From a Closure Rule to a Topology
A topology can be described by its open sets, by its closed sets, or by the closure operation it assigns to every subset. The last description is useful only if the closure rule obeys the right axioms. We will show that those axioms are enough: the subsets left unchanged by the rule are precisely the closed sets of a unique topology.
Let \(X\) be a set, which may be empty, and let \(c:\mathcal{P}(X)\to\mathcal{P}(X)\) be a map. A fixed set of \(c\) is a subset \(C\subseteq X\) such that \(c(C)=C\). The construction below starts with this map alone; no topology on \(X\) is assumed.
- \(c(\varnothing)=\varnothing\);
- \(A\subseteq c(A)\);
- \(c(c(A))=c(A)\);
- \(c(A\cup B)=c(A)\cup c(B)\).
One useful consequence is monotonicity: enlarging the input cannot make its image smaller. This is not an additional axiom; it follows from the four conditions.
Proof. Since \(A\subseteq B\), we have \(A\cup B=B\). Preservation of finite unions gives \(c(B)=c(A\cup B)=c(A)\cup c(B)\). Every element of \(c(A)\) therefore belongs to \(c(B)\cup c(A)=c(B)\), proving \(c(A)\subseteq c(B)\). \(\square\)
The Fixed Sets Produce the Closed Sets
Let \(\mathcal{F}=\{C\subseteq X:c(C)=C\}\), the family of fixed sets. To obtain a topology, we first verify the closed-set conditions: \(\varnothing\) and \(X\) must belong to \(\mathcal{F}\), arbitrary intersections of members must belong to \(\mathcal{F}\), and finite unions of members must belong to \(\mathcal{F}\).
Proof. We first show that \(\mathcal{F}\) satisfies the closed-set conditions. The axiom \(c(\varnothing)=\varnothing\) says that \(\varnothing\in\mathcal{F}\). Extensivity gives \(X\subseteq c(X)\), and \(c(X)\subseteq X\) because \(c(X)\) is a subset of \(X\). Thus \(c(X)=X\), so \(X\in\mathcal{F}\). This also holds when \(X=\varnothing\).
Let \((C_i)_{i\in I}\) be any family in \(\mathcal{F}\). If \(I\) is nonempty, put \(C=\bigcap_{i\in I}C_i\). Extensivity gives \(C\subseteq c(C)\). For every \(i\in I\), \(C\subseteq C_i\), so monotonicity gives \(c(C)\subseteq c(C_i)=C_i\). Therefore \(c(C)\subseteq\bigcap_{i\in I}C_i=C\), and hence \(c(C)=C\). If \(I\) is empty, the intersection is \(X\), which is already fixed. Thus \(\mathcal{F}\) is closed under arbitrary intersections.
If \(C,D\in\mathcal{F}\), then \(c(C\cup D)=c(C)\cup c(D)=C\cup D\), so \(C\cup D\in\mathcal{F}\). The empty union is \(\varnothing\), also in \(\mathcal{F}\); consequently \(\mathcal{F}\) is closed under finite unions.
Now take complements. Since \(\varnothing,X\in\mathcal{F}\), the sets \(X,\varnothing\) belong to \(\mathcal{T}_c\). If \(U_j=X\setminus C_j\) are in \(\mathcal{T}_c\), then \[ \bigcup_{j\in J}U_j=X\setminus\bigcap_{j\in J}C_j. \] The intersection on the right belongs to \(\mathcal{F}\), so this union belongs to \(\mathcal{T}_c\). For two members \(U=X\setminus C\) and \(V=X\setminus D\), \[ U\cap V=X\setminus(C\cup D). \] Since \(C\cup D\in\mathcal{F}\), the intersection belongs to \(\mathcal{T}_c\). These are the topology axioms, so \(\mathcal{T}_c\) is a topology, and its closed sets are exactly the members of \(\mathcal{F}\).
It remains to identify its closure operation. Idempotence says \(c(A)\) is fixed, and extensivity says \(A\subseteq c(A)\). If \(C\) is any fixed set containing \(A\), monotonicity gives \(c(A)\subseteq c(C)=C\). Thus \(c(A)\) is the smallest closed set in \(\mathcal{T}_c\) containing \(A\), which is the definition of closure. Finally, any topology with precisely \(\mathcal{F}\) as its closed sets must have as its open sets precisely their complements. Its open sets therefore equal \(\mathcal{T}_c\), proving uniqueness. \(\square\)
Worked Reconstructions
Worked Example: Closing Up Equivalence Classes
Let \(X=\{a,b,c,d\}\), partitioned into \(P=\{a,b\}\) and \(Q=\{c,d\}\). For \(A\subseteq X\), define \(c(A)\) to be the union of every block \(P\) or \(Q\) that intersects \(A\). For example, \(c(\{a,c\})=P\cup Q=X\), because the input meets both blocks.
We verify the axioms. The empty set meets neither block, so \(c(\varnothing)=\varnothing\). Every element of \(A\) lies in a block met by \(A\), so \(A\subseteq c(A)\). Once all blocks met by \(A\) have been included, applying \(c\) again adds no block; hence \(c(c(A))=c(A)\). Finally, a block meets \(A\cup B\) exactly when it meets \(A\) or meets \(B\). Thus \(c(A\cup B)=c(A)\cup c(B)\).
