with second countability resolves this: second countability is inherited by subspaces, and in a metrizable space, second countability is equivalent to separability. This fact is used constantly in functional analysis: if $X$ is a separable Banach space, then every closed subspace of $X$ is separable. For instance, $H^1_0(U) \subseteq H^1(U) \subseteq L^2(U)$, and since $L^2(U)$ is separable, so are all the Sobolev spaces embedded in it. ## Separability in [Banach Spaces](/page/Banach%20Space) The role of separability in functional analysis goes far beyond the abstract topological definition. In Banach spaces, separability interacts with duality, weak topologies, and compactness in ways that have no counterpart in general topology. ### The Landscape of Separable Banach Spaces The concrete Banach spaces of analysis divide cleanly into separable and non-separable families. The following result establishes the separability of the most important function spaces. [quotetheorem:548] The proof for $L^p$ ($p < \infty$) constructs a countable dense subset from rational linear combinations of indicator functions of rational-endpoint boxes. This is a two-stage approximation: first approximate an arbitrary $L^p$ function by a simple function (which is standard measure theory), then approximate the simple function by one with rational coefficients and rational-box supports. Each stage introduces only a countable set of approximants. The non-separability of $L^\infty$ has a geometric flavour: the indicator functions $\{\mathbb{1}_{(0,t)}\}_{t \in (0,1)}$ form an uncountable set of mutually distance-$1$ elements, and any dense subset must contain a point near each one. This packing argument is the standard technique for proving non-separability: find an uncountable set whose pairwise distances are bounded below. The same classification extends to Sobolev spaces: $W^{k,p}(U)$ is separable for $1 \le p < \infty$ (since the $W^{k,p}$ norm involves only $L^p$ norms of [derivatives](/page/Derivative), and a countable dense subset can be constructed from smooth functions with rational coefficients and compact support). The space $W^{k,\infty}(U)$ is not separable. ### Separability and Duality The relationship between the separability of a Banach space and the separability of its dual is asymmetric: the dual can be "larger" (less separable) than the space itself. [quotetheorem:546] The proof constructs a countable dense subset of $X$ by selecting, for each element of a countable dense subset of $X^*$, a unit vector in $X$ that nearly realises its norm. The resulting set is shown to be dense by a Hahn-Banach separation argument: if its closed span missed some element of $X$, there would be a nonzero functional in $X^*$ vanishing on the span, but this functional would then be far from every element of the countable dense subset of $X^*$, contradicting density. The converse fails, and the failure is instructive: [example:Separability Does Not Ascend To The Dual] The sequence space $\ell^1(\mathbb{N})$ is separable (eventually-zero sequences with rational entries are countable and dense). However, $(\ell^1)^* = \ell^\infty(\mathbb{N})$, which is not separable (as shown above by the uncountable packing $\{\mathbb{1}_S\}_{S \subseteq \mathbb{N}}$). This asymmetry is not an isolated pathology. It reflects a general principle: the dual of a "small" (separable) space can be "large" (non-separable) because the space of bounded linear functionals on a countably-generated space can have uncountable complexity. The dual $X^*$ is separable only when $X$ has enough structure to force the functionals to be countably determined — for instance, when $X$ is reflexive and separable, in which case $X^{**} = X$ is separable, and by the theorem above, $X^*$ must be separable too. [/example] This gives a useful criterion: **if $X$ is reflexive and separable, then $X^*$ is separable.