LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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.

A point lies in the closure of $A \subseteq R$ iff some sequence in $A$ converges to it, so a subset of $R$ is closed iff it is sequentially closed

Statement

Let $A \subseteq R$ and $x \in R$ , with closure as in Interior, closure, boundary and exterior of a subset of $R$ and sequences and convergence as in Sequences of reals: bounded, eventually, frequently, tails, subsequences and Limits and Cauchy sequences of reals. Then

$x \in \overline{A} ⟺ there is a sequence (a_{k})_{k \in N} with a_{k} \in A for every k and a_{k} \to x .$

Consequently $A$ is closed if and only if it is sequentially closed: whenever a sequence with all its terms in $A$ converges, its limit lies in $A$ .

The right-to-left direction is choice free; the left-to-right direction spends ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Producing a sequence from a point of the closure requires selecting one point of $A$ from each of the countably many sets $N_{1 / (k + 1)} (x) \cap A$ , and this library has no canonical rule for that selection, so the axiom of countable choice is invoked explicitly at step 2.2 and nowhere else.

Facts & Assumptions

Given: A subset $A \subseteq R$ and a real $x$ . Sequences are functions on $N$ , which contains $0$ , so a sequence is $(a_{k})_{k \in N}$ and the radii used below are $1 / (k + 1)$ rather than $1 / k$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

$\overline{A}$ is exactly the set of adherent points of $A$ , that is, of points every neighbourhood of which meets $A$ ; and $A$ is closed exactly when $A = \overline{A}$ (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L2]

$N_{ε} (x) = {y \in R : ∣ y - x ∣ < ε}$ for real $ε > 0$ (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L3]

$(a_{k}) \to x$ means: for every rational $ε > 0$ there is $K \in N$ with $∣ a_{k} - x ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

Strictly between any two reals lies a rational; in particular for every real $ε > 0$ there is a rational $q$ with $0 < q < ε$ (The rationals embed densely in the reals).

[L5]

Reciprocal Archimedean property: for every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean).

[L6]

Canonical naturals: $n \cdot 1_{R} > 0$ for $n \geq 1$ and $m \leq n$ in $N$ gives $m \cdot 1_{R} \leq n \cdot 1_{R}$ (Canonical naturals are positive and strictly increasing); a positive element has a positive inverse and $0 < a \leq b$ gives $0 < 1 / b \leq 1 / a$ (Inverses of positives are positive, and reciprocation reverses order). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

[L7]

Countable choice: for every family $(X_{k})_{k \in N}$ of nonempty sets there is a function $f$ with domain $N$ such that $f (k) \in X_{k}$ for every $k$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1assume-hypL3

For the right-to-left implication, assume $a_{k} \in A$ for every $k \in N$ and $a_{k} \to x$ , and let $ε > 0$ be an arbitrary real.

1.2assume-hypL1L2L6

For the left-to-right implication, assume $x \in \overline{A}$ ; then for every $k \in N$ the radius $1 / (k + 1)$ is a positive real and the set $X_{k} : = N_{1 / (k + 1)} (x) \cap A$ is nonempty, because $x$ is an adherent point of $A$ by [L1].

2.1step 1.1L1L2L3L4choose

Fix a rational $q$ with $0 < q < ε$ by [L4], and then $K \in N$ with $∣ a_{k} - x ∣ < q$ for all $k \geq K$ by [L3]; in particular $∣ a_{K} - x ∣ < q < ε$ , so $a_{K} \in N_{ε} (x) \cap A$ and that intersection is nonempty. As $ε$ was an arbitrary positive real, $x$ is an adherent point of $A$ , hence $x \in \overline{A}$ by [L1].

2.2step 1.2L2L7choose

Apply [L7] to the family $(X_{k})_{k \in N}$ of step 1.2 and fix $f$ with $f (k) \in X_{k}$ for every $k$ ; putting $a_{k} : = f (k)$ gives a sequence with $a_{k} \in A$ and $∣ a_{k} - x ∣ < 1 / (k + 1)$ for every $k \in N$ .

3.1step 2.2L3L5L6choose

That sequence converges to $x$ : let $ε > 0$ be rational, fix by [L5] a natural $n \geq 1$ with $1 / n < ε$ , and put $K : = n - 1$ , a natural number since $n \geq 1$ ; for every $k \geq K$ one has $k + 1 \geq n \geq 1$ , hence $0 < 1 / (k + 1) \leq 1 / n < ε$ by [L6], and therefore $∣ a_{k} - x ∣ < 1 / (k + 1) < ε$ .

4.1step 2.1step 2.2step 3.1

Step 2.1 gives the implication from right to left and steps 2.2 and 3.1 give it from left to right, so $x \in \overline{A}$ holds exactly when some sequence with all terms in $A$ converges to $x$ .

4.2step 2.1step 2.2step 3.1L1

Sequential closedness: if $A$ is closed and a sequence with all terms in $A$ converges to some $y$ , then $y \in \overline{A}$ by step 2.1 and $\overline{A} = A$ by [L1], so $y \in A$ ; conversely, if every convergent sequence with terms in $A$ has its limit in $A$ , then any $y \in \overline{A}$ is the limit of the sequence produced by steps 2.2 and 3.1, hence lies in $A$ , so $\overline{A} \subseteq A$ , and with $A \subseteq \overline{A}$ this gives $A = \overline{A}$ , that is, $A$ is closed.

5.1step 4.1step 4.2∎

Both assertions of the statement are proved, namely the sequential description of the closure in step 4.1 and the equivalence of closedness with sequential closedness in step 4.2.

Remarks

Where the choice is spent, and why it cannot be avoided here. Step 2.2 is the only appeal to The Axiom of Countable Choice ( ${A C}_{ω}$ ). A canonical selection would require a rule picking a distinguished element of an arbitrary nonempty subset of $R$ , and $R$ carries no well-ordering that this library has constructed, so this library has no such rule to offer. Contrast Every subset of an at most countable set is at most countable and A nonempty set is at most countable iff it is a surjective image of $N$ , where the selection is from subsets of $N$ and the least element is canonical.
The choice is genuinely confined to one direction. Step 2.1 selects a single rational $q$ and a single index $K$ for one $ε$ at a time, and finitely many selections need no choice principle. So "the limit of a convergent sequence in a closed set lies in the set" is a theorem of ZF, and only the production of a sequence out of a point of the closure is not.
The indices start at $0$ . Since $N$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), the shrinking radii are $1 / (k + 1)$ and not $1 / k$ ; the latter is undefined at $k = 0$ . The threshold in step 3.1 is $K = n - 1$ for the same reason, and $n \geq 1$ is exactly what makes $K$ a natural number.

A point lies in the closure of $A \subseteq R$ iff some sequence in $A$ converges to it, so a subset of $R$ is closed iff it is sequentially closed

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A point lies in the closure of A⊆RA⊆R iff some sequence in AA converges to it, so a subset of RR is closed iff it is sequentially closed

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A point lies in the closure of $A \subseteq R$ iff some sequence in $A$ converges to it, so a subset of $R$ is closed iff it is sequentially closed