limsup, liminf, and Subsequential Limits

Fix two objects $-\infty$ and $+\infty$, distinct from one another and neither of them a real number (The real numbers), and set

$\overline{\mathbb{R}} := \mathbb{R} \cup \{-\infty, +\infty\}.$

This is a new object, introduced here explicitly with its own order and its own partial arithmetic. It is not an enlargement of the field $\mathbb{R}$, and no operation of $\mathbb{R}$ (Complete ordered field (least-upper-bound property)) is redefined by anything below.

The order. For $a, b \in \overline{\mathbb{R}}$ declare

$a \le b \quad :\Longleftrightarrow \quad a = -\infty \ \text{ or } \ b = +\infty \ \text{ or } \ \big(a, b \in \mathbb{R} \text{ and } a \le b \text{ in } \mathbb{R}\big),$

with $\mathbb{R}$ ordered as in Order on the reals, and write $a < b$ for "$a \le b$ and $a \ne b$" as usual (Partial order and partially ordered set).

$(\overline{\mathbb{R}}, \le)$ is a totally ordered set, and the inclusion of $\mathbb{R}$ preserves and reflects the order. All four checks are immediate from the displayed clauses.

• Reflexive. For $a = \pm\infty$ one of the first two clauses applies; for $a \in \mathbb{R}$ the third does, since $a \le a$ in $\mathbb{R}$.
• Antisymmetric. Suppose $a \le b$ and $b \le a$. If $a = -\infty$ then $b \le a$ forces $b = -\infty$, since the clause $a = +\infty$ fails and $b, a$ are not both real. Symmetrically $b = -\infty$ forces $a = -\infty$, and $a = +\infty$ or $b = +\infty$ forces the other to be $+\infty$. In the one remaining situation $a$ and $b$ are both real and antisymmetry is that of $\mathbb{R}$.
• Transitive. Let $a \le b \le c$. If $a = -\infty$ or $c = +\infty$ the conclusion is one of the first two clauses. Otherwise $a \ne -\infty$ forces, in $a \le b$, either $b = +\infty$ or $a, b \in \mathbb{R}$; and $c \ne +\infty$ forces, in $b \le c$, either $b = -\infty$ or $b, c \in \mathbb{R}$. The value $b = +\infty$ is incompatible with the second alternative pair, so $b$ is real, hence so are $a$ and $c$, and transitivity is that of $\mathbb{R}$.
• Total. If $a = -\infty$ or $b = +\infty$ then $a \le b$; if $b = -\infty$ or $a = +\infty$ then $b \le a$; otherwise both are real and the order of $\mathbb{R}$ is total.
• Preserved and reflected. For $a, b \in \mathbb{R}$ the first two clauses fail, so $a \le b$ in $\overline{\mathbb{R}}$ says exactly $a \le b$ in $\mathbb{R}$.

In particular $-\infty$ is the least and $+\infty$ the greatest element of $\overline{\mathbb{R}}$, and $-\infty < x < +\infty$ for every $x \in \mathbb{R}$.

Reflection. Extend negation by

$-(+\infty) := -\infty, \qquad -(-\infty) := +\infty,$

keeping the field negative on $\mathbb{R}$. The resulting map $\nu : \overline{\mathbb{R}} \to \overline{\mathbb{R}}$, $\nu(a) = -a$, satisfies $\nu(\nu(a)) = a$ and

$a \le b \iff -b \le -a \qquad (a, b \in \overline{\mathbb{R}}).$

For $a$ and $b$ real this is the elementwise order reversal in $\mathbb{R}$: translation invariance (Order is preserved by adding a constant and by adding inequalities) applied with the constant $-a-b$ turns $a < b$ into $-b < -a$ and, applied with the constant $a+b$, turns it back, while $a = b$ holds exactly when $-a = -b$. In every other case both sides are decided by the first two clauses of the order: $a = -\infty$ makes both sides true, as does $b = +\infty$, and if $a \ne -\infty$, $b \ne +\infty$ and $a, b$ are not both real then one of $a = +\infty$, $b = -\infty$ holds and both sides are false.

Partial addition. For $a, b \in \overline{\mathbb{R}}$ the sum $a + b$ is defined by

• $a + b$ = the field sum, when $a, b \in \mathbb{R}$;
• $a + b := +\infty$ when $a = +\infty$ and $b \ne -\infty$, or $b = +\infty$ and $a \ne -\infty$;
• $a + b := -\infty$ when $a = -\infty$ and $b \ne +\infty$, or $b = -\infty$ and $a \ne +\infty$;

and the two sums $(+\infty) + (-\infty)$ and $(-\infty) + (+\infty)$ are left undefined. Addition is commutative where defined, and

$-(a + b) = (-a) + (-b),$

each side being defined exactly when the other is: the excluded pairs $\{+\infty, -\infty\}$ are exchanged by $\nu$, and the three clauses above are exchanged accordingly.

Partial multiplication. For $a, b \in \overline{\mathbb{R}}$ the product $ab$ is defined by

• $ab$ = the field product, when $a, b \in \mathbb{R}$;
• $ab := +\infty$ when one of $a, b$ is $\pm\infty$, the other is $\ne 0$, and both are $> 0$ or both are $< 0$;
• $ab := -\infty$ when one of $a, b$ is $\pm\infty$, the other is $\ne 0$, and one is $> 0$ and the other $< 0$;

and every product with one factor $0$ and the other $\pm\infty$ is left undefined. The comparisons $> 0$ and $< 0$ here are taken in the order above, under which $+\infty > 0 > -\infty$.

Nothing else is defined. There is no subtraction, no division, no exponentiation and no absolute value on $\overline{\mathbb{R}}$ in this library; where such an expression is wanted it is written out in the two defined operations, and where a case falls in the undefined list the statement carries an explicit hypothesis saying so.

• $\overline{\mathbb{R}}$ is not a field, and not an ordered field. It has no additive inverse for $+\infty$: $(+\infty) + b$ is $+\infty$ whenever it is defined and is never $0$. So none of the field axioms (Complete ordered field (least-upper-bound property)) is available here, and no algebraic manipulation valid in $\mathbb{R}$ may be transported to $\overline{\mathbb{R}}$ without a separate justification.

• Why the excluded cases are excluded. The three defined clauses of each operation are exactly the cases in which the value is forced by the limiting behaviour of the sequences involved, and the excluded cases are exactly the ones in which it is not. For the product this is proved on the companion page: Null times divergent has no rule: $x_k = 1/k$ with $y_k = ck$ gives product limit $c$, and with $y_k = k^2$ gives divergence ↗ exhibits a null sequence and two sequences diverging to $+\infty$ whose products behave differently, so no value assigned to $0 \cdot (+\infty)$ could be compatible with products of limits. The same phenomenon rules out a value for $(+\infty) + (-\infty)$: with $a_k = k$ and $b_k = -k$ the sum is constantly $0$, while with $b_k = -2k$ it diverges to $-\infty$. Leaving them undefined is not squeamishness, it is the only option that keeps every later statement about limits true without a side condition hidden inside the arithmetic.

• This is the separate introduction that Conventions: $\sup \emptyset$, unbounded sets, and the extended reals points to. That remark refuses the conventions $\sup S = +\infty$ and $\inf \emptyset = +\infty$ inside $\mathbb{R}$, and records that the extended real line is introduced explicitly here, with its own order and its own partial arithmetic kept separate from $\mathbb{R}$ rather than quietly extending it. This is that introduction. The suprema and infima of Complete ordered field (least-upper-bound property), Greatest lower bound (infimum) and the whole suprema page remain real numbers with their nonempty and bounded hypotheses intact; what is new is a separate supremum operation, taken in $\overline{\mathbb{R}}$ and named as such, supplied by Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$.

• The symbols were already in circulation, and this definition does not change what they meant. Divergence to $+\infty$ and to $-\infty$ defines the single phrase "$x_k \to +\infty$" as an abbreviation for a condition on $(x_k)$, and says in as many words that it does not define an object named $+\infty$. That reading is still correct: nothing in Divergence to $+\infty$ and to $-\infty$ is restated or reinterpreted here, and 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 $L = \pm\infty$ is where the two are related, by a definition that quotes the old one rather than replacing it. Likewise the interval notation $(a, \infty)$ of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length is notation for a condition on one side, not an endpoint, and stays that way.

• Why the order is defined by three clauses rather than by a picture. The clauses are what the verifications above actually use, and they make the two facts that later proofs lean on immediate: every element is $\le +\infty$ and every element is $\ge -\infty$, with no case analysis at the point of use.

