Checking a Topology on a Finite Set

A Finite Checklist for Candidate Topologies

A collection of subsets is not a topology merely because its members look orderly or because it contains the whole set. It must satisfy the topology axioms, including closure under unions of arbitrarily many of its members. On a finite underlying set, the candidate collection is itself finite, so there is a practical way to check that union condition using only binary unions. We will establish that test, then apply the axioms in examples where the first failure is different and in one example where every condition holds.

Let \(X\) be a finite set and let \(\mathcal{T}\subseteq\mathcal{P}(X)\) be a candidate collection. Its members are subsets of \(X\); they are not automatically open sets unless \(\mathcal{T}\) satisfies the topology axioms.

The two membership checks in (1) and (2) are separate: the empty set and \(X\) must both be present. The union condition applies to families of every size, not just pairs. The intersection condition applies to finite families; it does not require closure under arbitrary intersections.

Why Binary Checks Are Enough on a Finite Set

Proof. If \(\mathcal{T}\) is a topology, it contains \(\varnothing\) and \(X\) by conditions (1) and (2). Binary unions are unions of two members, so condition (3) gives closure under binary unions. Binary intersections are finite intersections, so condition (4) gives closure under binary intersections.

Conversely, suppose \(\varnothing,X\in\mathcal{T}\) and \(\mathcal{T}\) is closed under binary unions and binary intersections. Consider any family \((U_i)_{i\in J}\) of members of \(\mathcal{T}\), where the index set \(J\) may be infinite. Although the family may repeat members, it has only finitely many distinct values because \(\mathcal{T}\subseteq\mathcal{P}(X)\) and \(X\) is finite. If there are no distinct values, \(J\) is empty and the union is \(\varnothing\in\mathcal{T}\). Otherwise write the distinct values as \(V_1,\ldots,V_m\). Their union equals the union of the original family, and repeated use of binary union closure shows \(V_1\cup\cdots\cup V_m\in\mathcal{T}\).

Now take a finite family of members of \(\mathcal{T}\). If the family is empty, its intersection is \(X\in\mathcal{T}\). If it has \(n\geq 1\) members, repeated use of binary intersection closure shows that its intersection belongs to \(\mathcal{T}\). Thus all four topology conditions hold. \(\square\)

The finiteness used here is finiteness of the candidate collection, which follows from finiteness of \(X\): there are only finitely many subsets of \(X\). The theorem gives a convenient check: after confirming \(\varnothing\) and \(X\), test every pair of members for both their union and their intersection.

Finding the First Failed Condition

A useful testing habit is to examine the conditions in order and stop at the first failure. Once a condition fails, the collection is not a topology, even if it satisfies all the other conditions. The next examples isolate failures of the required elements, union closure, and intersection closure.

A Shortcut for Collections Ordered by Inclusion

Sometimes the members of a candidate collection are linearly ordered by inclusion: for any two members \(U,V\), either \(U\subseteq V\) or \(V\subseteq U\). On a finite underlying set this makes the pair checks especially quick, because for comparable sets the union is the larger set and the intersection is the smaller set.

Proof. Take any two members \(U,V\in\mathcal{T}\). If \(U\subseteq V\), then \(U\cup V=V\in\mathcal{T}\) and \(U\cap V=U\in\mathcal{T}\). If \(V\subseteq U\), then \(U\cup V=U\in\mathcal{T}\) and \(U\cap V=V\in\mathcal{T}\). Thus \(\mathcal{T}\) is closed under binary unions and intersections. It contains \(\varnothing\) and \(X\), so the finite-set test proves it is a topology. \(\square\)

The hypothesis that \(X\) is finite matters. Here is a concrete reason not to apply the proposition to an infinite set. Let \(X=\{0,1,2,\ldots\}\cup\{z\}\) and let \(\mathcal{C}=\{\varnothing,X\}\cup\{\{0,1,\ldots,n\}:n=0,1,2,\ldots\}\). These sets are linearly ordered by inclusion, and \(\mathcal{C}\) contains \(\varnothing\) and \(X\). But the union of the family of all the finite initial segments is \(\{0,1,2,\ldots\}\), which is not in \(\mathcal{C}\): it is neither \(\varnothing\), nor \(X\), nor a finite initial segment. So \(\mathcal{C}\) is not a topology.

Common pitfall. A tempting but incorrect argument is: “The sets are linearly ordered, so the union of any family is its largest member.” This assumes a largest member exists. The infinite family of initial segments above has no largest member, and its union is missing from the collection. On a finite \(X\), any family of members has only finitely many distinct values; if it is nonempty and linearly ordered, those values have a largest member. That finite-set fact is what makes the shortcut valid here.

A Practical Test to Keep

For a candidate collection on a finite set, the finite-set test turns the definition into a finite procedure. Confirm membership of \(\varnothing\) and \(X\), then check each pair \(U,V\in\mathcal{T}\) for both \(U\cup V\in\mathcal{T}\) and \(U\cap V\in\mathcal{T}\). The theorem proves that these checks cover families of every size where required. If a check fails, record the particular pair and the resulting set; that identifies the failed axiom rather than relying on an impression that the collection is “almost” a topology.