Why Subspaces Keep These Countability Properties
A subspace may have a different collection of open sets from its ambient space: its open sets are obtained by intersecting ambient open sets with the subspace. This makes it natural to ask whether countability properties based on open sets survive when we pass to a subspace. For first and second countability, the answer is yes. The construction in each case is an intersection, but the kind of basis being counted is different.
Let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\). The subspace topology on \(A\) is \(\mathcal{T}_A=\{U\cap A:U\in\mathcal{T}\}\). All statements that \(A\) is first-countable or second-countable refer to this topology.
The distinction matters: first countability asks for a separate countable collection near each point, while second countability asks for one countable collection that works across the entire space. A second-countable space is first-countable, since the basis members containing any fixed point form a local base; the inheritance arguments below, however, work directly with each definition.
First Countability Passes to Subspaces
Proof. If \(A=\varnothing\), there are no points at which a local base must be supplied, so the assertion holds. Suppose \(A\ne\varnothing\), and fix \(a\in A\). Since \(X\) is first-countable, choose a finite or countable local base \(\mathcal{U}_a\) at \(a\) in \(X\). Consider \(\mathcal{U}_{a,A}=\{U\cap A:U\in\mathcal{U}_a\}\). Each member is open in \(A\) and contains \(a\). Let \(O\) be any open neighborhood of \(a\) in \(A\). By the definition of the subspace topology, \(O=V\cap A\) for some open set \(V\) in \(X\). Since \(a\in V\), the local base property gives \(U\in\mathcal{U}_a\) with \(a\in U\subseteq V\). Therefore \(a\in U\cap A\subseteq V\cap A=O\). The collection \(\mathcal{U}_{a,A}\) is a local base at \(a\); it is finite or countable because it is the image of a finite or countable collection under \(U\mapsto U\cap A\). This holds for every \(a\in A\), so \(A\) is first-countable. \(\square\)
The proof is local: it starts with a local base at the particular point \(a\). Intersections that happen to be empty can be discarded; in fact, none of these intersections is empty because each \(U\) contains \(a\in A\).
Worked Example: A Local Base on a Subinterval
Take \(X=\mathbb{R}\) with its standard topology and \(A=(0,1)\) with the subspace topology. Fix \(a\in A\). For each \(n\in\mathbb{N}\), let \(U_n=(a-1/n,a+1/n)\cap A\). Each \(U_n\) is an open neighborhood of \(a\) in \(A\). To check the local base condition, let \(O\) be any open neighborhood of \(a\) in \(A\). Write \(O=V\cap A\) for an open \(V\subseteq\mathbb{R}\) containing \(a\). There is \(\varepsilon>0\) such that \((a-\varepsilon,a+\varepsilon)\subseteq V\). Choose \(n\in\mathbb{N}\) with \(1/n<\varepsilon\). Then \(U_n\subseteq V\cap A=O\). Thus \((U_n)_{n\in\mathbb{N}}\) is a countable local base at \(a\). The same construction works for every \(a\in(0,1)\), so this subspace is first-countable.
Check: the set \(U_n\) includes only points \(x\in(0,1)\) satisfying \(a-1/n<x<a+1/n\). In particular, the endpoints of the interval are not included, as required by the open-ball neighborhoods used in the construction.
Second Countability Passes to Subspaces
Proof. If \(A=\varnothing\), the empty collection is a basis for its topology, so the claim holds. Otherwise let \(\mathcal{B}\) be a finite or countable basis for \(X\), and set \(\mathcal{B}_A=\{B\cap A:B\in\mathcal{B}\}\). Every member of \(\mathcal{B}_A\) is open in \(A\). To verify the basis condition, take an open set \(O\) in \(A\) and a point \(a\in O\). Write \(O=V\cap A\) with \(V\) open in \(X\). The basis property of \(\mathcal{B}\) gives \(B\in\mathcal{B}\) such that \(a\in B\subseteq V\). Hence \(a\in B\cap A\subseteq V\cap A=O\). Thus \(\mathcal{B}_A\) is a basis for \(A\). It is finite or countable because it is the image of \(\mathcal{B}\) under \(B\mapsto B\cap A\); repeated sets do not increase its size. Therefore \(A\) is second-countable. \(\square\)
Unlike the first-countability argument, this construction uses the same family of intersections for every point of \(A\). It also directly explains why one should intersect basis members with \(A\), rather than require the original basis members to lie inside \(A\).
Worked Example: An Inherited Basis on the Rationals
Let \(X=\mathbb{R}\) with the standard topology and \(A=\mathbb{Q}\) with the subspace topology. The intervals \((p,q)\) with rational endpoints \(p<q\) form a countable basis for \(\mathbb{R}\): there are countably many rational pairs, and between any two distinct real numbers there are rational numbers. Theorem 2 gives the basis \(\{(p,q)\cap\mathbb{Q}:p,q\in\mathbb{Q},\ p<q\}\) for \(A\).
For a direct check, let \(O\) be relatively open in \(\mathbb{Q}\), and choose \(a\in O\). There is an open \(V\subseteq\mathbb{R}\) with \(O=V\cap\mathbb{Q}\). Because \(a\in V\), some interval around \(a\) lies in \(V\). Choose rational numbers \(p<a<q\) close enough to \(a\) that \((p,q)\subseteq V\). Then \(a\in(p,q)\cap\mathbb{Q}\subseteq O\). This verifies the basis condition at every point of every relatively open set. Hence \(\mathbb{Q}\) is second-countable.