Let $A \subseteq \overline{\mathbb{R}}$ be any subset of the extended real line (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$, its order, and the arithmetic that is left undefined) and write $A_{\mathbb{R}} := A \cap \mathbb{R}$. Then $A$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$ (Upper bound, least upper bound, and strict upper bound), each unique, which we write $\sup A$ and $\inf A$ with the ambient set always $\overline{\mathbb{R}}$. Explicitly:

• $\sup A = +\infty$ if $+\infty \in A$, or if $A_{\mathbb{R}}$ is not bounded above in $\mathbb{R}$;
• $\sup A = -\infty$ if $+\infty \notin A$ and $A_{\mathbb{R}} = \emptyset$;
• $\sup A$ is the real supremum $\sup A_{\mathbb{R}}$ (Complete ordered field (least-upper-bound property)) if $+\infty \notin A$ and $A_{\mathbb{R}}$ is nonempty and bounded above in $\mathbb{R}$;

and dually, with $-\infty$ and $+\infty$ exchanged and "above" replaced by "below", for $\inf A$ (Greatest lower bound (infimum), Every nonempty set bounded below has an infimum).

Agreement. If $A \subseteq \mathbb{R}$ is nonempty and bounded above in $\mathbb{R}$ (Lower bound, bounded below, bounded set) then $\sup A$ computed in $\overline{\mathbb{R}}$ is the real number $\sup A$ of Complete ordered field (least-upper-bound property); if $A \subseteq \mathbb{R}$ is nonempty and bounded below then $\inf A$ computed in $\overline{\mathbb{R}}$ is the real number $\inf A$ of Every nonempty set bounded below has an infimum. In particular the notation is unambiguous on the sets for which the real supremum and infimum are defined, and $\sup \emptyset = -\infty$, $\inf \emptyset = +\infty$ in $\overline{\mathbb{R}}$.

No hypothesis is placed on $A$. This is exactly what the real supremum cannot do, and it is why every $\limsup$ statement on this page holds for every sequence rather than for bounded ones only. It is also not a weakening of the discipline this library keeps around suprema: the operation supplied here is a different operation, taken in a different ordered set, and the agreement clause records exactly where the two coincide.

• What makes this work is that $\overline{\mathbb{R}}$ has a top and a bottom. The three defining clauses of the order (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$, its order, and the arithmetic that is left undefined) put $+\infty$ above everything and $-\infty$ below everything, and every case above is settled by one of those two facts or by the least-upper-bound property of $\mathbb{R}$ applied to the real part. Nothing else about $\overline{\mathbb{R}}$ is used, and in particular no arithmetic is used at all.

• The two exceptional cases of FALSE: every subset of $\mathbb{R}$ has a supremum are not repaired, they are relocated. That false statement records that in $\mathbb{R}$ a set may fail to have a supremum, and it stays true; the discipline of Conventions: $\sup \emptyset$, unbounded sets, and the extended reals, which refuses to write $\sup S = +\infty$ inside $\mathbb{R}$, also stays in force. What is proved here is a statement about a different ordered set.

• The empty set is not an exception here, and that is the point of the ambient set. In $\mathbb{R}$ the empty set has no supremum, because every real is an upper bound and there is no least one. In $\overline{\mathbb{R}}$ every element is still an upper bound of $\emptyset$, but now there is a least one, namely $-\infty$. The two statements are about different ordered sets and neither contradicts the other.

• Where this is consumed. Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ needs the supremum of a tail range of an arbitrary real sequence, which may be unbounded, and then the infimum of the resulting family, which may contain $+\infty$; both are supplied here and by nothing earlier in the library. Fourteen items on this page depend on it, and five more on the companion page of examples.

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.

Let $(x_k)$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). For $n \in \mathbb{N}$ let

$T_n \;:=\; \{\, x_k : k \in \mathbb{N},\ k \ge n \,\} \subseteq \mathbb{R}$

be the $n$-th tail range of $(x_k)$, a nonempty subset of $\mathbb{R}$ since $x_n \in T_n$. Regard $T_n$ as a subset of $\overline{\mathbb{R}}$ (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$, its order, and the arithmetic that is left undefined) and put

$s_n \;:=\; \sup T_n \in \overline{\mathbb{R}}, \qquad i_n \;:=\; \inf T_n \in \overline{\mathbb{R}},$

the supremum and infimum taken in $\overline{\mathbb{R}}$, which exist for every $n$ and for every sequence by Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$. The limit superior and limit inferior of $(x_k)$ are then

$\limsup_{k} x_k \;:=\; \inf \{\, s_n : n \in \mathbb{N} \,\}, \qquad \liminf_{k} x_k \;:=\; \sup \{\, i_n : n \in \mathbb{N} \,\},$

again taken in $\overline{\mathbb{R}}$ and again existing by Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$, since $\{s_n : n \in \mathbb{N}\}$ and $\{i_n : n \in \mathbb{N}\}$ are subsets of $\overline{\mathbb{R}}$ on which no hypothesis is needed. Both are elements of $\overline{\mathbb{R}}$, and either may be $+\infty$ or $-\infty$. The notations $\limsup_{k \to \infty} x_k$, $\varlimsup_k x_k$ and $\overline{\lim}_k x_k$ all denote the first of them elsewhere; this library writes $\limsup_k x_k$.

