From One Countable Basis to Many Local Bases
A basis describes open sets throughout a space, while a local base records only the neighborhoods relevant to one point. The link between these ideas is direct: from a countable basis, keep just the basis elements that contain the point. This tutorial proves that construction works and then shows how it applies to every subspace of a second-countable space.
Let \((X,\mathcal{T})\) be a topological space. A family \(\mathcal{B}\) of open subsets of \(X\) is a basis if every point of \(X\) belongs to some member of \(\mathcal{B}\), and whenever \(x\in U\in\mathcal{T}\), there is \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\). A family is countable if it is finite or countably infinite. A space is second countable if it has a countable basis.
The inclusion \(V\subseteq U\) is essential: a local base must provide neighborhoods as small as any prescribed open neighborhood. The family need not consist of all open neighborhoods of the point; it only needs to refine each one.
The Main Construction
Proof. Fix \(x\in X\). Every member of \(\mathcal{B}_x\) is open and contains \(x\), by the definition of a basis. Also, \(\mathcal{B}_x\) is a subfamily of the countable family \(\mathcal{B}\), so it is countable. Now let \(U\) be any open neighborhood of \(x\). The basis property gives \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). Since \(x\in B\), this \(B\) belongs to \(\mathcal{B}_x\). Thus every open neighborhood \(U\) contains a member of \(\mathcal{B}_x\), which is exactly the local base condition. If \(X=\varnothing\), there are no points to check, so first countability holds vacuously. \(\square\)
The local base depends on the point: \(\mathcal{B}_x\) discards every basis element that does not contain \(x\). This point-by-point restriction is what turns a global basis into the required family of neighborhoods.
Worked Example: Neighborhoods of Zero in the Real Line
Give a countable local base at \(0\) in \(\mathbb{R}\) with its standard topology. A countable basis for this topology is the family of intervals \((p,q)\), where \(p,q\in\mathbb{Q}\) and \(p<q\). The rational numbers are dense in \(\mathbb{R}\), so every open neighborhood contains such an interval around each of its points. The family is countable because it is indexed by pairs from the countable set \(\mathbb{Q}\).
By the theorem, all rational-endpoint intervals containing \(0\) form a countable local base at \(0\). A smaller, explicit local base is \(W_n=(-1/n,1/n)\), for \(n\in\mathbb{N}\). Its first three members are \(W_1=(-1,1)\), \(W_2=(-1/2,1/2)\), and \(W_3=(-1/3,1/3)\). Each is open and contains \(0\).
To check the local base condition, let \(U\) be any open neighborhood of \(0\). By the standard topology on \(\mathbb{R}\), there is \(\varepsilon>0\) such that \((-\varepsilon,\varepsilon)\subseteq U\). Choose \(n\in\mathbb{N}\) with \(n>1/\varepsilon\). Then \(1/n<\varepsilon\), so \(W_n=(-1/n,1/n)\subseteq(-\varepsilon,\varepsilon)\subseteq U\). Thus the displayed family is a countable local base. As a check, for \(U=(-1/4,1/4)\), \(W_5=(-1/5,1/5)\subseteq U\), since \(1/5<1/4\).
Taking Traces on a Subspace
Let \(A\subseteq X\). The subspace topology on \(A\) consists of the sets \(A\cap U\) for \(U\in\mathcal{T}\). In particular, an open set in \(A\) is understood relative to \(A\); it need not be open in \(X\).
Proof. Each \(B\cap A\) is open in \(A\), since \(B\) is open in \(X\). Let \(G\) be open in \(A\) and \(x\in G\). By the definition of the subspace topology, there is an open \(U\subseteq X\) such that \(G=A\cap U\). Since \(x\in U\) and \(\mathcal{B}\) is a basis for \(X\), choose \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). Then \(x\in B\cap A\subseteq U\cap A=G\). This proves the basis condition for \(\mathcal{B}|_A\). If \(\mathcal{B}\) is countable, its family of traces is countable, because taking intersections with \(A\) sends each basis element to one trace; different elements may give the same trace, which cannot increase the size of the family. Thus \(A\) is second countable, and the theorem shows that it is first countable. If \(A=\varnothing\), the basis condition has no points to check and the conclusion is again vacuous. \(\square\)
There is also a direct local version of the construction. For \(x\in A\), the sets \(A\cap B\) with \(B\in\mathcal{B}\) and \(x\in B\) form a countable local base at \(x\) in \(A\). Indeed, if \(G=A\cap U\) is a relative open neighborhood of \(x\), the basis property in \(X\) supplies \(B\) with \(x\in B\subseteq U\), and then \(x\in A\cap B\subseteq G\).
Worked Example: A Subspace of the Real Line
Let \(A=\mathbb{Q}\cap[0,1]\), with the subspace topology inherited from the standard topology on \(\mathbb{R}\). Taking traces of the rational-endpoint interval basis gives a countable basis for \(A\). We can make the local base at \(1/2\) explicit by defining \(V_n=A\cap(1/2-1/n,1/2+1/n)\), for \(n\in\mathbb{N}\).
Substitution gives \(V_1=A\cap(-1/2,3/2)=A\), since the interval contains all of \([0,1]\). Also, \(V_2=A\cap(0,1)=\mathbb{Q}\cap(0,1)\), and \(V_3=A\cap(1/6,5/6)=\mathbb{Q}\cap(1/6,5/6)\). In each case \(1/2\in V_n\). Every \(V_n\) is open in \(A\), since it is the intersection of \(A\) with an open interval in \(\mathbb{R}\).
Let \(G\) be any open neighborhood of \(1/2\) in \(A\). There is an open set \(U\subseteq\mathbb{R}\) such that \(G=A\cap U\). Since \(1/2\in U\), some \(\varepsilon>0\) satisfies \((1/2-\varepsilon,1/2+\varepsilon)\subseteq U\). Choose \(n\in\mathbb{N}\) with \(1/n<\varepsilon\). Then \(V_n=A\cap(1/2-1/n,1/2+1/n)\subseteq A\cap U=G\). This verifies the local base condition at \(1/2\). The check is relative to \(A\): the conclusion needed is \(V_n\subseteq G\), not that \(V_n\) contains an interval of real numbers outside \(A\).
Worked Example: Local Bases in a Finite Space
Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). This topology has the finite basis \(\mathcal{B}=\{\{a\},X\}\): each listed set is open, and each nonempty open set is itself one of the basis elements.
At \(a\), the basis elements containing the point are \(\mathcal{B}_a=\{\{a\},X\}\). The open neighborhoods of \(a\) are exactly \(\{a\}\) and \(X\). For the neighborhood \(\{a\}\), the basis member \(\{a\}\) is contained in it; for the neighborhood \(X\), either basis member is contained in it. At \(b\), the only basis element containing the point is \(X\), so \(\mathcal{B}_b=\{X\}\). The only open neighborhood of \(b\) is \(X\), and it contains this local base member. Thus both points have finite local bases, as the theorem predicts.
A Common Mistake and Why the Construction Matters
Common pitfall. From a basis, it is not enough to choose one basis neighborhood of \(x\) and declare that singleton family a local base. In \(\mathbb{R}\), choose the basis element \(B=(-1,1)\) at \(0\). The proposed family \(\{B\}\) fails the local base test for the open neighborhood \(U=(-1/2,1/2)\): its only member is not contained in \(U\), because, for example, \(3/4\in B\) but \(3/4\notin U\). The correction is to retain all basis elements containing \(0\), as the theorem does, or to choose a family that keeps refining neighborhoods, such as \(\{(-1/n,1/n):n\in\mathbb{N}\}\).
The practical point is that second countability supplies a countable collection of neighborhood choices at every point, even when the space is not presented in metric terms. For subspaces, traces preserve this control: open neighborhoods become relative open sets, and the same refinement argument still works. The construction is useful whenever a question concerns local behavior but the available information is a global countable basis.
Check Your Understanding
For each question, identify the relevant basis or local base condition and check the required inclusion.
- For a countable basis \(\mathcal{B}\), explain why the family of basis elements containing a fixed point \(x\) is countable.
- In the standard topology on \(\mathbb{R}\), verify directly that \(\{(-1/n,1/n):n\in\mathbb{N}\}\) is a local base at \(0\).
- Why does taking traces \(B\cap A\) of a basis element \(B\) give an open set in the subspace \(A\)?
- For \(A=\mathbb{Q}\cap[0,1]\), write the first four members of the family \(V_n=A\cap(1/2-1/n,1/2+1/n)\).
- Give a neighborhood of \(0\) in \(\mathbb{R}\) that shows why the single set \((-1,1)\) is not a local base at \(0\).