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CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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The limit inferior is the least subsequential limit in R‾

Statement

Let (xk) be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). Then lim inf⁡kxk∈SL⁡‾(x) and lim inf⁡kxk≤L for every L∈SL⁡‾(x) (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾, Convergence in R‾ and the extended subsequential limit set: L∈R‾ is an extended subsequential limit when some subsequence converges to L, or diverges to L=±∞).

So the extended subsequential limit set of any real sequence has a least element as well as a greatest one, and the two are lim inf⁡kxk and lim sup⁡kxk respectively (The limit superior is itself a subsequential limit in R‾ and is the greatest one). Every extended subsequential limit lies between them.

Facts & Assumptions

Given: A sequence (xk) of reals, and its reflection yk:=−xk.

[L1]

Reflection on R‾: a↦−a satisfies −(−a)=a and a≤b if and only if −b≤−a (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined).

[L2]
[L3]

For every real sequence the extended subsequential limit set is nonempty and has greatest element the limit superior (The limit superior is itself a subsequential limit in R‾ and is the greatest one).

[L5]

Scalar multiples of convergent sequences: zj→z in R implies czj→cz (Algebra of limits: sums, scalar multiples, products and quotients).

[L6]

Divergence to ±∞, and order reversal: zj>M is equivalent to −zj<−M, and M runs over all reals exactly when −M does (Divergence to +∞ and to −∞, Order is preserved by adding a constant and by adding inequalities).

Proof

technique · direct
1.1

Put yk:=−xk, a sequence of reals; then −yk=xk for every k, by the involution property of the reflection.

givenL1L4
1.2

Let L∈R‾ and let n:N→N be strictly increasing with (xnj) converging to L in R‾.

givenL4
1.3

By [L3] applied to the sequence (yk), the set SL⁡‾(y) is nonempty and has greatest element N0:=lim sup⁡kyk, and N0=−lim inf⁡kxk by [L2].

givenL2L3L7
2.1

The reflected subsequence (ynj)=(−xnj) converges to −L in R‾. If L is real this is the scalar rule with c=−1. If L=+∞ then for every real M there is J with xnj>M for all j≥J, hence ynj<−M for all such j; since −M runs over all reals as M does, ynj→−∞=−L. If L=−∞ the same argument with the inequalities exchanged gives ynj→+∞=−L.

step 1.2L1L4L5L6
3.1

Hence L∈SL⁡‾(x) implies −L∈SL⁡‾(y), the same index map serving. Applying that implication to the sequence (yk), whose reflection is (xk), gives conversely that N∈SL⁡‾(y) implies −N∈SL⁡‾(x). So SL⁡‾(x)={ −N:N∈SL⁡‾(y) }.

step 2.1step 1.1L1L4
4.1

Therefore −N0∈SL⁡‾(x), and −N0=−(−lim inf⁡kxk)=lim inf⁡kxk; and for any L∈SL⁡‾(x) the element −L lies in SL⁡‾(y), so −L≤N0 by maximality, whence lim inf⁡kxk=−N0≤L by order reversal. Thus lim inf⁡kxk is the least element of SL⁡‾(x).

step 3.1step 1.3L1L2∎

Remarks

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