Check: this family is countable because it is indexed by pairs from the countable set \(\mathbb{Q}\). Intersecting with \(\mathbb{Q}\) cannot create more sets than there are indexing pairs.
Worked Example: A Convergent Sequence with Its Limit
Let \(X=\mathbb{R}\) with the standard topology and \(A=\{0\}\cup\{1/k:k\in\mathbb{N}\}\). At \(0\), take \(U_n=(-1/n,1/n)\cap A\) for \(n\in\mathbb{N}\). Since \(1/k\) is positive, it belongs to \(U_n\) exactly when \(1/k<1/n\), which is equivalent to \(k>n\). Thus \(U_n=\{0\}\cup\{1/k:k>n\}\). If \(O\) is relatively open and contains \(0\), some interval \((-\varepsilon,\varepsilon)\) satisfies \((-\varepsilon,\varepsilon)\cap A\subseteq O\). Choose \(n\) with \(1/n<\varepsilon\); then \(U_n\subseteq O\). So the \(U_n\) form a countable local base at \(0\).
Every other point of \(A\) is isolated in \(A\). For \(m=1\), the interval \((1-1/4,1+1/4)\) intersects \(A\) only in \(\{1\}\), since the next point is \(1/2\), at distance \(1/2\), and all remaining points lie between \(0\) and \(1/2\). For \(m\ge2\), the distance from \(1/m\) to the next smaller point \(1/(m+1)\) is \(1/[m(m+1)]\). The distance to \(1/(m-1)\) is \(1/[m(m-1)]\), which is larger, and the distance to \(0\) is \(1/m\), also larger than \(1/[m(m+1)]\). Therefore the interval centered at \(1/m\) with radius \(1/[2m(m+1)]\) intersects \(A\) only in \(\{1/m\}\). The singleton \(\{1/m\}\) is consequently an open neighborhood in \(A\), and it alone is a local base there. Together these local bases verify first countability at every point of \(A\).
Check: for \(m\ge2\), the chosen radius is half the distance \(1/[m(m+1)]\) to \(1/(m+1)\), so the interval excludes that point and every other point of \(A\). At \(m=1\), the chosen radius \(1/4\) is strictly less than the distance \(1/2\) to \(1/2\).
Worked Example: The Integers as a Discrete Subspace
Take \(X=\mathbb{R}\) with the standard topology and \(A=\mathbb{Z}\). For each \(k\in\mathbb{Z}\), the interval \((k-1/2,k+1/2)\) is open in \(\mathbb{R}\). If \(j\in\mathbb{Z}\) and \(j\ne k\), then \(|j-k|\ge1>1/2\), so \(j\) is not in this interval. Its intersection with \(\mathbb{Z}\) is therefore exactly \(\{k\}\). The singleton sets \(\{\{k\}:k\in\mathbb{Z}\}\) form a basis for the subspace: their union covers \(\mathbb{Z}\), and every subset of \(\mathbb{Z}\) is a union of these singletons, hence open in the subspace topology. Since \(\mathbb{Z}\) is countable, this is a countable basis, so \(\mathbb{Z}\) is second-countable (and also first-countable).
Check: the strict inequality \(|j-k|\ge1>1/2\) applies to every distinct integer \(j\), while \(k\) itself has distance \(0<1/2\). Thus the intersection contains \(k\) and excludes every other integer.
A Common Basis Mistake
Common pitfall. One might try to form a subspace basis by keeping only ambient basis members that are contained in the subspace. For \(X=\mathbb{R}\) with its standard topology and \(A=\{0\}\), this goes wrong: no nonempty open interval of \(\mathbb{R}\) is contained in \(\{0\}\), so this rule produces no basis member containing \(0\). But \(\{0\}\) is open in the subspace topology, since \((-1,1)\cap A=\{0\}\). The correct construction intersects ambient basis members with \(A\). For example, \((-1,1)\cap A=\{0\}\), exactly the neighborhood needed.
The two inheritance results are useful because they let us work inside a subspace without rebuilding countability from scratch. Their hypotheses concern the ambient space, and their conclusions concern the subspace topology; neither proof requires \(A\) to be open or closed. Keep the constructions distinct: to get a local base at \(a\), intersect the members of a local base at \(a\) with \(A\); to get a basis for all of \(A\), intersect every member of one countable ambient basis with \(A\).
Check Your Understanding
Use the definitions and intersection constructions to answer the following questions.
- Let \(X\) be first-countable and \(A\subseteq X\). Describe a countable local base at \(a\in A\) using a local base in \(X\).
- Explain why the family \(\{B\cap A:B\in\mathcal{B}\}\) is a basis for \(A\) when \(\mathcal{B}\) is a basis for \(X\).
- In \(\mathbb{R}\) with the standard topology, give a local base at \(0\) for the subspace \(A=\{0\}\cup\{1/k:k\in\mathbb{N}\}\).
- Why does selecting only ambient basis elements contained in \(A=\{0\}\) fail, and what intersection supplies a basis member containing \(0\)?
- Which part of each inheritance proof needs modification when \(A=\varnothing\)?