Dense Sets and Separability

Separability is the standing hypothesis behind countless approximation arguments: in functional analysis, a Hilbert space is separable if and only if it admits a countable orthonormal basis, which is the setting for Fourier series and quantum mechanics' state spaces (separable Hilbert space $L^2$). In probability theory, separability of the underlying metric space underlies the existence of regular conditional distributions and the theory of weak convergence of measures (Prokhorov's theorem). In numerical analysis and approximation theory, the density of polynomials in $C[a,b]$ (Weierstrass approximation theorem) or of simple functions in $L^p$ spaces is exactly a density statement, enabling algorithms to approximate arbitrary continuous or integrable functions by simple, computable ones. In general topology and set-theoretic topology, nowhere-dense sets are the atomic pieces of meager sets, and the question of when a space cannot be a countable union of nowhere-dense sets is answered by the Baire Category Theorem, a cornerstone used to prove the existence of continuous nowhere-differentiable functions and the Open Mapping / Closed Graph theorems in functional analysis. In computer science, dense countable subsets underlie the theoretical justification for floating-point and rational approximations of real-valued computations, and separability assumptions appear in the theory of Polish spaces used in descriptive set theory and in machine learning's treatment of continuous state spaces.

Ask the tutor: How does the notion of a meager (first category) set, built from countable unions of nowhere-dense sets, set up the proof of the Baire Category Theorem?

Think of the real line with its usual topology. Around every real number, no matter how small a neighborhood you draw, there is a rational number inside it — this is exactly the statement that $\mathbb{Q}$ is dense in $\mathbb{R}$. The rationals leave no "gap" open anywhere: every open interval, however tiny, is hit. This is very different from saying the rationals are most of $\mathbb{R}$ in a measure-theoretic sense — Lebesgue measure of $\mathbb{Q}$ is zero — density is a purely topological notion about intersecting open sets, not about size. Contrast this with the set $\mathbb{Z}$ of integers inside $\mathbb{R}$. Its closure is itself (it's already closed), and that closure has empty interior: no open interval is contained entirely in $\mathbb{Z}$. So $\mathbb{Z}$ is nowhere dense. A nowhere-dense set is "thin" in the strongest topological sense — you can always find an open sub-neighborhood of any open set that entirely avoids it (after passing to the closure). Nowhere-dense sets are the atoms out of which countable unions build "meager" (topologically negligible) sets, foreshadowing the Baire Category Theorem, where a complete metric space is proven to be not a countable union of nowhere-dense sets.

Separability is the statement that a single countable set can serve as this dense skeleton for the entire space. Countability is the crucial economizing feature: it means the space, however large and complicated (even uncountable, like $\mathbb{R}^n$), can be approximated by a set we can enumerate — $d_1, d_2, d_3, \ldots$. This is what allows proofs by "check it on a countable dense set, then pass to the limit," a technique that recurs throughout Continuity arguments, throughout the construction of measures, and throughout functional analysis (e.g. separable Hilbert spaces admit countable orthonormal bases).

Example 1: The dyadic rationals are dense in $[0,1]$, and are nowhere dense as a subset of $\mathbb{R}\setminus[0,1]$ considerations aside — but let's do something less standard: showing a non-uniform, asymmetric dense set. Let $X = \mathbb{R}$ with the usual topology, and let

be the set of dyadic rationals that are not integers (an intentionally asymmetric, "punctured" variant of the usual dyadic rationals, to make sure density survives removing a nontrivial infinite subset). Claim: $D$ is dense in $\mathbb{R}$. Let $U = (a,b)$ be any nonempty open interval, $a < b$. We must find an element of $D$ inside $(a,b)$. Choose $n$ large enough that $2^{-n} < (b-a)/2$; this is possible since $2^{-n} \to 0$ as $n \to \infty$ (Archimedean property of $\mathbb{R}$). Consider the grid of points $\{m/2^n : m \in \mathbb{Z}\}$, spaced $2^{-n}$ apart. Since the interval $(a,b)$ has length $b - a > 2\cdot 2^{-n}$, it must contain at least one grid point $m_0/2^n$ with room to spare: concretely, let $m_0 = \lceil a \cdot 2^n \rceil + 1$; then $m_0/2^n > a$ automatically, and

