TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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 set if and only if a net in the set converges to it

Statement

For $A \subseteq X$ and $p \in X$ , one has $p \in \overline{A}$ if and only if there is a net in $A$ converging to $p$ .

Facts & Assumptions

Given: A subset $A$ of a topological space $X$ and a point $p \in X$ .

[L1]

$p \in \overline{A}$ exactly when every neighbourhood of $p$ meets $A$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set).

[L2]

Finite intersections of neighbourhoods of $p$ are neighbourhoods of $p$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[L3]

A net converges exactly when it is eventually in every neighbourhood (Convergence and cluster points of a net in a topological space).

Proof

technique · constructive

1.1L1construct

Suppose $p \in \overline{A}$ . Let $E = {(N, a) : N \in N (p), a \in N \cap A}$ , ordered by $(N, a) ⪯ (M, b)$ when $M \subseteq N$ , and put $x_{(N, a)} = a$ .

1.2L1L3

Conversely, if a net $x$ in $A$ converges to $p$ , every neighbourhood $N$ of $p$ contains some eventual value $x_{d} \in A$ , so $N \cap A \neq \emptyset$ and $p \in \overline{A}$ .

2.1step 1.1L1L2

The index set is directed: for $(N, a), (M, b)$ , the set $N \cap M$ is a neighbourhood and meets $A$ ; for $c \in (N \cap M) \cap A$ , the pair $(N \cap M, c)$ is above both.

2.2step 1.1L3

Given a neighbourhood $N$ of $p$ , choose $(N, a) \in E$ . Every later pair has its second coordinate in a subset of $N$ , so $x$ is eventually in $N$ and therefore converges to $p$ .

A point lies in the closure of a set if and only if a net in the set converges to it

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources