Countable Density and Countable Bases
A dense set gives points that can be found arbitrarily close to every point of a space. A base gives open sets that can be found inside every open neighborhood. In a metrizable space, these two kinds of countability are equivalent: metric balls let us pass from one to the other. The constructions also show exactly where the metric is needed.
A space is metrizable if its topology is induced by some metric \(d\): its open sets are unions of open balls \(B_d(x,r)=\{y\in X:d(x,y)<r\}\), where \(r>0\). The metric in the proof can be any metric inducing the given topology; the conclusion is about the topology, not about a preferred metric.
From a Countable Dense Set to a Countable Base
Proof. If \(X=\varnothing\), the empty family is a countable base. Suppose \(X\ne\varnothing\), and let \(D\subseteq X\) be a countable dense set. Fix a metric \(d\) inducing \(\mathcal{T}\), and let \(\mathbb{Q}_{>0}\) denote the positive rational numbers. Define \(\mathcal{B}=\{B_d(a,q):a\in D,\ q\in\mathbb{Q}_{>0}\}\). This family is countable because it is indexed by the product of two countable sets.
We check the base condition. Let \(x\in U\), where \(U\) is open. Since the topology is induced by \(d\), there is \(\varepsilon>0\) such that \(B_d(x,\varepsilon)\subseteq U\). Density gives \(a\in D\cap B_d(x,\varepsilon/3)\), so \(d(x,a)<\varepsilon/3\). The interval \((d(x,a),\,\varepsilon-d(x,a))\) is nonempty: indeed, \(d(x,a)<\varepsilon/3\) implies \(2d(x,a)<\varepsilon\). Choose a positive rational \(q\) in this interval. Then \(d(x,a)<q\), so \(x\in B_d(a,q)\). If \(y\in B_d(a,q)\), the triangle inequality gives \(d(x,y)\le d(x,a)+d(a,y)<d(x,a)+q<\varepsilon\). Thus \(x\in B_d(a,q)\subseteq B_d(x,\varepsilon)\subseteq U\), as required. Therefore \(\mathcal{B}\) is a countable base. \(\square\)
The rational radii are essential to this countability argument: the center is chosen from the countable set \(D\), and the radius from the countable positive rationals. Allowing every positive real radius would not give this direct countability conclusion.
Worked Example: A Countable Base for the Real Line
Give \(\mathbb{R}\) its standard topology and usual metric \(d(x,y)=|x-y|\). The rationals \(\mathbb{Q}\) are countable and dense in \(\mathbb{R}\): every nonempty open interval contains a rational. The construction in Theorem 1 therefore gives the family \(\{(a-q,a+q):a\in\mathbb{Q},\ q\in\mathbb{Q}_{>0}\}\).
For a specific base-element check, take \(x=\sqrt{2}\) and \(U=(1,2)\). We have \(1<\sqrt{2}<2\), so \(x\in U\). Choose the rational center \(a=3/2\) and rational radius \(q=1/3\). The distance from the center to \(x\) is \(|\sqrt{2}-3/2|=(3/2)-\sqrt{2}\), which is less than \(1/3\), since \(\sqrt{2}>7/6\). The ball is \((3/2-1/3,3/2+1/3)=(7/6,11/6)\), and \(1<7/6<11/6<2\). Hence \(x\in(7/6,11/6)\subseteq U\).
Check: its center \(3/2\) is rational, its radius \(1/3\) is positive rational, and the computed ball is contained in \(U\). For arbitrary \(x\in U\), the theorem’s argument supplies such a ball, so this family is a base, not merely a collection of open sets.
From a Countable Base to a Countable Dense Set
Proof. Let \(\mathcal{B}\) be a countable base for \(X\). If \(X=\varnothing\), then \(\varnothing\) is a countable dense subset. Otherwise, for each nonempty \(B\in\mathcal{B}\), choose one point \(x_B\in B\), and set \(D=\{x_B:B\in\mathcal{B},\ B\ne\varnothing\}\). This set is countable, since there is at most one selected point for each member of a countable family. To prove density, let \(U\) be any nonempty open set and choose \(x\in U\). The base property gives \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). This \(B\) is nonempty, so its selected point \(x_B\) belongs to \(D\cap U\). Every nonempty open set therefore meets \(D\), proving that \(D\) is dense. \(\square\)
The selection in this proof is just one point from each of countably many nonempty sets, as usual in elementary topology. The argument does not construct a metric and does not use one.
Worked Example: Selecting a Dense Set from a Base on the Rationals
Give \(X=\mathbb{Q}\) the subspace topology inherited from standard \(\mathbb{R}\). The family \(\mathcal{B}=\{\mathbb{Q}\cap(a,b):a,b\in\mathbb{Q},\ a<b\}\) is a countable base: there are countably many pairs of rational endpoints, and any relatively open neighborhood of a rational \(x\) contains \(\mathbb{Q}\cap(a,b)\) for rational \(a<x<b\).
