TheoremStatement: AI-adaptedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The squeeze theorem

Statement

Let $(a_k)$ , $(x_k)$ and $(b_k)$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with

$a_k \le x_k \le b_k \quad \text{eventually},$

and suppose $(a_k)$ and $(b_k)$ both converge to the same real $L$ (Limits and Cauchy sequences of reals). Then $(x_k)$ converges to $L$ .

The middle sequence is not assumed to converge; that is the point of the theorem, and it is why the squeeze is a tool for establishing convergence rather than for computing a limit already known to exist.

Facts & Assumptions

Given: Sequences $(a_k)$ , $(x_k)$ , $(b_k)$ of reals, an index $K_0 \in \mathbb{N}$ with $a_k \le x_k \le b_k$ for every $k \ge K_0$ , and a real $L$ such that $(a_k)$ converges to $L$ and $(b_k)$ converges to $L$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L1]

Convergence, quantified over rational $\varepsilon > 0$ (Limits and Cauchy sequences of reals).

[L2]

Absolute value: for $c > 0$ , $|u| < c$ if and only if $-c < u < c$ (Basic properties of the absolute value).

[L3]

Order arithmetic in $\mathbb{R}$ : adding a constant preserves $<$ and $\le$ ; $u < v \le w$ and $u \le v < w$ both give $u < w$ ; and $\le$ is transitive (Order is preserved by adding a constant and by adding inequalities, Complete ordered field (least-upper-bound property), Ordered field).

[L4]

The order on $\mathbb{N}$ is total and transitive, so three indices $K_0, K_1, K_2$ admit an index $K$ with $K \ge K_0$ , $K \ge K_1$ and $K \ge K_2$ , and then $k \ge K$ gives $k \ge K_0$ , $k \ge K_1$ , $k \ge K_2$ ( $\le$ is a linear order on $\mathbb{N}$ ).

Proof

technique · direct

1.1

Let $\varepsilon > 0$ be rational. By convergence of $(a_k)$ there is $K_1$ with $|a_k - L| < \varepsilon$ for all $k \ge K_1$ , and by convergence of $(b_k)$ there is $K_2$ with $|b_k - L| < \varepsilon$ for all $k \ge K_2$ .

2.1

By [L2], $|a_k - L| < \varepsilon$ gives $-\varepsilon < a_k - L$ , that is $L - \varepsilon < a_k$ , for all $k \ge K_1$ ; and $|b_k - L| < \varepsilon$ gives $b_k - L < \varepsilon$ , that is $b_k < L + \varepsilon$ , for all $k \ge K_2$ .

step 1.1L2L3

3.1

Choose $K$ with $K \ge K_0$ , $K \ge K_1$ and $K \ge K_2$ . For every $k \ge K$ the hypothesis gives $a_k \le x_k \le b_k$ , so $L - \varepsilon < a_k \le x_k \le b_k < L + \varepsilon$ , hence $L - \varepsilon < x_k < L + \varepsilon$ , hence $-\varepsilon < x_k - L < \varepsilon$ , hence $|x_k - L| < \varepsilon$ .

step 2.1givenL2L3L4choose

4.1

The rational $\varepsilon > 0$ was arbitrary, so for every rational $\varepsilon > 0$ there is an index $K$ beyond which $|x_k - L| < \varepsilon$ ; that is, $(x_k)$ converges to $L$ .

step 3.1L1∎

Remarks

The proof is direct from the definition and does not route through Limits preserve non-strict inequalities. It could not: that lemma assumes the middle sequence converges, which is what is being proved here.
Both bounding sequences must have the same limit. With $a_k = -1$ , $b_k = 1$ and any $(x_k)$ taking values in $[-1, 1]$ the hypothesis $a_k \le x_k \le b_k$ holds and nothing whatever follows, since a bounded sequence need not converge (FALSE: every bounded sequence converges).
The most common use is with $a_k = -c_k$ and $b_k = c_k$ for a null sequence $(c_k)$ : if $|x_k| \le c_k$ eventually and $c_k \to 0$ , then $x_k \to 0$ . That special case also follows from A null sequence times a bounded sequence is null whenever $x_k$ is presented as a product of a null sequence and a bounded one, but the squeeze needs no such presentation.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 43 results over 12 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

J. K. Hunter, An Introduction to Real Analysis, Ch. 3 (standard reference, not scraped)
Squeeze theorem (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.4 (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)