** The proof is immediate: $X^{**} = X$ is separable, so by the [Separability of a Banach Space from Separability of its Dual](/theorems/546) applied to $X^*$ (whose dual is $X^{**} = X$), $X^*$ is separable. This applies to $L^p$ for $1 < p < \infty$ (which is reflexive), confirming that $(L^p)^* = L^{p'}$ is separable for $1 < p < \infty$. ### Weak* Metrizability: The Payoff of Separability The most consequential application of separability in functional analysis is the metrizability of the weak* topology on bounded subsets of the dual. This is the result that makes sequential weak* compactness work — and sequential weak* compactness is the engine of the direct method in the calculus of variations and of a large part of PDE existence theory. The Banach-Alaoglu theorem states that $B_{X^*}$ is compact in the weak* topology. In general, this is topological compactness (every open cover has a finite subcover), which does not guarantee sequential compactness. But when $X$ is separable, the following theorem restores the sequential picture. [quotetheorem:547] The metric is explicit: take a countable dense subset $\{x_n\}$ of $X$ and define $d(f, g) = \sum 2^{-n} |f(x_n) - g(x_n)| / (1 + |f(x_n) - g(x_n)|)$. The geometric weights ensure convergence, and the density of $\{x_n\}$ ensures that $d$ separates points (two bounded functionals agreeing on a dense set must agree everywhere, by continuity). The consequences are immediate and powerful: 1. **Sequential weak* compactness.** Since $B_{X^*}$ is weak* compact (Banach-Alaoglu) and metrizable (by the theorem), it is sequentially compact: every sequence in $B_{X^*}$ has a weak* convergent subsequence. This is the statement used in PDE theory hundreds of times — typically phrased as "since $\{u_n\}$ is bounded in $X^*$, there exists a subsequence $u_{n_k} \overset{*}{\rightharpoonup} u$." 2. **Diagonal extraction.** When working with multiple bounded sequences in duals of separable spaces, metrizability allows the standard diagonal argument: extract a subsequence converging in the first space, then a sub-subsequence converging in the second, and so on. This would not work with nets instead of sequences. 3. **Continuity and closedness via sequences.** In a metrizable space, sequential characterisations of continuity, closedness, and compactness are valid. This means one can verify that a set is weakly* closed by checking it against sequences rather than arbitrary nets — a significant simplification. ### Why $L^\infty$ is Problematic The non-separability of $L^\infty$ is not merely an abstract curiosity — it has concrete analytical consequences. Since $L^1$ is separable and $(L^1)^* = L^\infty$, the weak* topology on bounded subsets of $L^\infty$ *is* metrizable (by the theorem, applied with $X = L^1$). However, $L^\infty$ itself is non-separable, so: 1. The weak topology on $L^\infty$ (not the weak* topology inherited from being the dual of $L^1$, but the weak topology as a Banach space in its own right) on bounded sets is *not* guaranteed to be metrizable, since $(L^\infty)^*$ is non-separable and strictly larger than $L^1$. 2. Standard reflexivity-based arguments fail: $L^\infty$ is not reflexive, so the weak and weak* topologies differ, and one cannot freely interchange them. 3. Extracting [weakly convergent](/page/Weak%20Convergence) subsequences in $L^\infty$ (as opposed to weak* convergent subsequences) requires additional hypotheses beyond mere boundedness. This is why PDE theory works primarily in $L^p$ spaces with $p < \infty$ and their Sobolev relatives, reserving $L^\infty$ for pointwise estimates rather than as an ambient variational space. ## Separability in [Hilbert Spaces](/page/Hilbert%20Space) In a Hilbert space, separability acquires a clean geometric characterisation: it is equivalent to the existence of a countable orthonormal basis. This connects the abstract topological condition (countable dense subset) to the concrete algebraic-geometric structure (Fourier expansions with respect to countably many basis vectors). The [Orthogonal System](/page/Orthogonal%20System) page develops this connection in full detail. The key result is the [Existence of Orthonormal Bases in Separable Hilbert Spaces](/theorems/543): a Hilbert space $H$ is separable if and only if it admits a countable (finite or countably infinite) complete orthonormal system $\{e_k\}$. The proof is constructive: take a countable dense subset, extract a linearly independent subcollection, and apply [Gram-Schmidt Orthonormalisation](/theorems/542). The consequence is the [Isometric Isomorphism of Separable Hilbert Spaces with Sequence Spaces](/theorems/544): every separable infinite-dimensional Hilbert space is isometrically