LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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.

Bounded finite rank operators are compact

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field and let $T : X \to Y$ be a bounded linear operator (A bounded linear operator between normed spaces) whose range $T (X)$ admits an ordered basis of finite length (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). Then $T$ is compact (Compact linear operator).

Facts & Assumptions

[A1]

$T$ is compact exactly when $\overline{T ({\overline{B}}_{X})}$ is compact, where ${\overline{B}}_{X} = {x \in X : ∥ x ∥ \leq 1}$ (Compact linear operator); the operator norm satisfies $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ for every $x$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

[A2]

A subspace $W$ of a normed space $V$ admitting an ordered basis of finite length is a closed subset of $V$ (A finite-dimensional normed subspace is closed, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis), and its closed unit ball ${w \in W : ∥ w ∥ \leq 1}$ is compact (The closed unit ball is compact if and only if the normed space is finite-dimensional).

[A3]

A continuous image of a compact set is compact (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism), and scalar multiplication is continuous on a normed space (Vector addition and scalar multiplication are continuous in a normed space).

[A4]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded); a closed subset of a compact topological space is a compact subset (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact).

Proof

technique · direct

Given: Normed spaces $X, Y$ over one scalar field and a bounded linear $T : X \to Y$ whose range $R : = T (X)$ admits an ordered basis of finite length.

1.1A1algebra

Every point of $T ({\overline{B}}_{X})$ lies in $R$ , and $∥ T x ∥ \leq ∥ T ∥$ for every $x \in {\overline{B}}_{X}$ by [A1]; writing $s : = ∥ T ∥$ , this says $T ({\overline{B}}_{X}) \subseteq s {\overline{B}}_{R}$ , where ${\overline{B}}_{R} = {v \in R : ∥ v ∥ \leq 1}$ .

1.2A2

By [A2] the set ${\overline{B}}_{R}$ is compact and $R$ is closed in $Y$ .

2.1step 1.2A3A4

The set $s {\overline{B}}_{R}$ is the image of the compact set ${\overline{B}}_{R}$ under the continuous map $v \mapsto s v$ , so it is a compact subset of $Y$ by [A3]; it is therefore closed in $Y$ by [A4].

3.1step 1.1step 2.1A4

Since $T ({\overline{B}}_{X}) \subseteq s {\overline{B}}_{R}$ by [step 1.1], the closure $\overline{T ({\overline{B}}_{X})}$ is contained in the closed set $s {\overline{B}}_{R}$ ; being a closed subset of the compact space $s {\overline{B}}_{R}$ , it is compact by [A4].

4.1step 3.1A1∎

By [A1] compactness of $\overline{T ({\overline{B}}_{X})}$ is exactly compactness of $T$ , so $T$ is compact.

Bounded finite rank operators are compact

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources