Set Sequences and Tail Operations

Looking at All Sets from a Chosen Index Onward

Given a sequence of sets, one can focus not only on its first few terms but on everything from a chosen index onward. The union of that tail records the points contained in at least one of its sets; the intersection records the points shared by every set in the tail. As the starting index moves forward, these two operations change in opposite directions. Their behavior is especially simple for nested and periodic sequences.

Throughout, let $X$ be any set and let $(A_n)_{n\ge1}$ be a sequence of subsets of $X$. These are purely set-theoretic definitions: no measure or measurability assumption is needed. The index $n$ is a positive integer, and the tail beginning at $n$ includes $A_n$ itself.

Definition: For $n\ge1$, the tail union and tail intersection beginning at $n$ are $$ U_n=\bigcup_{k=n}^{\infty}A_k, \qquad V_n=\bigcap_{k=n}^{\infty}A_k. $$ Thus $x\in U_n$ exactly when $x\in A_k$ for at least one $k\ge n$, while $x\in V_n$ exactly when $x\in A_k$ for every $k\ge n$.

The two operations answer different questions about the same tail. The union collects anything that appears somewhere in the tail; the intersection keeps only what is present throughout it. In particular, neither operation is a finite union or intersection of the first $n$ sets. It uses all indices $k\ge n$.

How the Tails Change with the Starting Index

Proposition: For every sequence of subsets of $X$, the tail unions decrease with $n$, and the tail intersections increase with $n$: $$ U_{n+1}\subseteq U_n \qquad\text{and}\qquad V_n\subseteq V_{n+1}. $$

Proof. If $x\in U_{n+1}$, then $x\in A_k$ for some $k\ge n+1$. This same index satisfies $k\ge n$, so $x\in U_n$. Hence $U_{n+1}\subseteq U_n$. If $x\in V_n$, then $x\in A_k$ for every $k\ge n$. In particular, $x\in A_k$ for every $k\ge n+1$, which means $x\in V_{n+1}$. Hence $V_n\subseteq V_{n+1}$. Both arguments apply even if some sets are empty or equal to $X$. $\square$

Advancing the starting index removes one set from consideration. Removing a set can only shrink a union or enlarge an intersection. This explains the opposite directions in the proposition without any assumption that the sets $A_n$ themselves are nested.

Nested Sequences

A sequence is increasing if $A_n\subseteq A_{n+1}$ for every $n$. It is decreasing if $A_{n+1}\subseteq A_n$ for every $n$. These conditions describe inclusion between successive sets; they do not mean that the sets are numerical quantities.

Theorem: If $(A_n)$ is increasing, then for every $n\ge1$, $$ U_n=\bigcup_{k=1}^{\infty}A_k, \qquad V_n=A_n. $$ If $(A_n)$ is decreasing, then for every $n\ge1$, $$ U_n=A_n, \qquad V_n=\bigcap_{k=1}^{\infty}A_k. $$

Proof. First suppose the sequence is increasing. Since $A_n\subseteq A_k$ whenever $k\ge n$, every point of $A_n$ belongs to the tail intersection $V_n$. Thus $A_n\subseteq V_n$. Also $V_n\subseteq A_n$, because $A_n$ is one of the sets being intersected. Therefore $V_n=A_n$. For the tail union, $U_n\subseteq\bigcup_{k=1}^{\infty}A_k$ follows because the tail indices are among all positive indices. Conversely, if $x\in\bigcup_{k=1}^{\infty}A_k$, choose $j\ge1$ with $x\in A_j$. If $j\ge n$, then $x\in U_n$. If $j<n$, increasingness gives $A_j\subseteq A_n$, so again $x\in U_n$. This proves the first pair of identities.

Now suppose the sequence is decreasing. Every $A_k$ with $k\ge n$ is contained in $A_n$, so $U_n\subseteq A_n$; and $A_n\subseteq U_n$ because $A_n$ is included in its own tail. Thus $U_n=A_n$. For the intersection, $\bigcap_{k=1}^{\infty}A_k\subseteq V_n$, since satisfying every set condition implies satisfying all the tail conditions. Conversely, if $x\in V_n$, then $x\in A_k$ for every $k\ge n$, in particular $x\in A_n$. For any $j<n$, decreasingness implies $A_n\subseteq A_j$, so $x\in A_j$ as well. Thus $x$ belongs to every $A_j$, and $V_n\subseteq\bigcap_{k=1}^{\infty}A_k$. The second pair follows. $\square$

Worked Example: An Increasing Sequence of Initial Segments

Take $X=\mathbb{N}=\{1,2,3,\ldots\}$ and $A_n=\{1,2,\ldots,n\}$. Since every element of $A_n$ is also in $A_{n+1}$, this sequence is increasing. Compute the tail union and intersection beginning at $n=3$, then give the formulas for general $n$.

The first sets in the tail are $A_3=\{1,2,3\}$, $A_4=\{1,2,3,4\}$, and $A_5=\{1,2,3,4,5\}$. Their union, extended over all later indices, contains every positive integer: for any $j\in\mathbb{N}$, the set $A_k$ contains $j$ whenever $k\ge j$ and $k\ge3$. Therefore $U_3=\mathbb{N}$. For the intersection, a point must belong to every set in the tail, including $A_3$; and each of $1,2,3$ belongs to all $A_k$ for $k\ge3$. Hence $V_3=\{1,2,3\}$.

For an arbitrary starting index $n$, the same reasoning gives $U_n=\mathbb{N}$ and $V_n=\{1,2,\ldots,n\}$. Check: at $n=3$, the theorem for increasing sequences gives $U_3=\bigcup_{k\ge1}A_k=\mathbb{N}$ and $V_3=A_3=\{1,2,3\}$, matching the direct calculation.

