Open Sets Can Be Described by Smaller Ones
A topology may contain many open sets, and listing all of them is often inconvenient. A basis provides a more economical description: it is a collection of open sets such that every point in every open set lies in a basis element contained within that open set. The basis elements act as a supply of small open neighborhoods. The topology is still the collection of all its open sets; the basis is a way to describe those sets, not usually a topology in its own right.
This idea uses the local characterization of open sets established in Open Sets in Topological Spaces: a set is open if and only if each of its points lies in an open set contained in it. A basis makes that test more specific by requiring the smaller open neighborhoods to come from a chosen collection.
The condition includes the requirement that every basis element is open, since \(\mathcal{B}\subseteq\mathcal{T}\). It also ensures that the basis elements cover \(X\): apply it to the open set \(X\), so that each \(x\in X\) belongs to some \(B\in\mathcal{B}\). If \(X\) is empty, this covering condition is vacuous, and the empty collection is a basis for the unique topology on \(X\).
Every Open Set Is a Union of Basis Elements
The neighborhood condition has an equivalent set-theoretic form. An open set is exactly a union of basis elements. In one direction, the basis condition supplies enough basis elements to cover each open set. In the other, any union of basis elements is open because each basis element is open and a topology is closed under arbitrary unions.
Proof. Suppose \(\mathcal{B}\) is a basis, and let \(U\in\mathcal{T}\). Consider all basis elements contained in \(U\). Their union is contained in \(U\), and every point \(x\in U\) belongs to one of them by the basis condition. Therefore $$ U=\bigcup\{B\in\mathcal{B}:B\subseteq U\}. $$ If \(U\) is empty, the displayed union is empty, so the equality still holds.
Conversely, suppose every open set is a union of basis elements. Let \(U\in\mathcal{T}\) and \(x\in U\). Write \(U\) as a union of members of \(\mathcal{B}\). Since \(x\) belongs to that union, it belongs to at least one of its members, say \(B\). This member satisfies \(x\in B\subseteq U\), as required. Thus \(\mathcal{B}\) is a basis. \(\square\)
This characterization explains why a basis can be much smaller than the topology it describes. A single open set may be a union of many basis elements, and many different collections can represent the same topology. It also clarifies what the basis condition does not say: it does not require every open set to be a basis element.
Worked Example: Rational Intervals Form a Basis for the Usual Topology
In the usual topology on \(\mathbb{R}\), let $$ \mathcal{B}=\{(p,q):p,q\in\mathbb{Q}\text{ and }p<q\}. $$ Each member is open in the usual topology. To check the basis condition, let \(U\) be open and let \(x\in U\). By openness, there is an \(\varepsilon>0\) such that $$ (x-\varepsilon,x+\varepsilon)\subseteq U. $$ By the density of the rational numbers in \(\mathbb{R}\), choose \(p,q\in\mathbb{Q}\) with $$ x-\varepsilon<p<x<q<x+\varepsilon. $$ Then \(x\in(p,q)\subseteq U\). Hence \(\mathcal{B}\) is a basis for the usual topology.
For instance, if \(U=(0,2)\) and \(x=1\), then \((1/2,3/2)\) is a basis element containing \(x\) and contained in \(U\). The point \(1\) is not special: the same refinement argument works for every point of every usual open set. The rational endpoints make the basis countable, even though the usual topology has many more open sets.
Worked Example: A Basis Need Not Be a Topology
Let \(\mathcal{B}\) be the rational-interval basis for the usual topology on \(\mathbb{R}\). Its members are open intervals with rational endpoints. The set \(\mathbb{R}\) is open, and the basis characterization gives $$ \mathbb{R}=\bigcup_{B\in\mathcal{B}}B. $$ But \(\mathbb{R}\) is not itself a member of \(\mathcal{B}\), since it is not a bounded interval with rational endpoints. Thus \(\mathcal{B}\) does not contain every open set, and in particular it does not contain \(\mathbb{R}\). It is a basis for a topology without being a topology itself.
There is no conflict with the topology axioms. Those axioms apply to the full collection \(\mathcal{T}\), which includes arbitrary unions of basis elements. They do not require the smaller collection \(\mathcal{B}\) to include all such unions. The theorem that constructs a topology from a collection satisfying basis conditions, established earlier in Open Sets in Topological Spaces, concerns that full collection of unions rather than the basis collection alone.
Comparing Topologies Through Their Bases
Suppose two topologies on the same set have different bases. To determine whether one topology is finer than the other, it is enough to ask whether the basis elements of the coarser topology can be locally refined by basis elements of the finer one. Here, \(\mathcal{T}_1\subseteq\mathcal{T}_2\) means that every \(\mathcal{T}_1\)-open set is also \(\mathcal{T}_2\)-open.
Proof. First suppose \(\mathcal{T}_1\subseteq\mathcal{T}_2\). If \(B\in\mathcal{B}_1\), then \(B\in\mathcal{T}_1\), so \(B\in\mathcal{T}_2\). For any \(x\in B\), apply the basis condition for \(\mathcal{B}_2\) to the \(\mathcal{T}_2\)-open set \(B\). It gives \(C\in\mathcal{B}_2\) with \(x\in C\subseteq B\).
Conversely, suppose the stated refinement condition holds. Let \(U\in\mathcal{T}_1\), and take any \(x\in U\). The basis condition for \(\mathcal{B}_1\) gives \(B\in\mathcal{B}_1\) such that \(x\in B\subseteq U\). The assumed condition then gives \(C\in\mathcal{B}_2\) with \(x\in C\subseteq B\subseteq U\). Thus every point of \(U\) lies in a \(\mathcal{T}_2\)-open set contained in \(U\). The local characterization of open sets shows that \(U\in\mathcal{T}_2\). Therefore \(\mathcal{T}_1\subseteq\mathcal{T}_2\). If \(U\) is empty, it is already open in \(\mathcal{T}_2\), so that case is included as well. \(\square\)
The refinement must be checked at each point. It is not enough to find one \(\mathcal{B}_2\)-element inside each \(\mathcal{B}_1\)-element: different points of the larger basis element may require different smaller neighborhoods. When the two topologies are equal, the condition works in both directions. This gives a useful way to compare topologies without listing all their open sets.
Worked Example: Comparing the Usual and Discrete Topologies
On \(\mathbb{R}\), the usual topology has as a basis all open intervals, while the discrete topology has as a basis all singletons \(\{x\}\). Every usual-open interval \(I\) and every \(x\in I\) satisfy $$ x\in\{x\}\subseteq I. $$ The comparison theorem therefore gives \(\mathcal{T}_{\mathrm{usual}}\subseteq\mathcal{T}_{\mathrm{dis}}\). In the reverse direction, take the discrete basis element \(\{0\}\). No usual-open set containing \(0\) is contained in \(\{0\}\): every usual-open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\), and that interval contains nonzero points. The refinement criterion fails, so the discrete topology is not contained in the usual topology.
This comparison uses a pointwise test rather than an attempted comparison of all open subsets of \(\mathbb{R}\). It also reflects the intuitive distinction between the topologies: the discrete topology can isolate each point, whereas the usual topology cannot.
Restricting a Basis to a Subspace
A basis also adapts naturally when the space is replaced by a subspace. If \(A\subseteq X\), the sets open in \(A\) have the form \(A\cap O\), where \(O\) is open in \(X\), by the subspace characterization in Open Sets in Topological Spaces. Intersecting ambient basis elements with \(A\) therefore gives a basis for the subspace topology.
Proof. Every set \(A\cap B\), with \(B\in\mathcal{B}\), is open in \(A\) by the definition of the subspace topology. Let \(V\) be open in \(A\), and take \(x\in V\). There is an open set \(O\in\mathcal{T}\) such that \(V=A\cap O\). Since \(x\in O\) and \(\mathcal{B}\) is a basis, some \(B\in\mathcal{B}\) satisfies \(x\in B\subseteq O\). Consequently, $$ x\in A\cap B\subseteq A\cap O=V. $$ Thus every point of every subspace-open set lies in a member of \(\mathcal{B}_A\) contained in that set. This is the basis condition, so \(\mathcal{B}_A\) is a basis for \(A\). \(\square\)
Worked Example: A Basis for a Half-Open Subspace
Give \(\mathbb{R}\) its usual topology and let \(A=[0,2)\). The rational open intervals form a basis for \(\mathbb{R}\), so the theorem says that sets of the form $$ [0,2)\cap(p,q),\qquad p,q\in\mathbb{Q},\quad p<q, $$ form a basis for \(A\). For example, taking \(p=-1\) and \(q=1\) gives $$ [0,2)\cap(-1,1)=[0,1). $$ This set is open in the subspace \(A\), even though it is not open in \(\mathbb{R}\).
At the endpoint \(0\), a basis neighborhood in \(A\) can be one-sided in appearance: for example, \([0,1)\) contains \(0\) and is contained in \(A\). If \(V\) is any subspace-open neighborhood of \(0\), then \(V=A\cap O\) for some usual-open \(O\) containing \(0\). A sufficiently small rational-endpoint interval \((p,q)\) with \(p<0<q\) lies inside \(O\), and its intersection with \(A\) contains \(0\) and lies in \(V\). The intersection construction therefore handles boundary points without requiring a separate kind of ambient basis element.
Using Bases Carefully
A basis is most useful when it makes local checks simpler. Instead of testing a property against every open set, one can often work with the smaller neighborhoods supplied by the basis. For example, openness can be checked by finding a basis element around each point, and topology inclusion can be checked by the refinement theorem. The subspace result similarly lets one reuse a known basis rather than reconstruct the subspace topology from scratch.
Keep three distinctions clear. First, a basis element is open, but an open set need not itself be a basis element. Second, the basis describes a topology by unions; it is not generally the topology. Third, a basis for a subspace consists of intersections with the subspace, not merely the ambient basis elements that happen to lie entirely inside it. At a boundary point, intersecting an ambient neighborhood with the subspace can produce the needed relative neighborhood even when that intersection is not open in the ambient space.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- State the pointwise condition that defines a basis for a topology.
- How can an open set be expressed as a union of basis elements?
- What pointwise refinement condition characterizes inclusion of two topologies when bases for both are known?
- How is a basis for a subspace \(A\) obtained from a basis for the ambient space \(X\)?
- Why does a basis for a topology generally fail to be a topology itself?