TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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If each $y_j$ is a subsequential limit of $(x_k)$ and $y_j \to y \in \mathbb{R}$ , then $y$ is a subsequential limit of $(x_k)$

Statement

Let $(x_k)$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and let $(y_j)$ be a sequence of reals with

$y_j \in \operatorname{SL}(x)$ for every $j \in \mathbb{N}$ (Subsequential limit of a real sequence, and the subsequential limit set), and
$y_j \to y$ for some $y \in \mathbb{R}$ (Limits and Cauchy sequences of reals).

Then $y \in \operatorname{SL}(x)$ .

In words: the subsequential limit set of a real sequence contains the limit of every convergent sequence of its own elements. When the topology of $\mathbb{R}$ arrives, that property is what it calls sequential closedness; that sequential closedness is in turn equivalent to closedness for subsets of $\mathbb{R}$ is a theorem there and not a matter of naming, and the half of that equivalence running from sequential closedness to closedness spends the axiom of countable choice. Here the property is stated and proved purely in terms of sequences, with no choice principle and no topological notion used or needed.

Facts & Assumptions

Given: A sequence $(x_k)$ of reals; a sequence $(y_j)$ of reals with $y_j \in \operatorname{SL}(x)$ for every $j$ ; and a real $y$ with $y_j \to y$ .

[L1]

Subsequential limits and convergence: $y_j \in \operatorname{SL}(x)$ means that some strictly increasing $m : \mathbb{N} \to \mathbb{N}$ has $x_{m_i} \to y_j$ ; convergence of a sequence of reals is the rational- $\varepsilon$ condition of Limits and Cauchy sequences of reals, and to establish convergence it suffices to produce a threshold for every real $\varepsilon > 0$ , by the remark of Sequences of reals: bounded, eventually, frequently, tails, subsequences (Subsequential limit of a real sequence, and the subsequential limit set, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L2]

Index maps: a strictly increasing $m$ satisfies $m_i \ge i$ , and an index map with $n_j < n_{j+1}$ for every $j$ is strictly increasing (A strictly increasing index map satisfies $n_k \ge k$ ).

[L3]

Well-ordering principle: every nonempty subset of $\mathbb{N}$ has a least element (The well-ordering principle).

[L4]

Recursion theorem: for a set $A$ , an element $a \in A$ and $f : A \to A$ there is a unique $g : \mathbb{N} \to A$ with $g_0 = a$ and $g_{j+1} = f(g_j)$ (The recursion theorem).

[L5]

Absolute value and the triangle inequality: $|a + b| \le |a| + |b|$ , and $|a| < c$ if and only if $-c < a < c$ for $c > 0$ (The triangle inequality, Basic properties of the absolute value).

[L6]

Canonical naturals and reciprocals: for a natural $q \ge 1$ the element $q \cdot 1_{\mathbb{R}}$ is positive and invertible, $\big(2(n+1)\big) \cdot 1_{\mathbb{R}} = 2 \big((n+1) \cdot 1_{\mathbb{R}}\big)$ , and $0 < a \le b$ gives $0 < 1/b \le 1/a$ ; moreover for every real $\eta > 0$ there is a natural $m \ge 1$ with $1/m < \eta$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean).

[L7]

The order on $\mathbb{N}$ is total, so any two indices have a common upper bound (Order on the natural numbers, $\le$ is a linear order on $\mathbb{N}$ ).

[L8]

Below every positive real lies a positive rational, which is how a convergence hypothesis stated for rational $\varepsilon$ is instantiated at a real threshold (The rationals embed densely in the reals).

Proof

technique · constructive

1.1

For $n \in \mathbb{N}$ put $q_n := 2(n+1)$ , a natural number $\ge 1$ . Then $q_n \cdot 1_{\mathbb{R}} = 2\big((n+1) \cdot 1_{\mathbb{R}}\big) > 0$ is invertible, $1/q_n > 0$ , and $1/q_n + 1/q_n = 2/q_n = 1/(n+1)$ .

givenL6algebra

1.2

By hypothesis $(y_j)$ converges to $y$ and every $y_j$ lies in $\operatorname{SL}(x)$ , so for each $j$ there is a strictly increasing $m$ with $x_{m_i} \to y_j$ .

givenL1

2.1

For every $n \in \mathbb{N}$ there is $k > n$ with $|x_k - y| < 1/(n+1)$ . Indeed, take a rational $\varepsilon_1$ with $0 < \varepsilon_1 < 1/q_n$ and instantiate the convergence $y_j \to y$ at $\varepsilon_1$ to obtain an index $J$ with $|y_J - y| < \varepsilon_1 < 1/q_n$ . Since $y_J \in \operatorname{SL}(x)$ , fix a strictly increasing $m$ with $x_{m_i} \to y_J$ , take a rational $\varepsilon_2$ with $0 < \varepsilon_2 < 1/q_n$ and an index $I$ with $|x_{m_i} - y_J| < \varepsilon_2$ for all $i \ge I$ , and let $i$ be an index at least as large as both $I$ and $n+1$ . Then $k := m_i$ satisfies $k = m_i \ge i \ge n + 1 > n$ and, by the triangle inequality applied to $x_k - y = (x_k - y_J) + (y_J - y)$ , $|x_k - y| \le |x_k - y_J| + |y_J - y| < 1/q_n + 1/q_n = 1/(n+1)$ .

step 1.1step 1.2L1L2L5L7L8

3.1

Define $f : \mathbb{N} \to \mathbb{N}$ by letting $f(n)$ be the least element of the set $G_n := \{\, k \in \mathbb{N} : k > n \text{ and } |x_k - y| < 1/(n+1) \,\}$ , which is nonempty by step 2.1. Then $f(n) > n$ and $|x_{f(n)} - y| < 1/(n+1)$ for every $n$ .

step 2.1L3construct

4.1

The recursion theorem applied to $\mathbb{N}$ , the element $f(0)$ and the function $f$ gives $n : \mathbb{N} \to \mathbb{N}$ with $n_0 = f(0)$ and $n_{j+1} = f(n_j)$ . Then $n_j < n_{j+1}$ for every $j$ , so $n$ is strictly increasing and $n_j \ge j$ ; moreover $|x_{n_{j+1}} - y| < 1/(n_j + 1) \le 1/(j+1)$ for every $j$ , using that $n_j + 1 \ge j + 1$ .

step 3.1L2L4L6

5.1

The subsequence $(x_{n_j})$ converges to $y$ : given a real $\varepsilon > 0$ , take a natural $m \ge 1$ with $1/m < \varepsilon$ ; every $j \ge m$ satisfies $j \ge 1$ , so step 4.1 applied at $j - 1$ gives $|x_{n_j} - y| < 1/j \le 1/m < \varepsilon$ . Producing such a threshold for every real $\varepsilon > 0$ establishes convergence, and $n$ is strictly increasing, so $y \in \operatorname{SL}(x)$ .

step 4.1L1L2L6discharge-construct∎

Remarks

The diagonal is where the two approximations are balanced. Step 2.1 spends half of the allowance $1/(n+1)$ on getting from $y$ to some $y_J$ and the other half on getting from $y_J$ to a term of $(x_k)$ arbitrarily far out. Splitting the allowance is what the natural number $q_n = 2(n+1)$ is for; nothing is halved in $\mathbb{R}$ , so no divisibility fact about the field is needed.
Choice is not used, for the same reason as in The limit superior is itself a subsequential limit in $\overline{\mathbb{R}}$ and is the greatest one: the index map is built by taking least elements of explicitly described nonempty subsets of $\mathbb{N}$ (The well-ordering principle) and applying The recursion theorem. The subsequences witnessing $y_J \in \operatorname{SL}(x)$ are used one at a time inside a single existence argument, never selected simultaneously for all $J$ .
The hypothesis that $y$ be real is essential to the statement, not to the proof technique. $\operatorname{SL}(x)$ is a set of real numbers by Subsequential limit of a real sequence, and the subsequential limit set, so a limit outside $\mathbb{R}$ could not be asserted to lie in it. The extended set $\overline{\operatorname{SL}}(x)$ always contains its greatest and least elements (The limit superior is itself a subsequential limit in $\overline{\mathbb{R}}$ and is the greatest one, The limit inferior is the least subsequential limit in $\overline{\mathbb{R}}$ ), which is the corresponding statement at the two ends.
Consequence: a nonempty $\operatorname{SL}(x)$ that is bounded above contains its supremum. Write $\sigma := \sup \operatorname{SL}(x)$ , a real number under those two hypotheses. For each $j$ the element $\sigma - 1/(j+1)$ fails to bound $\operatorname{SL}(x)$ above (Epsilon characterisation of the supremum), so some $y_j \in \operatorname{SL}(x)$ satisfies $\sigma - 1/(j+1) < y_j \le \sigma$ , and such a sequence converges to $\sigma$ by the reciprocal Archimedean property and the squeeze (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , The squeeze theorem). The theorem then puts $\sigma$ in $\operatorname{SL}(x)$ . This is a second route to the finite case of The limit superior is itself a subsequential limit in $\overline{\mathbb{R}}$ and is the greatest one, available only when the set is already known to be nonempty and bounded above; the route taken there is direct and covers the infinite cases too, which this one cannot.

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Sources

Subsequential limit (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (3.7) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.4 (standard reference, not scraped)
D. Auroux, Math 112 notes, 5 March 2019 (standard reference, not scraped)

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