Finite Closure as a Set-System Principle
Many families of sets are designed to remain stable under the operations that combine finitely many events or regions. An algebra of sets formalizes this stability: starting with any of its members, one may take complements, unions, or intersections a finite number of times and remain in the family. The word “algebra” here refers to a collection of subsets, not to an algebraic structure with addition and multiplication.
Throughout this tutorial, \(X\) is an arbitrary set and \(\mathcal{A}\) is a family of subsets of \(X\), so \(\mathcal{A}\subseteq\mathcal{P}(X)\). No measure is assumed. For \(E\subseteq X\), its complement is always relative to \(X\): \(E^c=X\setminus E\). Finite closure does not assert closure under countable unions; that stronger requirement is part of the definition of a sigma-algebra.
Condition (3) is stated for two sets, but it gives closure under every union of finitely many members. The definition also implies closure under intersections and differences. These consequences are useful both when verifying an example and when using an algebra once it has been verified.
Consequences of the Definition
Proof. First, \(\varnothing\in\mathcal{A}\) because \(X\in\mathcal{A}\) and complements belong to \(\mathcal{A}\). For a union of \(n\geq 1\) members, repeated application of the two-set union rule gives \(E_1\cup\cdots\cup E_n\in\mathcal{A}\). The union of zero sets is \(\varnothing\), which is also in \(\mathcal{A}\).
For intersections, take \(E,F\in\mathcal{A}\). Their complements are in \(\mathcal{A}\), so \(E^c\cup F^c\in\mathcal{A}\). Taking its complement and using the set identity \((E^c\cup F^c)^c=E\cap F\) shows \(E\cap F\in\mathcal{A}\). Repeating this argument gives closure under intersections of any positive finite number of members; the intersection of zero sets is \(X\in\mathcal{A}\).
Next, \(E\setminus F=E\cap F^c\). Both \(E\) and \(F^c\) belong to \(\mathcal{A}\), and closure under intersections therefore gives \(E\setminus F\in\mathcal{A}\). Finally, \(E\triangle F=(E\setminus F)\cup(F\setminus E)\), so closure under differences and finite unions gives \(E\triangle F\in\mathcal{A}\). Each conclusion follows from the defining rules and the displayed set identities. \(\square\)
The converse is also useful: the defining checks can be replaced by checks for closure under finite unions and complements, or under finite intersections and complements, provided \(X\) is included. In practice, the three conditions in the definition are usually the shortest verification.
Worked Examples
Worked Example: Unions of Blocks in a Finite Partition
Let \(X=\{1,2,3,4,5,6\}\), and divide it into the disjoint blocks \(P_1=\{1,2\}\), \(P_2=\{3\}\), and \(P_3=\{4,5,6\}\). Let \(\mathcal{A}\) consist of all unions of these three blocks, including the empty union. Verify that \(\mathcal{A}\) is an algebra, then compute some operations.
The whole set is the union of all three blocks, so \(X\in\mathcal{A}\). Any member is obtained by selecting some of the blocks. Because the blocks are disjoint and together cover \(X\), its complement is exactly the union of the blocks not selected. Thus the complement is again in \(\mathcal{A}\). The union of two members selects the blocks selected by either member, and is also in \(\mathcal{A}\). This checks the definition.
For a numerical check, take \(A=P_1\cup P_3=\{1,2,4,5,6\}\) and \(B=P_2\cup P_3=\{3,4,5,6\}\). Then \(A^c=P_2=\{3\}\), \(A\cup B=P_1\cup P_2\cup P_3=X\), and \(A\cap B=P_3=\{4,5,6\}\). Each computed set is a union of whole blocks. Check: the intersection contains precisely the elements appearing in both lists, namely \(4,5,6\); the union contains every element from \(1\) through \(6\).
Worked Example: Finite Unions of Half-Open Interval Cells
On \(X=\mathbb{R}\), let \(\mathcal{A}\) consist of finite unions of cells formed by finitely many real cut points. More precisely, for cut points \(t_1<\cdots<t_k\), the cells are \((-\infty,t_1)\), \([t_1,t_2),\ldots,[t_{k-1},t_k)\), and \([t_k,\infty)\). If there are no cut points, the only cell is \(\mathbb{R}\). Verify that this family is an algebra.
Consider any two members \(A,B\in\mathcal{A}\). Each is described using a finite set of cut points. Take the union \(T\) of both finite sets of cut points, and list the distinct points in increasing order. This gives a common refinement of the two cell partitions. Every cell from either original partition is a union of cells in the refined partition: inserting an extra cut point only splits a cell into smaller half-open cells.