using $2/2^n < b - a$ from our choice of $n$. So $m_0/2^n \in (a,b)$. Now if $m_0/2^n$ happens to be an integer (i.e. $2^n \mid m_0$), replace $m_0$ by $m_0+1$: since the grid spacing $2^{-n}$ is strictly less than $b-a$ divided by 2, and we have at least $2$ grid spacings of room inside $(a,b)$ by construction (as $b-a > 2\cdot2^{-n}$), the next grid point $(m_0+1)/2^n$ still lies in $(a,b)$ if $m_0/2^n$ did with strict room, which holds by the strict inequality above with slack. Concretely, since $m_0/2^n < a + 2/2^n$ and $2/2^n < b-a$, we get $(m_0+1)/2^n = m_0/2^n + 2^{-n} < a + 2/2^n + 2^{-n} < a + (b-a) + 2^{-n}$, and choosing $n$ even larger if necessary so that also $3\cdot 2^{-n} < b-a$ guarantees two consecutive grid points both lie in $(a,b)$, at most one of which is an integer (integers among dyadic points with denominator $2^n$ occur every $2^n$ grid steps apart, so two consecutive grid points cannot both be integers once $n \ge 1$). Hence at least one of $m_0/2^n$, $(m_0+1)/2^n$ is a non-integer dyadic rational lying in $(a,b)$, i.e. lies in $D \cap (a,b)$. Since $(a,b)$ was an arbitrary nonempty basic open interval and the topology of $\mathbb{R}$ has these intervals as a base, and any nonempty open $U$ contains some such interval, $D$ meets every nonempty open set. By (b) $\Rightarrow$ (a) of the theorem, $D$ is dense in $\mathbb{R}$. Since $D$ is a subset of the countable set of all dyadic rationals, $D$ is countable, exhibiting $\mathbb{R}$'s separability via a deliberately non-obvious countable dense set.

Example 2: A nowhere-dense set that is uncountable — the standard Cantor set $\mathcal{C} \subset [0,1]$. Recall $\mathcal{C} = \bigcap_{n=0}^\infty C_n$ where $C_0 = [0,1]$ and $C_{n+1}$ is obtained from $C_n$ by deleting the open middle third of each closed interval comprising $C_n$. Each $C_n$ is a finite union of $2^n$ closed intervals each of length $3^{-n}$. Claim: $\mathcal{C}$ is closed and nowhere dense in $\mathbb{R}$, despite being uncountable. First, $\mathcal C$ is closed: each $C_n$ is a finite union of closed intervals, hence closed, and $\mathcal C = \bigcap_n C_n$ is an intersection of closed sets, hence closed. Since $\mathcal{C}$ is closed, $\overline{\mathcal C} = \mathcal C$, so nowhere-density reduces to showing $\operatorname{int}(\mathcal C) = \varnothing$. Suppose for contradiction that $\operatorname{int}(\mathcal C) \neq \varnothing$; then there is a nonempty open interval $(a,b) \subseteq \mathcal C$ with $b - a = \varepsilon > 0$. Since $\mathcal{C} \subseteq C_n$ for every $n$, we'd have $(a,b) \subseteq C_n$ for every $n$. But $C_n$ consists of $2^n$ intervals each of length $3^{-n}$, so no subinterval of $C_n$ has length exceeding $3^{-n}$ unless it's contained within one of these length-$3^{-n}$ pieces — in particular $(a,b) \subseteq C_n$ forces $\varepsilon = b-a \le 3^{-n}$ (an interval of length $> 3^{-n}$ cannot fit inside a single piece of $C_n$, and different pieces of $C_n$ are separated by gaps, so no interval can span two pieces either). Choosing $n$ large enough that $3^{-n} < \varepsilon$ (possible by the Archimedean property, since $3^{-n} \to 0$) contradicts $\varepsilon \le 3^{-n}$. Hence no such $(a,b)$ exists, so $\operatorname{int}(\mathcal C) = \varnothing$. Therefore $\mathcal C$ is nowhere dense — despite the well-known fact that $\mathcal C$ is uncountable (in bijection with $\{0,1\}^{\mathbb N}$ via ternary expansions using only digits $0,2$). This shows nowhere-density is a statement about the arrangement of a set relative to open sets, entirely independent of cardinality: a nowhere-dense set can be uncountable, and (as Example 1's cousin, $\mathbb Q$, shows) a dense set can be countable.