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LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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The tail suprema of any real sequence are nonincreasing in R\overline{\mathbb{R}}, so the limit superior exists for every sequence

Statement

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

  1. Monotonicity of the extended bounds under inclusion. If ABRA \subseteq B \subseteq \overline{\mathbb{R}} (The extended real line R=R{,+}\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}, its order, and the arithmetic that is left undefined) then supAsupBandinfBinfA,\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 R\overline{\mathbb{R}} has a least upper bound and a greatest lower bound in R\overline{\mathbb{R}}, agreeing with the real supremum and infimum on nonempty sets bounded in R\mathbb{R}. No hypothesis is placed on AA or BB; in particular AA may be empty.
  2. The tail bounds are monotone. TmTnT_m \subseteq T_n whenever nmn \le m, and hence smsnandinim(nm).s_m \le s_n \qquad \text{and} \qquad i_n \le i_m \qquad (n \le m). In particular sn+1sns_{n+1} \le s_n and inin+1i_n \le i_{n+1} for every nn, and insni_n \le s_n for every nn.
  3. Existence. lim supkxk\limsup_k x_k and lim infkxk\liminf_k x_k exist in R\overline{\mathbb{R}} for every sequence of reals, bounded or not.

Claim 1 is the tool the rest of this page uses whenever two extended suprema are compared. It is proved here, from the definition of a least upper bound, rather than quoted from the suprema page, for the reason given in the remarks below.

Facts & Assumptions

Given: A sequence (xk)(x_k) of reals, its tail ranges Tn={xk:kn}T_n = \{x_k : k \ge n\}, and the extended bounds sn=supTns_n = \sup T_n, in=infTni_n = \inf T_n (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limit superior and limit inferior of a real sequence as infnsupknxk\inf_n \sup_{k \ge n} x_k and supninfknxk\sup_n \inf_{k \ge n} x_k in R\overline{\mathbb{R}}).

[L1]

Every subset of R\overline{\mathbb{R}} has a least upper bound and a greatest lower bound in R\overline{\mathbb{R}}, with no hypothesis on the subset (Every subset of R\overline{\mathbb{R}} has a least upper bound and a greatest lower bound in R\overline{\mathbb{R}}, agreeing with the real supremum and infimum on nonempty sets bounded in R\mathbb{R}).

[L2]

Least upper bound and greatest lower bound in a poset: supA\sup A is an upper bound of AA that is \le every upper bound of AA, and infA\inf A is a lower bound that is \ge every lower bound; each is unique when it exists (Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L4]

The order on N\mathbb{N} is total and transitive (Order on the natural numbers, \le is a linear order on N\mathbb{N}).

Proof

technique · direct
1.1

Let ABRA \subseteq B \subseteq \overline{\mathbb{R}} be arbitrary. By [L1] the four elements supA\sup A, supB\sup B, infA\inf A, infB\inf B of R\overline{\mathbb{R}} all exist and are uniquely determined.

givenL1L2
1.2

Let nmn \le m in N\mathbb{N}. Every element of TmT_m has the form xkx_k with kmk \ge m, and then knk \ge n by transitivity, so xkTnx_k \in T_n; hence TmTnT_m \subseteq T_n.

givenL4
1.3

For every nn the tail range TnT_n contains xnx_n, so inxni_n \le x_n because ini_n is a lower bound of TnT_n, and xnsnx_n \le s_n because sns_n is an upper bound of TnT_n; transitivity gives insni_n \le s_n.

givenL1L2L3
2.1

Since supB\sup B is an upper bound of BB and ABA \subseteq B, every element of AA is supB\le \sup B, so supB\sup B is an upper bound of AA; as supA\sup A is the least of the upper bounds of AA, this gives supAsupB\sup A \le \sup B. Dually infB\inf B is a lower bound of BB, hence of AA, and as infA\inf A is the greatest of the lower bounds of AA this gives infBinfA\inf B \le \inf A. Claim 1 is proved.

step 1.1L1L2
3.1

Applying claim 1 to the inclusion TmTnT_m \subseteq T_n valid for nmn \le m gives smsns_m \le s_n and inimi_n \le i_m; the special case m=n+1m = n + 1 gives sn+1sns_{n+1} \le s_n and inin+1i_n \le i_{n+1}. Together with insni_n \le s_n this is claim 2.

step 1.2step 1.3step 2.1
4.1

The families {sn:nN}\{s_n : n \in \mathbb{N}\} and {in:nN}\{i_n : n \in \mathbb{N}\} are subsets of R\overline{\mathbb{R}}, so [L1] applies to them with no hypothesis, and lim supkxk=inf{sn}\limsup_k x_k = \inf\{s_n\} and lim infkxk=sup{in}\liminf_k x_k = \sup\{i_n\} exist in R\overline{\mathbb{R}} for every sequence of reals. This is claim 3.

step 3.1L1L2

Remarks

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 53 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources