Comparing Topologies Locally
Two topologies on the same set are compared by asking which sets they declare open. In practice, a topology may contain many open sets, while a basis records them through smaller building blocks. The basis-refinement criterion turns the global question of inclusion into a local one: around each point of a basic open set in the coarser topology, can we find a basic open set in the finer topology that stays inside it?
Throughout, all topologies being compared are on the same set \(X\). A basis \(\mathcal{B}\) for a topology \(\mathcal{T}\) is a collection of subsets of \(X\) such that every point of \(X\) belongs to some member of \(\mathcal{B}\), and whenever \(x\in B_1\cap B_2\) for \(B_1,B_2\in\mathcal{B}\), there is \(B_3\in\mathcal{B}\) with \(x\in B_3\subseteq B_1\cap B_2\). The open sets of \(\mathcal{T}\) are precisely the unions of members of \(\mathcal{B}\); the empty union is \(\varnothing\).
Suppose \(\mathcal{B}_1\) is a basis for \(\mathcal{T}_1\), and \(\mathcal{B}_2\) is a basis for \(\mathcal{T}_2\). The relevant local condition starts with a basis element from the coarser topology. Each of its points must lie in a finer basis element contained in it. The finer basis element may depend on both the point and the coarser basis element.
The Basis-Refinement Criterion
Proof. First suppose \(\mathcal{T}_2\subseteq\mathcal{T}_1\). Fix \(V\in\mathcal{B}_2\) and \(x\in V\). Since \(V\) is open in \(\mathcal{T}_2\), the inclusion makes it open in \(\mathcal{T}_1\). Because \(\mathcal{B}_1\) is a basis for \(\mathcal{T}_1\), the basis property applied to this open set and the point \(x\) gives \(U\in\mathcal{B}_1\) with \(x\in U\subseteq V\).
Conversely, suppose the stated local condition holds. Let \(O\) be any open set in \(\mathcal{T}_2\). If \(O=\varnothing\), then it is open in \(\mathcal{T}_1\). Otherwise, take any \(x\in O\). Since \(O\) is a union of \(\mathcal{B}_2\)-elements, there is \(V\in\mathcal{B}_2\) with \(x\in V\subseteq O\). The local condition supplies \(U\in\mathcal{B}_1\) with \(x\in U\subseteq V\subseteq O\). This works for every \(x\in O\), so \(O\) is the union of the \(\mathcal{B}_1\)-elements that are contained in \(O\). Therefore \(O\) is open in \(\mathcal{T}_1\), and \(\mathcal{T}_2\subseteq\mathcal{T}_1\). If \(X=\varnothing\), the condition is vacuous and both topologies consist only of \(\varnothing\), so the equivalence also holds in that case. \(\square\)
For equality of topologies, apply the criterion in both directions: each \(\mathcal{B}_2\)-element must be locally refined by \(\mathcal{B}_1\)-elements, and each \(\mathcal{B}_1\)-element must be locally refined by \(\mathcal{B}_2\)-elements. The refinement is local, not a requirement that every element of one basis contain or be contained in a whole element of the other basis.
Worked Example: The Usual and Discrete Topologies on \(\mathbb{R}\)
Let \(\mathcal{T}_{\mathrm{u}}\) be the usual topology on \(\mathbb{R}\), with basis \(\mathcal{B}_{\mathrm{u}}=\{(a,b):a,b\in\mathbb{R},\ a<b\}\). Let \(\mathcal{T}_{\mathrm{d}}\) be the discrete topology, whose open sets are all subsets of \(\mathbb{R}\), with basis \(\mathcal{B}_{\mathrm{d}}=\{\{x\}:x\in\mathbb{R}\}\). We test which topology is finer.
To test whether \(\mathcal{T}_{\mathrm{d}}\) is finer than \(\mathcal{T}_{\mathrm{u}}\), take any usual basis element \(V=(a,b)\) and any \(x\in V\), so \(a<x<b\). The discrete basis element \(U=\{x\}\) satisfies \(x\in U\subseteq V\). The criterion therefore gives \(\mathcal{T}_{\mathrm{u}}\subseteq\mathcal{T}_{\mathrm{d}}\).
For the reverse direction, use the discrete basis element \(V=\{0\}\) and its point \(x=0\). Any usual basis element \(U=(a,b)\) containing \(0\) has \(a<0<b\). Set \(\delta=\min(-a,b)/2\). Both \(-a\) and \(b\) are positive, so \(\delta>0\), and \(0<\delta<b\), while \(a<0<\delta\). Thus \(\delta\in(a,b)\) and \(\delta\ne0\); no such \(U\) is contained in \(\{0\}\). The criterion fails in this direction. Hence the discrete topology is strictly finer than the usual topology.
Check: the inclusion also follows directly because every subset of \(\mathbb{R}\) is discrete-open, while not every subset is usual-open: \(\{0\}\) is not usual-open, since every usual open interval containing \(0\) contains points other than \(0\).
Finite Bases and Direction of Refinement
Worked Example: Two Topologies on a Three-Point Set
Let \(X=\{a,b,c\}\). Consider the bases \(\mathcal{B}_1=\{X,\{a,b\},\{a\}\}\) and \(\mathcal{B}_2=\{X,\{a,b\}\}\), and let \(\mathcal{T}_1,\mathcal{T}_2\) be their generated topologies. Each collection covers \(X\). For \(\mathcal{B}_1\), the only intersection of distinct nonempty basis elements that is not already one of them is \(\{a,b\}\cap\{a\}=\{a\}\); for \(\mathcal{B}_2\), intersections of its elements are again in \(\mathcal{B}_2\). Thus both collections satisfy the basis conditions.
The unions of members of \(\mathcal{B}_1\) give \(\mathcal{T}_1=\{\varnothing,\{a\},\{a,b\},X\}\). The unions of members of \(\mathcal{B}_2\) give \(\mathcal{T}_2=\{\varnothing,\{a,b\},X\}\). To use the criterion, start with each \(\mathcal{B}_2\)-element. For \(V=X\), each \(x\in X\) can use \(U=X\in\mathcal{B}_1\). For \(V=\{a,b\}\), the point \(a\) can use \(U=\{a\}\), and the point \(b\) can use \(U=\{a,b\}\). The local condition holds, so \(\mathcal{T}_1\) is finer than \(\mathcal{T}_2\).
For the reverse test, take \(V=\{a\}\in\mathcal{B}_1\) and \(x=a\). The members of \(\mathcal{B}_2\) containing \(a\) are \(X\) and \(\{a,b\}\); neither is a subset of \(\{a\}\), since both contain \(b\). The reverse local condition fails. This agrees with the explicit lists of open sets: \(\{a\}\in\mathcal{T}_1\) but \(\{a\}\notin\mathcal{T}_2\).
Different Bases Can Generate the Same Topology
Worked Example: Short Intervals and All Open Intervals
Let \(\mathcal{B}_{\mathrm{all}}\) consist of all open intervals \((a,b)\) with real endpoints \(a<b\). Let \(\mathcal{B}_{\mathrm{short}}\) consist of those intervals with \(b-a<1\). Both are bases for the usual topology on \(\mathbb{R}\). For the second collection, every \(x\in\mathbb{R}\) belongs, for example, to \((x-\tfrac14,x+\tfrac14)\), whose length is \(\tfrac14<1\). If \(x\) lies in the intersection of two short intervals, that intersection has the form \((L,R)\) with \(L<x<R\). Choose \(\delta=\min(x-L,R-x,\tfrac14)\). Then \(\delta>0\), and \((x-\delta,x+\delta)\) contains \(x\), lies inside \((L,R)\), and has length \(2\delta\le\tfrac12<1\). This verifies the basis intersection condition.
Compare the topology generated by \(\mathcal{B}_{\mathrm{short}}\) to the usual topology generated by \(\mathcal{B}_{\mathrm{all}}\). For any \(V=(a,b)\in\mathcal{B}_{\mathrm{all}}\) and any \(x\in V\), let \(\delta=\min(x-a,b-x,\tfrac14)\). Each entry in the minimum is positive, so \(\delta>0\); the interval \(U=(x-\delta,x+\delta)\) contains \(x\), is contained in \(V\), and has length \(2\delta\le\tfrac12<1\). Thus \(U\in\mathcal{B}_{\mathrm{short}}\), giving refinement in this direction. In the other direction, every short interval is already an element of \(\mathcal{B}_{\mathrm{all}}\); for any such \(V\) and any \(x\in V\), take \(U=V\). Refinement holds both ways, so the generated topologies are equal.
For a specific check, take \(V=(-2,3)\) and \(x=0\). The interval \((-\tfrac14,\tfrac14)\) contains \(0\), lies in \((-2,3)\), and has length \(\tfrac14-(-\tfrac14)=\tfrac12<1\). It is a valid short basis element refining \(V\) at \(0\).
A Common Pitfall: Comparing Basis Elements Globally
A mistaken test is to ask whether every element of the coarser basis is itself an element of the finer basis, or whether each entire coarser basis element is contained in one finer basis element. Neither requirement appears in the criterion. The criterion asks for a finer basis element around each point of each coarser basis element; different points may use different refinements.
Wrong calculation: in the short-interval example, \((-2,3)\) has length \(3-(-2)=5\), so \((-2,3)\notin\mathcal{B}_{\mathrm{short}}\). Concluding from this that the short-interval topology is not finer than the usual topology is an invalid global comparison. Correction: for each \(x\in(-2,3)\), choose \(\delta=\min(x+2,3-x,\tfrac14)\). This is positive, and \((x-\delta,x+\delta)\) is a short basis element containing \(x\) and contained in \((-2,3)\). The criterion tests exactly this pointwise refinement, not membership of the large interval in the finer basis.
The direction of the test is equally important: to prove that \(\mathcal{T}_1\) is finer, begin with basis elements of the proposed coarser topology \(\mathcal{T}_2\), then find refining elements from the proposed finer basis \(\mathcal{B}_1\). Reversing those roles tests the reverse inclusion instead.
Check Your Understanding
For each question, identify which basis is being used for the proposed finer topology and check the containment at the specified point.
- On \(\mathbb{R}\), compare the usual topology with the topology generated by all open intervals of length less than \(2\). What local interval can refine \((-3,4)\) at \(x=1\)?
- On \(X=\{p,q\}\), compare the topology generated by \(\{X,\{p\}\}\) with the indiscrete topology, whose only open sets are \(\varnothing\) and \(X\). Which is finer?
- State the basis-refinement condition needed to prove \(\mathcal{T}_1\subseteq\mathcal{T}_2\), using bases \(\mathcal{B}_1\) and \(\mathcal{B}_2\).
- Give a reason that inclusion of basis collections, by itself, is not the criterion for comparing the topologies they generate.