Subsequential limit of a real sequence, and the subsequential limit set

Let $(x_k)$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and let $L \in \mathbb{R}$. Then $L$ is a subsequential limit of $(x_k)$ when some subsequence of $(x_k)$ converges to $L$: that is, when there is a strictly increasing $n : \mathbb{N} \to \mathbb{N}$ such that

$x_{n_j} \longrightarrow L \qquad (j \to \infty)$

in the sense of Limits and Cauchy sequences of reals. The subsequential limit set of $(x_k)$ is

$\operatorname{SL}(x) \;:=\; \{\, L \in \mathbb{R} : L \text{ is a subsequential limit of } (x_k) \,\} \subseteq \mathbb{R}.$

Both pieces of the definition are already fixed elsewhere and are only combined here: strictly increasing and subsequence are Sequences of reals: bounded, eventually, frequently, tails, subsequences, and converges is Limits and Cauchy sequences of reals. Nothing about $(x_k)$ itself is assumed; in particular $(x_k)$ is not assumed to converge, and $\operatorname{SL}(x)$ may be empty, a single point, or larger.

A subsequence looks arbitrarily far out. A strictly increasing index map satisfies $n_j \ge j$ for every $j$ (A strictly increasing index map satisfies $n_k \ge k$), so the indices $n_j$ are cofinal in $\mathbb{N}$ and a subsequential limit is determined by the behaviour of $(x_k)$ at arbitrarily large indices. Consequently no finite initial segment of $(x_k)$ affects $\operatorname{SL}(x)$: a sequence and each of its tails have the same subsequential limits.

Terminology. Some texts say cluster point, limit point or accumulation value of the sequence for the same notion. This library says subsequential limit throughout, reserving limit point for the topological notion of a limit point of a set, which is a different thing: the set $\{x_k\}$ of values of the constant sequence $x_k = 0$ has no limit point, while $0$ is a subsequential limit of that sequence.

• A convergent sequence has exactly one subsequential limit, its limit. If $x_k \to x$ then every subsequence converges to $x$ (Subsequences inherit the limit), so every subsequential limit equals $x$ by uniqueness of limits (A sequence has at most one limit); and $x$ itself is one, taking $n_j = j$, which is strictly increasing. So $\operatorname{SL}(x) = \{x\}$. The converse fails: $\operatorname{SL}$ being a single point does not force convergence, as the unbounded sequence of The sequence $1, 1, 2, 1, 3, 1, 4, \dots$ is unbounded and has a convergent subsequence ↗ shows.

• The subsequential limit set can be empty. The sequence $x_k = k$ has no subsequential limit at all, since every subsequence is unbounded and an unbounded sequence does not converge (Every convergent sequence is bounded). Bolzano-Weierstrass (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence) says exactly that boundedness is what rules this out: for a bounded sequence, $\operatorname{SL}(x) \ne \emptyset$.

• It can also be large. The alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$, $s_{\sigma(k)} = -s_k$, which satisfies $|s_k| = 1$, $s \circ e \equiv 1$ and $s \circ o \equiv -1$ has subsequential limit set $\{-1, 1\}$, and The sequence $(-1)^k(1 + 1/k)$ is bounded with subsequential limit set exactly $\{-1, 1\}$ ↗ carries out that computation for a sequence that converges to neither. The systematic study of $\operatorname{SL}(x)$, in particular that it has a greatest and a least element for a bounded sequence, belongs to the $\limsup$ page and is not begun here.