Sets with Index Labels
When a collection of sets has more than a few members, listing every set can become cumbersome. An index set lets us refer to each member by a label: \(A_i\) is the set carrying label \(i\). The labels tell us which sets belong to the family, and indexed union and intersection turn the whole family into one set. Their definitions are membership tests: a point is in the union when it belongs to at least one set, and it is in the intersection when it belongs to every set.
Fix an ambient set \(X\), and let \(I\) be an index set. For each \(i\in I\), let \(A_i\subseteq X\). The collection \((A_i)_{i\in I}\) is an indexed family of subsets of \(X\). Different indices may label the same subset; the index labels specify which sets are being considered, not necessarily distinct sets. The ambient set matters especially when the index set is empty.
Why the Quantifiers Matter
The union uses an existential condition: at least one label must work. The intersection uses a universal condition: every label must work. This is exactly the difference between “there is an index” and “for each index.” The order of the quantifier and the membership condition must be kept together. In particular, belonging to many sets in a family is not enough for membership in the intersection if even one set is missed.
Proof. By definition, the indexed union consists precisely of those \(x\in X\) for which there is an index \(i\in I\) with \(x\in A_i\). Thus each side of the first equivalence asserts the same membership condition. By definition, the indexed intersection consists precisely of those \(x\in X\) for which \(x\in A_i\) holds for every \(i\in I\). Thus the two sides of the second equivalence also assert the same condition. \(\square\)
These definitions generalize the familiar union and intersection of a finite list of sets. For example, if \(I=\{1,2,3\}\), then \(\bigcup_{i\in I}A_i=A_1\cup A_2\cup A_3\), and \(\bigcap_{i\in I}A_i=A_1\cap A_2\cap A_3\). The index notation is useful because the same definitions continue to make sense when there are infinitely many labels.
Worked Example: A Finite Indexed Family
Take \(X=\{-1,0,1,2,3,4,5\}\), \(I=\{1,2,3,4\}\), and \(A_1=\{-1,0,2\}\), \(A_2=\{0,1,2\}\), \(A_3=\{2,3\}\), and \(A_4=\{0,2,4\}\). Find the indexed union and intersection.
For the union, include an element of \(X\) if it appears in at least one set. The elements appearing are \(-1\) from \(A_1\), \(0\) from \(A_1,A_2,A_4\), \(1\) from \(A_2\), \(2\) from all four sets, \(3\) from \(A_3\), and \(4\) from \(A_4\). The element \(5\) appears in none. Therefore \(\bigcup_{i\in I}A_i=\{-1,0,1,2,3,4\}\).
For the intersection, check which elements of \(X\) occur in every one of the four sets. The element \(2\) occurs in \(A_1,A_2,A_3,A_4\). The element \(0\) is absent from \(A_3\); \(-1\) is absent from \(A_2\); \(1\) is absent from \(A_1\); \(3\) is absent from \(A_1\); \(4\) is absent from \(A_1\); and \(5\) is absent from all four. Hence \(\bigcap_{i\in I}A_i=\{2\}\).
Check: \(2\in A_i\) for each of the four indices, so \(2\) is in the intersection. Each other element of the displayed union is missing from at least one of the four sets, so none of them belongs to the intersection.
Infinite Families: Intervals and Discrete Sets
For an infinite family, the same membership tests apply; the only difference is that there may be infinitely many conditions to check. A useful way to understand an intersection is to start with one candidate point and ask whether it meets all the conditions. To disprove membership in an intersection, it is enough to find one index whose set excludes the point. To prove membership in a union, it is enough to find one index whose set contains the point.
Worked Example: Nested Intervals
In \(X=\mathbb{R}\) with its standard topology, let \(I=\mathbb{N}=\{1,2,\dots\}\) and \(A_n=[0,1/n]\). Compute the union and intersection.
Since \(1\in\mathbb{N}\), the first set is \(A_1=[0,1]\). For every \(n\in\mathbb{N}\), \(1/n\leq 1\), so \(A_n\subseteq[0,1]\). Thus every member of the family is contained in \(A_1\), and \(A_1\) itself occurs in the family. It follows that \(\bigcup_{n\in\mathbb{N}}A_n=[0,1]\).
For the intersection, \(0\in A_n\) for every \(n\), so \(0\) belongs to it. If \(x\in\bigcap_{n\in\mathbb{N}}A_n\), then \(x\in[0,1/n]\) for every \(n\), so \(x\geq0\) and \(x\leq1/n\) for every positive integer \(n\). If \(x>0\), choose \(n\in\mathbb{N}\) with \(n>1/x\); then \(1/n<x\), contradicting \(x\leq1/n\). Therefore \(x=0\), and \(\bigcap_{n\in\mathbb{N}}A_n=\{0\}\).
Check: The proposed union contains \(A_1=[0,1]\), and no \(A_n\) extends beyond \([0,1]\). The proposed intersection contains \(0\) in every interval, while every \(x>0\) is excluded by any integer \(n>1/x\); any \(x<0\) is already excluded from \(A_1\).
Worked Example: Sets That Shift with the Index
Let \(X=\mathbb{Z}\), \(I=\mathbb{N}\), and \(A_n=\{n,n+1\}\). Find the union and intersection.
