Convergence in $\overline{\mathbb{R}}$ and the extended subsequential limit set: $L \in \overline{\mathbb{R}}$ is an extended subsequential limit when some subsequence converges to $L$, or diverges to $L = \pm\infty$

Let $(x_k)$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and let $L \in \overline{\mathbb{R}}$ (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$, its order, and the arithmetic that is left undefined). Say that $(x_k)$ converges to $L$ in $\overline{\mathbb{R}}$ when one of the following holds, according to which of the three kinds of element $L$ is:

• $L \in \mathbb{R}$ and $(x_k)$ converges to $L$ in the sense of Limits and Cauchy sequences of reals;
• $L = +\infty$ and $x_k \to +\infty$ in the sense of Divergence to $+\infty$ and to $-\infty$;
• $L = -\infty$ and $x_k \to -\infty$ in the sense of Divergence to $+\infty$ and to $-\infty$.

Then $L$ is an extended subsequential limit of $(x_k)$ when some subsequence of $(x_k)$ converges to $L$ in $\overline{\mathbb{R}}$: when there is a strictly increasing $n : \mathbb{N} \to \mathbb{N}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences) such that $(x_{n_j})_{j \in \mathbb{N}}$ converges to $L$ in the sense just given. The extended subsequential limit set of $(x_k)$ is

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

This extends the published Subsequential limit of a real sequence, and the subsequential limit set and does not replace it. That definition is finite by design: there $L$ ranges over $\mathbb{R}$ and $\operatorname{SL}(x) \subseteq \mathbb{R}$. Its clause is quoted verbatim as the first of the three clauses above, so

$\overline{\operatorname{SL}}(x) \cap \mathbb{R} = \operatorname{SL}(x),$

immediately from the definitions: a real $L$ lies in $\overline{\operatorname{SL}}(x)$ exactly when some subsequence converges to $L$ in the sense of Limits and Cauchy sequences of reals, which is exactly the condition $L \in \operatorname{SL}(x)$. The extended set is therefore $\operatorname{SL}(x)$ together with at most the two extra points $\pm\infty$, each present exactly when some subsequence diverges to it. Nothing about $\operatorname{SL}(x)$ is redefined, and every statement proved about $\operatorname{SL}(x)$ elsewhere in the library remains a statement about the same set.

Neither is Divergence to $+\infty$ and to $-\infty$ reinterpreted. The phrase "$x_k \to +\infty$" keeps exactly the meaning fixed there, an abbreviation for "for every real $M$, eventually $x_k > M$". What is new is only that the phrase is now allowed to appear as one of three clauses in a single definition whose parameter $L$ ranges over $\overline{\mathbb{R}}$, so that the three situations can be quantified over together. In particular the warning recorded there stands: a sequence diverging to $+\infty$ has no limit in $\mathbb{R}$, and none of the rules of Algebra of limits: sums, scalar multiples, products and quotients applies to it.

• An extended limit is unique. Suppose $(x_k)$ converges to $L$ and to $L'$ in $\overline{\mathbb{R}}$. If both are real, $L = L'$ by uniqueness of real limits (A sequence has at most one limit). If one is real and the other is $\pm\infty$, that is impossible, because a sequence diverging to $\pm\infty$ is unbounded and so does not converge, as Divergence to $+\infty$ and to $-\infty$ records. If $L = +\infty$ and $L' = -\infty$ then, taking $M = 0$ in both conditions, there are $K$ and $K'$ with $x_k > 0$ for $k \ge K$ and $x_k < 0$ for $k \ge K'$; any index at least as large as both gives $0 < x_k < 0$, which is impossible. So the three clauses are mutually exclusive and each determines $L$.

• Why the extended set is the right object for a $\limsup$ theorem. The greatest subsequential limit of an arbitrary real sequence need not be a real number: the sequence that alternates between $0$ and larger and larger values has $\operatorname{SL}(x) = \{0\}$, whose greatest element is $0$, while the behaviour that dominates it is a subsequence running off to $+\infty$. That is exactly the content of A sequence with $\limsup = +\infty$: the greatest subsequential limit exists only in $\overline{\mathbb{R}}$ ↗, and it is why The limit superior is itself a subsequential limit in $\overline{\mathbb{R}}$ and is the greatest one is stated for $\overline{\operatorname{SL}}(x)$ rather than for $\operatorname{SL}(x)$.

• A tail changes nothing. A strictly increasing index map satisfies $n_j \ge j$ (A strictly increasing index map satisfies $n_k \ge k$), so all three clauses depend only on the behaviour of $(x_k)$ at arbitrarily large indices, and a sequence and each of its tails have the same extended subsequential limit set. This is the same observation made for $\operatorname{SL}(x)$ in Subsequential limit of a real sequence, and the subsequential limit set, with the two divergence clauses added.