Worked Example: A Decreasing Sequence of Tails in the Natural Numbers

Let $X=\mathbb{N}$ and $A_n=\{j\in\mathbb{N}:j\ge n\}$. Compute $U_2$ and $V_2$, and then state the formulas for any $n$.

The sequence is decreasing: $A_{n+1}$ consists of the integers at least $n+1$, so $A_{n+1}\subseteq A_n$. Its first sets from index $2$ are $A_2=\{2,3,4,\ldots\}$, $A_3=\{3,4,5,\ldots\}$, and $A_4=\{4,5,6,\ldots\}$. The union of these sets and all later ones is $A_2$, because every later set is contained in $A_2$, while $A_2$ itself is included. Thus $U_2=\{2,3,4,\ldots\}$.

The intersection is empty. Indeed, for any candidate $j\in\mathbb{N}$, choose $k=\max(2,j+1)$. Then $k\ge2$, but $j<k$, so $j\notin A_k$. No $j$ belongs to every set in the tail, and therefore $V_2=\varnothing$. For general $n$, $U_n=A_n=\{n,n+1,\ldots\}$, while $V_n=\bigcap_{k\ge1}A_k=\varnothing$. Check: the decreasing-sequence identities give $U_2=A_2$ and $V_2=\bigcap_{k\ge1}A_k$; this last intersection is empty because for each $j$, the set $A_{j+1}$ excludes $j$.

Periodic Sequences

A sequence of sets has period $p$, where $p\ge1$ is an integer, if $A_{k+p}=A_k$ for every $k\ge1$. A tail of such a sequence contains repetitions of the same $p$ sets. Its union and intersection can therefore be computed using just one block of $p$ consecutive sets.

Proposition: If $A_{k+p}=A_k$ for every $k\ge1$, then for every $n\ge1$, $$ U_n=\bigcup_{r=0}^{p-1}A_{n+r}, \qquad V_n=\bigcap_{r=0}^{p-1}A_{n+r}. $$

Proof. Each index $k\ge n$ can be written uniquely as $k=n+r+qp$, where $r\in\{0,1,\ldots,p-1\}$ and $q$ is a nonnegative integer. Repeated use of the period condition gives $A_k=A_{n+r}$. Thus every set appearing in the tail is one of $A_n,A_{n+1},\ldots,A_{n+p-1}$, and each of these $p$ sets appears in the tail (take $q=0$). Repetitions do not change a union or an intersection. The two stated formulas follow. $\square$

Worked Example: A Three-Set Period

Let $X=\{a,b,c,d\}$, and repeat the sets $A_1=\{a,b\}$, $A_2=\{b,c\}$, and $A_3=\{b,d\}$ in that order. Compute the tail union and intersection for each possible starting position in one period.

For $n=1$, the tail begins $A_1,A_2,A_3,A_1,\ldots$. The union of one period is $\{a,b\}\cup\{b,c\}\cup\{b,d\}=\{a,b,c,d\}$, and the intersection is $\{a,b\}\cap\{b,c\}\cap\{b,d\}=\{b\}$. Therefore $U_1=\{a,b,c,d\}$ and $V_1=\{b\}$.

For $n=2$, one period is $A_2,A_3,A_1$. Its union is $\{b,c\}\cup\{b,d\}\cup\{a,b\}=\{a,b,c,d\}$, and its intersection is $\{b,c\}\cap\{b,d\}\cap\{a,b\}=\{b\}$. For $n=3$, one period is $A_3,A_1,A_2$. Its union is $\{b,d\}\cup\{a,b\}\cup\{b,c\}=\{a,b,c,d\}$, and its intersection is $\{b,d\}\cap\{a,b\}\cap\{b,c\}=\{b\}$. In each case, later terms repeat these same sets, so the computed one-period union and intersection give the full tails. Check: the three possible starting blocks contain the same three sets, only in a different order; union and intersection are unchanged by that reordering.

A Common Indexing Pitfall

A frequent error is to omit $A_n$ and start the tail at $A_{n+1}$. For example, with the increasing sequence $A_n=\{1,\ldots,n\}$, an incorrect calculation might say $V_3=A_4=\{1,2,3,4\}$, on the grounds that the sets continue to grow. But $V_3$ includes $A_3$, and every point in the intersection must belong to that set. The correct calculation is $V_3=A_3=\{1,2,3\}$.

The distinction can also affect a union. In the decreasing example $A_n=\{j:j\ge n\}$, omitting $A_n$ would give the union of the tail from $n+1$, namely $A_{n+1}$, rather than the requested $U_n=A_n$. Before computing either operation, write down the starting index explicitly and include its set.

Tail operations are useful because they summarize the behavior of a whole infinite sequence from each starting point. The nesting proposition applies to every sequence, while the formulas for increasing, decreasing, and periodic sequences provide direct computations in common cases. These operations also provide the set-theoretic building blocks used when studying limits of sequences of sets; the tail calculations themselves require no assumption that the sequence has a limit.

Check Your Understanding

For each question, identify exactly which indices enter the tail before simplifying the sets.

If $A_1=\{x\}$, $A_2=\{x,y\}$, $A_3=\{x,y,z\}$, and the sequence continues by adding one new element at each step, what are $U_3$ and $V_3$?
Prove directly from the definitions that $U_{n+1}\subseteq U_n$ for an arbitrary sequence of sets.
For a decreasing sequence, explain why $U_n=A_n$, including both set inclusions.
A sequence repeats $C,D,E,C,D,E,\ldots$. Express $V_2$ using only one period of sets.
For $A_n=\{j\in\mathbb{N}:j\ge n\}$, give a specific index $k\ge n$ showing that a fixed $j\in\mathbb{N}$ cannot belong to $V_n$.