Each \(A_n\) consists of positive integers, so \(\bigcup_{n\in\mathbb{N}}A_n\subseteq\mathbb{N}\). Conversely, \(1\in A_1=\{1,2\}\), and if \(k\in\mathbb{N}\) with \(k\geq2\), then \(k\in A_k=\{k,k+1\}\). Every positive integer therefore occurs in at least one set. Hence \(\bigcup_{n\in\mathbb{N}}A_n=\mathbb{N}\).
If an integer lies in the intersection, it must lie in both \(A_1=\{1,2\}\) and \(A_3=\{3,4\}\). These sets have no element in common, so no integer can lie in every \(A_n\), and \(\bigcap_{n\in\mathbb{N}}A_n=\varnothing\).
Check: The union includes \(1\) through \(A_1\), and each \(k\geq2\) through \(A_k\); all sets contain only positive integers. The intersection is empty because membership in every set would require membership in both \(A_1\) and \(A_3\), which is impossible.
Empty Index Sets and a Useful Inclusion Test
The definitions also determine what happens if \(I=\varnothing\). There is no index that can witness the existential condition for a union, so the union is empty. The universal condition for an intersection imposes no restrictions when there are no indices; consequently every element of the ambient set \(X\) satisfies it. This is why the ambient set is specified: the empty intersection is \(X\), not an unspecified set.
Worked Example: No Indices
Let \(X=\{a,b\}\) and \(I=\varnothing\). There are no sets \(A_i\) in the family. The condition “there exists \(i\in I\) with \(x\in A_i\)” is false for both \(a\) and \(b\), because there is no index to choose. Thus \(\bigcup_{i\in I}A_i=\varnothing\). The condition “for every \(i\in I,\ x\in A_i\)” is true for each \(x\in X\), since there is no index at which it could fail. Thus \(\bigcap_{i\in I}A_i=X=\{a,b\}\).
Check: The union contains no elements, while the intersection contains both elements of the stated ambient set. If the ambient set were different, the empty intersection would be that different ambient set.
An inclusion test gives another way to use the existential and universal conditions. It can be especially helpful when the family is infinite: instead of listing its members, check a condition for every index.
Proof. For the first equivalence, suppose \(\bigcup_{i\in I}A_i\subseteq B\). Fix \(i\in I\). Every \(x\in A_i\) belongs to the union, so \(x\in B\). Hence \(A_i\subseteq B\), for every \(i\). Conversely, suppose \(A_i\subseteq B\) for every \(i\in I\). If \(x\in\bigcup_{i\in I}A_i\), there is an index \(i\) with \(x\in A_i\). The assumed inclusion gives \(x\in B\), proving \(\bigcup_{i\in I}A_i\subseteq B\).
For the second equivalence, suppose \(B\subseteq\bigcap_{i\in I}A_i\). Fix \(i\in I\). Every \(x\in B\) belongs to the intersection and thus belongs to \(A_i\); so \(B\subseteq A_i\) for every \(i\). Conversely, suppose \(B\subseteq A_i\) for every \(i\in I\). Each \(x\in B\) then belongs to \(A_i\) for every index \(i\), so the membership test for intersection gives \(x\in\bigcap_{i\in I}A_i\). Thus \(B\subseteq\bigcap_{i\in I}A_i\). If \(I\) is empty, the first equivalence says \(\varnothing\subseteq B\) exactly when a condition required for every index holds vacuously; the second says \(B\subseteq X\), which holds by hypothesis. \(\square\)
A Common Quantifier Mistake
Common pitfall. Do not replace “for every index” with “for at least one index” when testing membership in an intersection. For the family \(A_n=\{n,n+1\}\) above, the incorrect calculation might say: “\(2\in A_1=\{1,2\}\), so \(2\in\bigcap_{n\in\mathbb{N}}A_n\).” This checks only one index. But \(A_3=\{3,4\}\), and \(2\notin A_3\), so the universal membership condition fails and the intersection is empty. The corrected test is: a candidate belongs to the intersection only if it belongs to every set in the family. Conversely, one successful index is all that is needed to establish membership in a union.
Indexed notation makes the distinction precise even when many sets are involved: the union collects points with at least one successful label, while the intersection collects points that pass every label’s test. Keeping that quantifier in view is the reliable first step in computing either set.
Check Your Understanding
For each question, state the membership condition you use before computing the set.
- Let \(X=\{0,1,2,3\}\), \(I=\{1,2\}\), \(A_1=\{0,1,2\}\), and \(A_2=\{1,2,3\}\). Find \(\bigcup_{i\in I}A_i\) and \(\bigcap_{i\in I}A_i\).
- For \(A_n=[-1/n,1/n]\subseteq\mathbb{R}\), \(n\in\mathbb{N}\), find the union and intersection.
- Let \(A_n=\{2n\}\subseteq\mathbb{Z}\), \(n\in\mathbb{N}\). Find the union. Is the intersection empty?
- If \(I=\varnothing\) and the ambient set is \(X\), what are \(\bigcup_{i\in I}A_i\) and \(\bigcap_{i\in I}A_i\)? Explain the different roles of “there exists” and “for every.”
- Suppose \(A_i\subseteq B\) for every \(i\in I\). What inclusion follows for the indexed union, and which membership condition justifies it?