For a direct computation, \(A=\{a,c\}\) meets \(P\) at \(a\) and \(Q\) at \(c\), so \(c(A)=\{a,b\}\cup\{c,d\}=X\). For \(B=\{b\}\), only \(P\) is met, so \(c(B)=\{a,b\}\).
The fixed sets are \(\varnothing\), \(P\), \(Q\), and \(X\): a fixed set must contain either all or none of each block. Their complements are, respectively, \(X\), \(Q\), \(P\), and \(\varnothing\). Therefore the recovered topology is \(\{\varnothing,P,Q,X\}\). Check: \(P\) and \(Q\) are disjoint, and \(P\cup Q=X\), so their intersection is \(\varnothing\) and their union is \(X\); all unions and finite intersections of the listed sets remain in the list.
Worked Example: A One-Way Closure Rule
Let \(X=\{1,2,3\}\), and define \[ c(A)= \begin{cases} A\cup\{2\},&1\in A,\\ A,&1\notin A. \end{cases} \] The rule adds \(2\) whenever the input contains \(1\), and makes no other additions.
The empty set stays empty, and every \(A\) is contained in \(c(A)\). Applying the rule twice adds no further elements: if \(1\in A\), the first application adds \(2\), but does not change whether \(1\) is present; if \(1\notin A\), neither application adds anything. Hence \(c(c(A))=c(A)\).
For the union axiom, \(1\in A\cup B\) holds exactly when \(1\in A\) or \(1\in B\). If neither contains \(1\), both sides of \(c(A\cup B)=c(A)\cup c(B)\) equal \(A\cup B\). If \(1\in A\), both sides include \(2\) as well as \(A\cup B\); the same conclusion holds if \(1\in B\). This verifies the final axiom in every case.
For \(A=\{1,3\}\), the rule gives \(c(A)=\{1,2,3\}=X\). For \(B=\{3\}\), it gives \(c(B)=\{3\}\). A fixed set is exactly a subset that contains \(2\) whenever it contains \(1\), so the fixed sets are \(\varnothing,\{2\},\{3\},\{2,3\},\{1,2\},X\). Taking complements gives the topology \[ \{\varnothing,\{1\},\{3\},\{1,2\},\{1,3\},X\}. \] Check: for the input \(\{1\}\), the smallest fixed set containing it is \(\{1,2\}\): the fixed supersets are \(\{1,2\}\) and \(X\), whose intersection is \(\{1,2\}\). This agrees with \(c(\{1\})=\{1,2\}\).
Worked Example: The Indiscrete Topology
Let \(X=\{p,q,r\}\) and set \(c(A)=\varnothing\) when \(A=\varnothing\), and \(c(A)=X\) when \(A\ne\varnothing\). The empty-set axiom holds by definition. Extensivity holds because the empty set is contained in itself and every nonempty \(A\) is contained in \(X\). Applying \(c\) twice leaves either \(\varnothing\) or \(X\) unchanged, so idempotence holds.
To check finite unions, if both \(A\) and \(B\) are empty, both sides of \(c(A\cup B)=c(A)\cup c(B)\) are empty. If at least one is nonempty, \(A\cup B\ne\varnothing\), so the left side is \(X\), and at least one of \(c(A),c(B)\) is \(X\), so the right side is also \(X\).
Only \(\varnothing\) and \(X\) are fixed: every nonempty proper subset maps to \(X\), not to itself. Their complements are \(X\) and \(\varnothing\), so the recovered topology is \(\{\varnothing,X\}\), the indiscrete topology. For instance, \(c(\{p\})=X\); the only fixed superset of \(\{p\}\) is \(X\), confirming the closure calculation.
Common Pitfall: Fixed Sets Are Closed, Not Necessarily Open
The construction takes complements of fixed sets to obtain open sets. It is generally wrong to declare the fixed sets themselves to be the open sets: the closed-set axioms require arbitrary intersections and finite unions, whereas open sets require arbitrary unions and finite intersections.
For a concrete failure, use the usual closure operator on \(\mathbb{R}\), with its standard topology. For each \(n\in\mathbb{N}\), let \(C_n=[1/n,1]\). Each \(C_n\) is closed and hence fixed. Their union is \((0,1]\): every \(x\) with \(0<x\leq1\) belongs to some \(C_n\), while \(0\) belongs to none. This union is not closed, since every open interval about \(0\) contains positive points of \((0,1]\), but \(0\notin(0,1]\). In fact, \(c(\bigcup_{n\in\mathbb{N}}C_n)=[0,1]\ne(0,1]\). Thus fixed sets need not be closed under arbitrary unions, and the complements—not the fixed sets—are the topology’s open sets.
The reconstruction rests on two distinct steps: the axioms make the fixed sets a valid family of closed sets, and taking their complements gives the topology. The smallest-fixed-superset argument then ensures that the original map is exactly closure in that topology, while taking complements also forces uniqueness.
Check Your Understanding
Use the four axioms and the fixed-set construction to answer the following.
- Which step in the theorem’s proof uses monotonicity, and how is monotonicity derived from the axioms?
- For \(X=\{u,v\}\), define \(c(A)=A\cup\{v\}\) if \(u\in A\), and \(c(A)=A\) otherwise. Find all fixed sets and the recovered topology.
- Why must \(X\) be a fixed set even when the axioms do not list \(c(X)=X\) separately?
- Explain why the intersection of an empty family of fixed sets causes no exception in the proof.
- If two topologies have the same fixed sets for their closure operators, why must the topologies be equal?