LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

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.

Uniformly convergent images of a compact set have compact union

Statement

Let $(X, d_{X})$ and $(Y, d_{Y})$ be metric spaces, let $K \subseteq X$ be compact, and let $f, f_{n} : X \to Y$ be continuous for every integer $n \geq 1$ . Suppose that for every $ε > 0$ there is $N \geq 1$ such that $d_{Y} (f_{n} (x), f (x)) < ε$ for all $n \geq N$ and all $x \in K$ . Then $W = f [K] \cup ⋃_{n = 1}^{\infty} f_{n} [K]$ is compact in $Y$ . The empty-set case is included, and the proof uses no choice principle.

Facts & Assumptions

Given: Metric spaces, a compact subset $K$ , continuous maps, and uniform convergence on $K$ as stated.

[F1]

A compact metric subspace has a finite subcover from every family of ambient open sets covering it; conversely this ambient-cover condition implies intrinsic compactness (Open cover, subcover, compact metric space, and compact subset of a metric space, A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

[F2]

A continuous map sends a compact subset of a metric space to a compact subset (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[F3]

In a metric space, open sets contain an open ball about each of their points, and open balls are open (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Proof

technique · open-cover argument with a uniform neighborhood of the limit image

1.1givenF1F2cases

If $K = \emptyset$ , then $W = \emptyset$ and it is compact. Otherwise let $V$ be any family of open subsets of $Y$ covering $W$ . By [F2], $f [K]$ is compact, so [F1] gives finitely many $V_{1}, \dots, V_{q} \in V$ covering $f [K]$ . Set $O = ⋃_{i = 1}^{q} V_{i}$ , an open set containing $f [K]$ .

2.1step 1.1F1F3algebra

Form the family of all balls $B_{Y} (y, r)$ with $y \in f [K]$ , $r > 0$ , and $B_{Y} (y, 2 r) \subseteq O$ . This family covers $f [K]$ by openness and [F3], so [F1] gives a finite subcover with centers $y_{1}, \dots, y_{s}$ and radii $r_{1}, \dots, r_{s}$ . Since $f [K] \neq \emptyset$ , $s \geq 1$ ; let $δ = \min_{i} r_{i} > 0$ . If $z \in Y$ and $d_{Y} (z, y) < δ$ for some $y \in f [K]$ , choose $i$ with $d_{Y} (y, y_{i}) < r_{i}$ ; then $d_{Y} (z, y_{i}) < δ + r_{i} \leq 2 r_{i}$ , so $z \in B_{Y} (y_{i}, 2 r_{i}) \subseteq O$ . Thus every point within distance $δ$ of some point of $f [K]$ lies in $O$ .

3.1givenstep 2.1algebra

Uniform convergence supplies $N \geq 1$ such that $d_{Y} (f_{n} (x), f (x)) < δ$ for every $x \in K$ and every $n \geq N$ . Since $f (x) \in f [K]$ , step 2.1 implies $f_{n} [K] \subseteq O$ for every $n \geq N$ .

4.1step 1.1step 3.1F1F2algebra∎

For each of the finitely many indices $1 \leq n < N$ , [F2] makes $f_{n} [K]$ compact, so [F1] supplies a finite subfamily of $V$ covering it. Together with $V_{1}, \dots, V_{q}$ from step 1.1, these finitely many members cover $f [K]$ , all $f_{n} [K]$ with $n \geq N$ , and each remaining $f_{n} [K]$ . Hence every open cover of $W$ has a finite subcover, so $W$ is compact by [F1]. The proof uses only finite subcovers and a finite minimum, not a choice axiom.

Depends on

Used by

Finite sampling reconstruction bridges compact-uniform raw quadrature limits Lemma

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Kovachki et al., Neural Operator: Learning Maps Between Function Spaces With Applications to PDEs (standard reference, not scraped)