Two Countability Conditions in Metric Spaces
Separability describes a space through a countable set of points: every nonempty open set must come close enough to contain one of them. Second countability describes it through a countable collection of open sets that can refine every neighborhood. In metric spaces, these descriptions are equivalent. The key construction is to center balls at a countable dense set and use positive rational radii.
Throughout, \((X,\mathcal{T})\) is a topological space. A family \(\mathcal{B}\) of open subsets of \(X\) is a basis if, whenever \(U\in\mathcal{T}\) and \(x\in U\), there is a \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\). The space is second countable if it has a finite or countable basis. A subset \(D\subseteq X\) is dense if every nonempty open subset of \(X\) meets \(D\). The space is separable if it has a finite or countable dense subset.
From a Countable Basis to a Dense Set
Proof. Let \(\mathcal{B}\) be a finite or countable basis for \((X,\mathcal{T})\). For each nonempty \(B\in\mathcal{B}\), choose one point \(x_B\in B\), and let \(D\) be the set of all chosen points. This uses only countable choice, namely the selection of one point from each member of a countable family of nonempty sets. Since there are at most countably many basis elements, \(D\) is at most countable.
To show that \(D\) is dense, let \(U\) be any nonempty open set and choose \(x\in U\). The basis property gives \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). This \(B\) is nonempty, so its chosen point \(x_B\) belongs to \(D\cap U\). Thus every nonempty open set meets \(D\), proving density. If \(X=\varnothing\), take \(D=\varnothing\); it is dense by the definition, and the empty family is a basis. \(\square\)
This direction does not require a metric. The metric is needed for the converse: distances let us approximate an arbitrary neighborhood by balls whose centers and radii come from countable sets.
From a Dense Set to a Countable Basis
Proof. First, every member of \(\mathcal{B}\) is open by the definition of the metric topology. The positive rationals are countable, and the product of two countable sets is countable, so the pairs \((a,r)\in D\times\mathbb{Q}_{>0}\) form a countable family. Consequently \(\mathcal{B}\) is countable.
It remains to verify the basis condition. Let \(U\) be open, let \(x\in U\), and choose \(\varepsilon>0\) such that \(B_d(x,\varepsilon)\subseteq U\). Density of \(D\) means that every open ball around \(x\) meets \(D\); in particular choose \(a\in D\cap B_d(x,\varepsilon/3)\). Then \(d(a,x)<\varepsilon/3\), so \(\varepsilon-d(a,x)>2\varepsilon/3\). The interval \((d(a,x),\varepsilon-d(a,x))\) is therefore nonempty. By density of the rational numbers in the real line, choose a positive rational \(r\) in that interval. We have \(d(a,x)<r\), so \(x\in B_d(a,r)\).
If \(y\in B_d(a,r)\), the triangle inequality gives \(d(y,x)\le d(y,a)+d(a,x)<r+d(a,x)<\varepsilon\). Thus \(B_d(a,r)\subseteq B_d(x,\varepsilon)\subseteq U\). We have found a member of \(\mathcal{B}\) containing \(x\) and contained in \(U\), as required. If \(X=\varnothing\), then \(D=\varnothing\) and \(\mathcal{B}=\varnothing\), which is a countable basis for the empty space. \(\square\)
Proof. If the metric space is second countable, Proposition 1 supplies a countable dense subset, so it is separable. If it is separable, it has a countable dense subset, and Proposition 2 constructs a countable basis, so it is second countable. These implications also cover the empty space, as addressed in the propositions. \(\square\)
Worked Constructions
Worked Example: The Real Line
Give a countable basis for \(\mathbb{R}\) with its standard metric. Take \(D=\mathbb{Q}\). The rationals are countable: each rational can be written as \(p/q\) for an integer \(p\) and a positive integer \(q\), and such pairs can be listed by increasing \(|p|+q\), with only finitely many pairs at each fixed value.
To check density, let \(x\in\mathbb{R}\) and let \(\varepsilon>0\). Choose \(n\in\mathbb{N}\) with \(1/n<\varepsilon\), and let \(m\) be the greatest integer less than or equal to \(nx\). Then \(0\le nx-m<1\), which, after division by \(n\), gives \(0\le x-m/n<1/n<\varepsilon\). Thus the rational \(m/n\) lies within \(\varepsilon\) of \(x\). Every open interval around every real point therefore meets \(\mathbb{Q}\).
Proposition 2 gives the countable family \(\{B(q,r):q\in\mathbb{Q},\ r\in\mathbb{Q},\ r>0\}\). In the standard metric, \(B(q,r)=(q-r,q+r)\). For instance, the interval \((-1/4,1/4)\) contains \(0\); the ball \(B(0,1/8)=(-1/8,1/8)\) is a member of the family, contains \(0\), and lies inside the given interval. The proposition proves the corresponding refinement for every open interval and every point in it.
Check: the family is countable because it is indexed by pairs of rationals. It is a basis because Proposition 2 verifies the refinement condition for an arbitrary open set, not just the interval used in this check.
Worked Example: The Rational Numbers as a Metric Space
Let \(X=\mathbb{Q}\) with the metric \(d(x,y)=|x-y|\) and its metric topology, which is the subspace topology inherited from the standard topology on \(\mathbb{R}\). The set \(D=\mathbb{Q}\) is countable and dense in \(X\): every nonempty open subset of \(X\) contains one of its points, and every point of \(X\) is already in \(D\).
