0.1Limit Superior and Inferior

The limit superior is the smallest of all upper bounds for each tail. Let $S_k = \sup_{n \ge k} x_n$. As the tail shrinks, the pool of numbers decreases, making $S_k$ a monotonically decreasing sequence. Thus, its limit as $k \to \infty$ is exactly its infimum. The limit inferior is the largest of all lower bounds for each tail.

For example, imagine $x_n = \sin(n)$. Each tail of $x_n$ has $1, 2, 3$ as upper bounds, but $1$ is the tightest upper bound across all tails. Moreover, each tail has $-1, -2, -3$ as lower bounds, with $-1$ as the tightest lower bound. Thus $\limsup x_n = 1$ and $\liminf x_n = -1$. The limit of a sequence exists if and only if the limit superior and inferior agree.

This structurally mirrors the real sequence definition, replacing the analytic $\sup / \inf$ with the set-theoretic $\bigcup / \bigcap$. The tail union $F_k = \bigcup_{n \ge k} E_n$ forms a monotonically shrinking sequence of sets, making its limit equivalent to its intersection.

If $x \in \limsup E_n$, then no matter how many terms you cut off from the start, $x$ is still contained in some subsequent set. We say that $x$ is in $E_n$ infinitely often. On the other hand, if $x \in \liminf E_n$, then $x$ is in all but finitely many sets in the sequence. We say that $x$ is eventually in $E_n$. Thus, $\limsup$ guarantees $x$ is in at least one set of any tail of the sequence, whereas $\liminf$ guarantees $x$ is in every element of one specific tail. A sequence of sets has a limit if and only if its limit superior and inferior agree.