Every quantity written here exists, and that is why the extended line was introduced. Each of the four operations above is an application of Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$ to a subset of $\overline{\mathbb{R}}$ carrying no hypothesis whatever. Written with the real supremum of Complete ordered field (least-upper-bound property) and the real infimum of Every nonempty set bounded below has an infimum instead, the definition would be available only for sequences that are bounded (Lower bound, bounded below, bounded set): $\sup T_n$ needs $T_n$ bounded above, and $\inf\{s_n\}$ needs $\{s_n\}$ nonempty, bounded below, and made of real numbers (Greatest lower bound (infimum)). None of those is automatic, and the discipline recorded in Conventions: $\sup \emptyset$, unbounded sets, and the extended reals forbids papering over the gap with a convention. The extended supremum is a different operation in a different ordered set, and it is total.

Values, when the sequence is bounded. If $(x_k)$ is bounded, say $|x_k| \le M$ for every $k$, then each $T_n$ is a nonempty subset of $\mathbb{R}$ bounded above by $M$ and below by $-M$, so by the agreement clause of Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$ each $s_n$ and each $i_n$ is the real supremum or infimum of $T_n$, and lies in $[-M, M]$. The family $\{s_n\}$ is then a nonempty set of reals bounded below by $-M$, so $\limsup_k x_k$ is likewise the real infimum of $\{s_n\}$ and lies in $[-M, M]$; dually for $\liminf_k x_k$. So for a bounded sequence both quantities are ordinary real numbers computed with the ordinary real supremum and infimum, and the extended line is doing no work. It is only for unbounded sequences that the values $\pm\infty$ occur.

• The order of the two operations is not symmetric and must be kept straight. $\limsup$ is an infimum of suprema and $\liminf$ a supremum of infima. Taking them in the other order gives $\sup_n \sup_{k \ge n} x_k = \sup_k x_k$ and $\inf_n \inf_{k \ge n} x_k = \inf_k x_k$, which are the extreme values of the whole sequence and carry no information about its behaviour at large indices. The point of the definition is that the inner operation looks at a tail and the outer one lets the tail recede.

• Why tails at all. Each $s_n$ is a bound on the whole tail from index $n$ on, so it forgets the first $n$ terms; letting $n$ grow forgets any fixed finite number of them. That is what makes $\limsup$ and $\liminf$ tail quantities in the sense of Convergence depends only on the tail, and it is the reason they can characterise convergence, which is itself a tail property.

• Neither quantity is a limit, and neither is claimed to be one. The symbols $\limsup$ and $\liminf$ are single pieces of notation for the two displayed expressions, exactly as "$x_k \to +\infty$" is a single abbreviation in Divergence to $+\infty$ and to $-\infty$. That the family $(s_n)$ does decrease to $\limsup_k x_k$ in a precise sense is a theorem, not part of this definition; the monotonicity half is The tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$, so the limit superior exists for every sequence.

• The tail ranges are sets, not sequences. $T_n$ is the range of the $n$-th tail, so repetitions and order are forgotten (Sequences of reals: bounded, eventually, frequently, tails, subsequences). That is harmless here, since a supremum depends only on the set of values, and it is what lets the whole definition be phrased with the order-theoretic operations of Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$ and nothing else.

Let $(x_k)$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), with tail ranges $T_n$ and extended tail bounds $s_n = \sup T_n$, $i_n = \inf T_n$ as in Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$.

1. Monotonicity of the extended bounds under inclusion. If $A \subseteq B \subseteq \overline{\mathbb{R}}$ (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$, its order, and the arithmetic that is left undefined) then $\sup A \le \sup B \qquad \text{and} \qquad \inf B \le \inf A,$ the four quantities being the extended bounds of Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$. No hypothesis is placed on $A$ or $B$; in particular $A$ may be empty.
2. The tail bounds are monotone. $T_m \subseteq T_n$ whenever $n \le m$, and hence $s_m \le s_n \qquad \text{and} \qquad i_n \le i_m \qquad (n \le m).$ In particular $s_{n+1} \le s_n$ and $i_n \le i_{n+1}$ for every $n$, and $i_n \le s_n$ for every $n$.
3. Existence. $\limsup_k x_k$ and $\liminf_k x_k$ exist in $\overline{\mathbb{R}}$ for every sequence of reals, bounded or not.