For each basis member \(\mathbb{Q}\cap(a,b)\), select its midpoint \((a+b)/2\). This midpoint is rational and lies strictly between \(a\) and \(b\), so it is a point of that basis member. The set of all selected midpoints is countable because it is indexed by the countable set of rational endpoint pairs. If \(U\subseteq\mathbb{Q}\) is nonempty and relatively open, choose \(x\in U\). The base property gives rational \(a<b\) with \(x\in\mathbb{Q}\cap(a,b)\subseteq U\). The selected midpoint \((a+b)/2\) then belongs to \(U\), proving that the selected set is dense.
For example, in the open set \(U=\mathbb{Q}\cap(1,2)\), the basis member \(\mathbb{Q}\cap(4/3,5/3)\) is contained in \(U\), and its selected point is \(((4/3)+(5/3))/2=3/2\). It is rational and lies strictly between \(1\) and \(2\). The same reasoning applies to every nonempty relatively open \(U\), so the selection is dense.
Worked Example: The Countable Discrete Space
Let \(X=\mathbb{N}\) with the discrete topology, in which every subset is open. The family \(\mathcal{B}=\{\{n\}:n\in\mathbb{N}\}\) is a countable base: for \(n\in U\), the singleton \(\{n\}\) satisfies \(n\in\{n\}\subseteq U\). Selecting the point \(n\) from the basis member \(\{n\}\) gives \(D=\mathbb{N}\), which is countable and dense. Indeed, every nonempty open set contains some \(n\), and therefore intersects \(D\).
This space is also metrizable by \(d(m,n)=0\) when \(m=n\) and \(d(m,n)=1\) when \(m\ne n\). For each \(n\), the ball \(B_d(n,1/2)\) is exactly \(\{n\}\), because \(d(n,n)=0<1/2\) while \(d(m,n)=1\not<1/2\) for \(m\ne n\). Thus the metric induces the discrete topology, and the example checks both properties directly.
The Equivalence and Its Scope
Proof. If the space is separable, Theorem 1 shows that it is second countable. If it is second countable, Theorem 2 shows that it is separable. These implications also cover the empty space, as noted in their proofs. \(\square\)
The metric hypothesis is needed for the implication from separability to second countability; it is not needed in the reverse direction. For a concrete illustration, give \(\mathbb{R}\) the lower limit topology, whose basis consists of intervals \([a,b)\) with \(a<b\). The rationals are dense because every nonempty basic interval \([a,b)\) contains a rational, so every nonempty open set intersects \(\mathbb{Q}\).
This space is not second countable. If \(\mathcal{B}\) were a base, then for every \(x\in\mathbb{R}\) the base property applied to \([x,x+1)\) would give \(B_x\in\mathcal{B}\) with \(x\in B_x\subseteq[x,x+1)\). Since \(B_x\) is open in the lower limit topology and contains \(x\), it contains some basic interval \([x,x+\delta)\) with \(\delta>0\). No point of \(B_x\) is less than \(x\), while \(x\in B_x\), so \(x\) is the least element of \(B_x\). For distinct \(x\) and \(y\), the sets \(B_x\) and \(B_y\) cannot be equal, because a set has only one least element. Thus the base would have to contain uncountably many distinct members, contradicting countability. This separable but non-second-countable example cannot be metrizable.
Common pitfall. Starting with a countable dense \(D\), one might claim that all balls centered in \(D\), with arbitrary positive real radii, form a countable base. That claim is unsupported: for each fixed center \(a\), the proposed family has a ball \(B_d(a,r)\) for every real \(r>0\), so the family of indices is not countable. The correction is to restrict the radii to \(\mathbb{Q}_{>0}\). The base proof still works because between the two distinct real numbers \(d(x,a)\) and \(\varepsilon-d(x,a)\) one can choose a rational radius. This restriction preserves the neighborhood-refinement argument while making the family countable.
Check Your Understanding
Use the definitions and the two constructions to answer the following.
- In Theorem 1, why must the rational radius \(q\) satisfy both \(d(x,a)<q\) and \(q<\varepsilon-d(x,a)\)?
- Explain why choosing one point from each nonempty member of a countable base produces a countable set.
- Does Theorem 2 require \(X\) to be metrizable? State the part of its proof that establishes your answer.
- For \(X=\mathbb{N}\) with the discrete topology, is the empty set dense? Is \(X\) dense? Justify both answers from the definition.
- Which implication in the equivalence can fail without metrizability, and what does the lower limit topology example show?