TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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Norm limit of compact operators is compact

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $X$ be a normed space, let $Y$ be a Banach space (Banach space), and let $T_{n} : X \to Y$ , $n \in N$ , be compact operators (Compact linear operator) with $∥ T_{n} - T ∥ \to 0$ for a bounded linear operator $T : X \to Y$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ). Then $T$ is compact.

Facts & Assumptions

[A1]

$S : X \to Y$ is compact exactly when $\overline{S ({\overline{B}}_{X})}$ is compact, where ${\overline{B}}_{X} = {x \in X : ∥ x ∥ \leq 1}$ (Compact linear operator); and $∥ S x ∥ \leq ∥ S ∥ ∥ x ∥$ 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 compact metric space $(M, d)$ has a finite subcover of every open cover; in particular, if every point of $M$ has a ball of the family ${M \cap B (y, ε) : y \in M}$ containing it, then finitely many of these balls cover $M$ (Open cover, subcover, compact metric space, and compact subset of a metric space, Open ball, closed ball and sphere in a metric space). Selecting one index from each of finitely many nonempty index sets is possible without choice (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

[A3]

For nonempty $A$ in a metric space, $\overline{A} = {y : d (y, A) = 0}$ (claim 1 of The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset); so for $y \in \overline{A}$ and real $δ > 0$ there is $a \in A$ with $d (y, a) < δ$ (Open ball, closed ball and sphere in a metric space).

[A4]

A subset $A$ of a metric space is totally bounded when for every real $ε > 0$ there are finitely many points $f_{0}, \dots, f_{m} \in A$ with $A \subseteq ⋃_{i \leq m} B (f_{i}, ε)$ (Finite $ε$ -net and totally bounded metric space, Open ball, closed ball and sphere in a metric space).

[A5]

A closed subset of a complete metric space is complete (claim 2 of Closed subspaces of complete metric spaces are complete; the converse under countable choice), and a Banach space is a complete normed space (Banach space); a complete and totally bounded metric space is compact under ${A C}_{ω}$ (A complete, totally bounded metric space is compact, proved from countable choice used exactly once).

Proof

technique · direct

Given: ${A C}_{ω}$ , a normed space $X$ , a Banach space $Y$ , compact operators $T_{n} : X \to Y$ with $∥ T_{n} - T ∥ \to 0$ , and $C_{n} : = \overline{T_{n} ({\overline{B}}_{X})}$ .

1.1A1

For every real $ε > 0$ there is $n$ with $∥ T - T_{n} ∥ < ε$ : this is exactly the convergence $∥ T_{n} - T ∥ \to 0$ in the norm metric of $B (X, Y)$ .

1.2A3

For every $c \in C_{n}$ and every real $δ > 0$ there is $x \in {\overline{B}}_{X}$ with $∥ c - T_{n} x ∥ < δ$ , because $c \in \overline{T_{n} ({\overline{B}}_{X})}$ , the set $T_{n} ({\overline{B}}_{X})$ is nonempty, and [A3] applies to $A = T_{n} ({\overline{B}}_{X}) \subseteq Y$ .

1.3A1

Each $C_{n}$ is compact by [A1].

1.4A3A4algebra

Whenever $A$ is a totally bounded nonempty subset of a metric space, its closure is totally bounded: given $ε > 0$ , [A4] gives finitely many points $f_{0}, \dots, f_{m} \in A$ with $A \subseteq ⋃_{i \leq m} B (f_{i}, ε / 2)$ ; for $y \in \overline{A}$ there is $a \in A$ with $d (y, a) < ε / 2$ by [A3], and $a \in B (f_{i}, ε / 2)$ for some $i$ , so $d (y, f_{i}) < ε$ ; hence the same finite set, whose points lie in $\overline{A}$ , is an $ε$ -net for $\overline{A}$ .

1.5A5

The set $\overline{T ({\overline{B}}_{X})}$ is closed in the complete space $Y$ , hence a complete metric space by [A5].

2.1step 1.1step 1.2step 1.3A1A2algebra

Fix a real $ε > 0$ and choose $n$ with $∥ T - T_{n} ∥ < ε / 4$ by [step 1.1]. The family of balls $C_{n} \cap B (T_{n} x, ε / 4)$ , $x \in {\overline{B}}_{X}$ , covers $C_{n}$ by [step 1.2]; since $C_{n}$ is compact by [step 1.3], finitely many indices $x_{0}, \dots, x_{m} \in {\overline{B}}_{X}$ satisfy $C_{n} \subseteq ⋃_{i \leq m} B (T_{n} x_{i}, ε / 4)$ , and one may pick these finitely many indices by [A2]. For every $x \in {\overline{B}}_{X}$ the point $T_{n} x$ lies in $C_{n}$ , so some $i$ has $∥ T_{n} x - T_{n} x_{i} ∥ < ε / 4$ ; writing $T x - T x_{i} = (T_{n} x - T_{n} x_{i}) + (T - T_{n}) (x - x_{i})$ and using $∥ x ∥, ∥ x_{i} ∥ \leq 1$ gives $∥ T x - T x_{i} ∥ \leq ∥ T_{n} x - T_{n} x_{i} ∥ + ∥ T - T_{n} ∥ ∥ x - x_{i} ∥ < ε / 4 + ε / 2 < ε$ ; thus ${T x_{0}, \dots, T x_{m}}$ is a finite $ε$ -net for $T ({\overline{B}}_{X})$ with centres in $T ({\overline{B}}_{X})$ , and $T ({\overline{B}}_{X})$ is totally bounded.

3.1step 1.4step 2.1

By [step 1.4] the closure $\overline{T ({\overline{B}}_{X})}$ is totally bounded as well, its finite nets having centres in that closure.

4.1step 1.5step 3.1A1A5∎

The space $\overline{T ({\overline{B}}_{X})}$ is complete by [step 1.5] and totally bounded by [step 3.1]; by [A5] it is compact, and therefore $T$ is compact by [A1].

Norm limit of compact operators is compact

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources