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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 (ACω)). Let X be a normed space, let Y be a Banach space (Banach space), and let Tn:XY, nN, be compact operators (Compact linear operator) with TnT0 for a bounded linear operator T:XY (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: xkx iff d(xk,x)0 in R). Then T is compact.

Facts & Assumptions

[A1]

S:XY is compact exactly when S(BX) is compact, where BX={xX:x1} (Compact linear operator); and SxSx 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 {MB(y,ε):yM} 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, 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 yA and real δ>0 there is aA 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 f0,,fmA with AimB(fi,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 ACω (A complete, totally bounded metric space is compact, proved from countable choice used exactly once).

Proof

technique · direct

Given: ACω, a normed space X, a Banach space Y, compact operators Tn:XY with TnT0, and Cn:=Tn(BX).

1.1

For every real ε>0 there is n with TTn<ε: this is exactly the convergence TnT0 in the norm metric of B(X,Y).

A1
1.2

For every cCn and every real δ>0 there is xBX with cTnx<δ, because cTn(BX), the set Tn(BX) is nonempty, and [A3] applies to A=Tn(BX)Y.

A3
1.3

Each Cn is compact by [A1].

A1
1.4

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 f0,,fmA with AimB(fi,ε/2); for yA there is aA with d(y,a)<ε/2 by [A3], and aB(fi,ε/2) for some i, so d(y,fi)<ε; hence the same finite set, whose points lie in A, is an ε-net for A.

A3A4algebra
1.5

The set T(BX) is closed in the complete space Y, hence a complete metric space by [A5].

A5
2.1

Fix a real ε>0 and choose n with TTn<ε/4 by [step 1.1]. The family of balls CnB(Tnx,ε/4), xBX, covers Cn by [step 1.2]; since Cn is compact by [step 1.3], finitely many indices x0,,xmBX satisfy CnimB(Tnxi,ε/4), and one may pick these finitely many indices by [A2]. For every xBX the point Tnx lies in Cn, so some i has TnxTnxi<ε/4; writing TxTxi=(TnxTnxi)+(TTn)(xxi) and using x,xi1 gives TxTxiTnxTnxi+TTnxxi<ε/4+ε/2<ε; thus {Tx0,,Txm} is a finite ε-net for T(BX) with centres in T(BX), and T(BX) is totally bounded.

step 1.1step 1.2step 1.3A1A2algebra
3.1

By [step 1.4] the closure T(BX) is totally bounded as well, its finite nets having centres in that closure.

step 1.4step 2.1
4.1

The space T(BX) 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].

step 1.5step 3.1A1A5

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