The construction gives the family \(B_X(q,r)=\{x\in\mathbb{Q}:|x-q|<r\}\), where \(q\) and \(r\) are rational and \(r>0\). Equivalently, these sets are \(\mathbb{Q}\cap(q-r,q+r)\). The family is countable because its possible centers and radii are both drawn from \(\mathbb{Q}\).
For a specific refinement, take the open set \(U=\mathbb{Q}\cap(-1,1)\) and its point \(x=1/2\). The ball \(B_X(1/2,1/4)\) contains \(1/2\), since \(|1/2-1/2|=0<1/4\). If \(y\) belongs to this ball, then \(|y| \le |y-1/2|+|1/2|<1/4+1/2=3/4<1\), so \(y\in U\). This verifies that the constructed family can refine an open set in the subspace topology.
Check: the radius \(1/4\) is positive and rational, the center \(1/2\) belongs to \(D\), and the displayed inequality shows the whole ball, not merely its center, lies in \(U\).
Worked Example: A Convergent Sequence with Its Limit
Let \(X=\{0\}\cup\{1/n:n\in\mathbb{N}\}\subseteq\mathbb{R}\), with the metric inherited from the standard metric on \(\mathbb{R}\). The set \(X\) is countable, and it is dense in itself: every nonempty open subset of \(X\) contains a point of \(X\), because it is a subset of \(X\) and is nonempty. Thus \(D=X\) is a countable dense set.
Proposition 2 gives a basis consisting of \(B_X(a,r)=\{y\in X:|y-a|<r\}\), with \(a\in X\) and positive rational \(r\). For example, consider the open neighborhood \(U=B_X(0,1/3)\) of \(0\). The ball \(B_X(0,1/4)\) is a member of the constructed family and lies inside \(U\): if \(|y|<1/4\), then \(|y|<1/3\). It contains \(0\) and also \(1/5\), since \(|1/5-0|=1/5<1/4\).
This example also shows why neighborhoods around the limit point need not be singletons. For every \(r>0\), choose \(n\in\mathbb{N}\) with \(n>1/r\). Then \(0<1/n<r\), so \(B_X(0,r)\) contains a point of \(X\) other than \(0\). The rational-radius construction still provides a countable basis even though these neighborhoods keep containing sequence terms.
Check: \(1/5<1/4<1/3\), so the displayed ball contains \(1/5\) and is contained in the stated neighborhood. Its center \(0\) belongs to \(D\), and its radius \(1/4\) is positive and rational.
Why the Metric Hypothesis Matters
The implication from separability to second countability is not true for all topological spaces. Here is a direct example. Let \(X\) be an uncountable set with the cofinite topology: its open sets are \(\varnothing\) and all subsets whose complements in \(X\) are finite.
Worked Example: A Separable Space That Is Not Second Countable
Choose a countably infinite subset \(D\subseteq X\). It is dense: a nonempty open set \(U\) has finite complement \(X\setminus U\), and an infinite set \(D\) cannot be contained in that finite complement. Therefore \(D\cap U\ne\varnothing\), proving that \(D\) is dense and \(X\) is separable.
Suppose, for contradiction, that \(X\) has a countable basis \(\mathcal{B}\). Every nonempty \(B\in\mathcal{B}\) is open and hence has finite complement. For each \(x\in X\), choose \(y\in X\setminus\{x\}\), which is possible because \(X\) is uncountable. The set \(X\setminus\{x\}\) is open and contains \(y\). The basis property gives \(B\in\mathcal{B}\) with \(y\in B\subseteq X\setminus\{x\}\). In particular, \(x\in X\setminus B\). It follows that the union of the complements \(X\setminus B\), over nonempty \(B\in\mathcal{B}\), covers \(X\).
There are at most countably many such basis elements, and each complement \(X\setminus B\) is finite. A countable union of finite sets is countable, so this cover would make \(X\) countable, contradicting its choice as uncountable. Thus \(X\) is not second countable, despite being separable.
Check: the argument uses the strict distinction between finite and infinite: removing the finite set \(X\setminus U\) cannot remove every point of the infinite set \(D\). The metric hypothesis in the equivalence theorem is therefore essential.
A Common Pitfall
A frequent mistake is to allow dense centers but use only one fixed radius. In \(\mathbb{R}\), balls with rational centers and radius \(1\) do not form a basis: the open set \((-1/4,1/4)\) contains \(0\), but any radius-\(1\) ball containing \(0\) also contains points outside this interval. For example, \(B(0,1)=(-1,1)\) is not contained in \((-1/4,1/4)\). The correction is to let the radius vary through all positive rational numbers. Those radii can be made small enough to fit inside any prescribed neighborhood, while the dense set of centers ensures a center can be chosen close to the point being refined.
Check Your Understanding
Use the definitions and the basis construction to answer the following.
- Explain why a countable basis yields a countable dense set, including how the set is chosen from the basis elements.
- In the proof from separability to second countability, why is it useful to choose a center within \(\varepsilon/3\) of \(x\)?
- For \(\mathbb{R}\) with its standard metric, describe the basis obtained from the dense set \(\mathbb{Q}\).
- Why would rational centers with radius exactly \(1\) fail to give a basis for \(\mathbb{R}\)?
- In the cofinite topology example, identify where uncountability contradicts the assumption of a countable basis.