Consequently, \(A\) and \(B\) are each unions of cells from the common refinement. The whole line is the union of all refined cells. The complement of a union of selected cells is the union of the remaining cells, and the union of two selected collections is again a collection of refined cells. Thus \(\mathbb{R}\in\mathcal{A}\), complements belong to \(\mathcal{A}\), and unions of two members belong to \(\mathcal{A}\). This proves the algebra property.
For an explicit computation, use cut points \(0\) and \(2\). The cells are \(C_1=(-\infty,0)\), \(C_2=[0,2)\), and \(C_3=[2,\infty)\). Set \(A=C_1\cup C_3\) and \(B=C_2\cup C_3\). The complement of \(A\) is \(C_2=[0,2)\); the union \(A\cup B\) is \(C_1\cup C_2\cup C_3=\mathbb{R}\); and the intersection \(A\cap B\) is \(C_3=[2,\infty)\). Check: the endpoint \(0\) is in \(C_2\), not \(C_1\), and \(2\) is in \(C_3\), not \(C_2\). The half-open convention therefore assigns each endpoint to exactly one cell, as required for the complement and intersection calculations.
Worked Example: Finite-or-Cofinite Subsets of the Integers
Let \(X=\mathbb{Z}\), and let \(\mathcal{A}\) consist of all finite subsets of \(\mathbb{Z}\) together with all cofinite subsets. Here “cofinite” means that the complement in \(\mathbb{Z}\) is finite. Verify the algebra rules and compute the operations for \(A=\{-2,0,3\}\) and \(B=\{0,5\}\).
The set \(\mathbb{Z}\) is cofinite because its complement is empty, so it belongs to \(\mathcal{A}\). Taking a complement switches finite sets and cofinite sets. For closure under unions, if both sets are finite, their union is finite. If at least one is cofinite, say \(E^c\) is finite and \(F\) is any set, then \((E\cup F)^c=E^c\cap F^c\subseteq E^c\), so the complement of \(E\cup F\) is finite. Thus the union is cofinite. All three algebra conditions hold.
For the specified finite sets, \(A\cup B=\{-2,0,3,5\}\) and \(A\cap B=\{0\}\). Their complements are \(\mathbb{Z}\setminus A\) and \(\mathbb{Z}\setminus B\), which are cofinite because the removed sets have respectively three and two elements. The difference \(A\setminus B=\{-2,3\}\). These results agree with the closure rules. Check: \(0\) is the only element listed in both \(A\) and \(B\), while \(-2\) and \(3\) are the elements of \(A\) that are not in \(B\).
How to Verify a Family—and a Common Pitfall
For a proposed family \(\mathcal{A}\subseteq\mathcal{P}(X)\), a reliable verification follows the three clauses of the definition separately. First check that the underlying set \(X\) is in the family. Next take an arbitrary member \(E\in\mathcal{A}\) and verify that its complement relative to \(X\) is also a member. Finally take arbitrary \(E,F\in\mathcal{A}\) and verify that \(E\cup F\) is a member. Showing the rules for a few sample sets is not enough; the argument must apply to every member or every pair of members.
A common mistake is to check closure under unions and intersections but forget complements. For a concrete counterexample, let \(X=\{1,2,3\}\) and \(\mathcal{C}=\{\varnothing,\{1\},X\}\). This family contains \(X\). It is closed under unions: unions of members are among \(\varnothing,\{1\},X\). It is also closed under intersections for the same reason. But \(\{1\}^c=\{2,3\}\notin\mathcal{C}\), so \(\mathcal{C}\) is not an algebra. The correction is to test complement closure explicitly; the other two operations cannot replace that requirement.
Check Your Understanding
For each question, use complements relative to the stated underlying set and distinguish a general closure argument from a calculation for particular sets.
- Let \(X=\{a,b,c,d\}\), partitioned into \(\{a,b\}\) and \(\{c,d\}\). List all members of the family of unions of partition blocks, and verify its complement rule.
- Suppose \(\mathcal{A}\) is an algebra on \(X\). Derive \(E\cap F\in\mathcal{A}\) from the complement and union axioms.
- On \(X=\mathbb{R}\), use cut points \(-1\) and \(1\) to identify the cell containing \(1\). Which cell contains \(1\) under the half-open convention?
- Give a family containing \(X\) and closed under complements but not under unions, or explain why your candidate fails one of those checks.
- For \(E,F\) in an algebra, write \(E\setminus F\) using only intersection and complement, and state why it belongs to the algebra.