isomorphic to $\ell^2(\mathbb{N})$. In other words, **separability determines the Hilbert space up to isometric isomorphism**. This is a far stronger conclusion than anything available in the Banach space setting, where separability is a necessary but far from sufficient condition for classification. ### Separability and the Spectral Theorem The [Spectral Theorem for Compact Self-Adjoint Operators](/theorems/538) asserts that a compact [self-adjoint operator](/page/Self-Adjoint%20Operators) $T$ on a Hilbert space $H$ has an orthonormal eigenbasis for $\overline{\operatorname{Range}(T)}$. When $H$ is separable, this eigenbasis is countable, and the spectral decomposition takes the form \begin{align*} Tx = \sum_{k=1}^\infty \mu_k\, (x, e_k)_H\, e_k, \end{align*} where $\{\mu_k\}$ is a sequence of real eigenvalues tending to zero and $\{e_k\}$ is the corresponding orthonormal eigenbasis. Without separability, the eigenvalue "sequence" might be an uncountable net, and the expansion would require uncountable summation — technically possible but analytically unnatural. ## Problems [problem] Let $(X, d)$ be a separable metric space and let $\{U_\alpha\}_{\alpha \in A}$ be an arbitrary collection of pairwise disjoint nonempty open subsets of $X$. Prove that $A$ is at most countable. [/problem] [solution] **Step 1: Set up the injection.** Let $D = \{d_n\}_{n=1}^\infty$ be a countable dense subset of $X$. Since each $U_\alpha$ is open and nonempty, $U_\alpha$ contains at least one element of $D$ (by density: $\overline{D} = X$ implies every nonempty open set meets $D$). **Step 2: Define the map.** For each $\alpha \in A$, choose $d_{n(\alpha)} \in D \cap U_\alpha$. This defines a function $\alpha \mapsto n(\alpha)$ from $A$ to $\mathbb{N}$. **Step 3: Verify injectivity.** If $\alpha \ne \beta$, then $U_\alpha \cap U_\beta = \varnothing$ (by the pairwise disjointness hypothesis). Since $d_{n(\alpha)} \in U_\alpha$ and $d_{n(\beta)} \in U_\beta$, we have $d_{n(\alpha)} \ne d_{n(\beta)}$, so $n(\alpha) \ne n(\beta)$. **Step 4: Conclude.** The map $\alpha \mapsto n(\alpha)$ is an injection from $A$ into $\mathbb{N}$, so $A$ is at most countable. [/solution] [problem] Let $X$ be a Banach space. Suppose $X$ is separable and reflexive. Prove that every bounded sequence in $X$ has a weakly convergent subsequence. [/problem] [solution] **Step 1: Identify the relevant duality.** Since $X$ is reflexive, $X^{**} = X$ (under the canonical embedding). The weak topology on $X$ coincides with the weak* topology on $X^{**} = X$ when $X$ is viewed as the dual of $X^*$. More precisely, a sequence $\{x_n\} \subseteq X$ converges weakly (i.e. $f(x_n) \to f(x)$ for all $f \in X^*$) if and only if the corresponding elements $\hat{x}_n \in X^{**}$ converge weak* in $X^{**}$. **Step 2: Establish separability of the dual.** Since $X$ is reflexive and separable, $X^* $ is separable. This follows from the [Separability of a Banach Space from Separability of its Dual](/theorems/546) applied to $X^*$: the dual of $X^*$ is $X^{**} = X$, which is separable by hypothesis, so $X^*$ is separable. **Step 3: Apply weak* metrizability.** Since $X^*$ is separable, the [Weak* Metrizability of the Dual Ball](/theorems/547) (applied with $X^*$ in place of $X$) gives that the closed unit ball $B_{X^{**}}$ is metrizable in the weak* topology. Under the identification $X^{**} = X$, this means $B_X$ is metrizable in the weak topology. **Step 4: Apply Banach-Alaoglu and extract a subsequence.** By the Banach-Alaoglu theorem, $B_{X^{**}}$ is weak* compact. Under reflexivity, $B_X$ is weakly compact. Since $B_X$ is weakly compact and metrizable (Step 3), it is sequentially compact. Now let $\{x_n\}$ be a bounded sequence in $X$, say $\|x_n\|_X \le M$. Then $\{x_n / M\} \subseteq B_X$, and by sequential compactness, there exists a subsequence $x_{n_k} / M$ converging weakly to some $y \in B_X$. Therefore $x_{n_k} \rightharpoonup My$ weakly in $X$. [/solution] ## References 1. J. Munkres, *Topology* (2000). 2. H. Brezis, *Functional Analysis, Sobolev Spaces and Partial Differential Equations* (2010). 3. W. Rudin, *Functional Analysis* (1991). 4. J. Conway, *A Course in Functional Analysis* (1990). 5. M. Reed and B. Simon, *Methods of Modern Mathematical Physics I: Functional Analysis* (1980).