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

Let $(x_k)$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and let $L\in \bar{\mathbb{R}}$ (The extended real line $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$, its order, and the arithmetic that is left undefined). Say that $(x_k)$ converges to $L$ in $\bar{\mathbb{R}}$ 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;
• $L=+\mathrm{\infty }$ and $x_k\to +\mathrm{\infty }$ in the sense of Divergence to $+\mathrm{\infty }$ and to $-\mathrm{\infty }$;
• $L=-\mathrm{\infty }$ and $x_k\to -\mathrm{\infty }$ in the sense of Divergence to $+\mathrm{\infty }$ and to $-\mathrm{\infty }$.

Then $L$ is an extended subsequential limit of $(x_k)$ when some subsequence of $(x_k)$ converges to $L$ in $\bar{\mathbb{R}}$: 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

$\stackrel{}{\mathrm{SL}}(x):=\{L\in \bar{\mathbb{R}}:L\text{ is an extended subsequential limit of }(x_k)\}\subseteq \bar{\mathbb{R}}\mathrm{.}$

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 $\mathrm{SL}(x)\subseteq \mathbb{R}$. Its clause is quoted verbatim as the first of the three clauses above, so

$\stackrel{}{\mathrm{SL}}(x)\cap \mathbb{R}=\mathrm{SL}(x),$

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

Neither is Divergence to $+\mathrm{\infty }$ and to $-\mathrm{\infty }$ reinterpreted. The phrase "$x_k\to +\mathrm{\infty }$" 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 $\bar{\mathbb{R}}$, so that the three situations can be quantified over together. In particular the warning recorded there stands: a sequence diverging to $+\mathrm{\infty }$ 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 $L^{\mathrm{\prime }}$ in $\bar{\mathbb{R}}$. If both are real, $L=L^{\mathrm{\prime }}$ by uniqueness of real limits (A sequence has at most one limit). If one is real and the other is $\pm \mathrm{\infty }$, that is impossible, because a sequence diverging to $\pm \mathrm{\infty }$ is unbounded and so does not converge, as Divergence to $+\mathrm{\infty }$ and to $-\mathrm{\infty }$ records. If $L=+\mathrm{\infty }$ and $L^{\mathrm{\prime }}=-\mathrm{\infty }$ then, taking $M=0$ in both conditions, there are $K$ and $K^{\mathrm{\prime }}$ with $x_k>0$ for $k\ge K$ and $x_k<0$ for $k\ge K^{\mathrm{\prime }}$; 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 $\mathrm{lim\hspace{0.17em}sup}$ 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 $\mathrm{SL}(x)=\{0\}$, whose greatest element is $0$, while the behaviour that dominates it is a subsequence running off to $+\mathrm{\infty }$. That is exactly the content of A sequence with $\mathrm{lim\hspace{0.17em}sup}=+\mathrm{\infty }$: the greatest subsequential limit exists only in $\bar{\mathbb{R}}$ ↗, and it is why The limit superior is itself a subsequential limit in $\bar{\mathbb{R}}$ and is the greatest one is stated for $\stackrel{}{\mathrm{SL}}(x)$ rather than for $\mathrm{SL}(x)$.

• 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 $\mathrm{SL}(x)$ in Subsequential limit of a real sequence, and the subsequential limit set, with the two divergence clauses added.