Approximable operator

Let $X$ and $Y$ be normed spaces over the same scalar field and let $\mathcal{B}(X,Y)$ be the space of bounded linear operators with the operator norm (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators, A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). Write

$\mathcal{F}(X,Y):=\{R\in \mathcal{B}(X,Y):R(X)\text{ admits an ordered basis of finite length}\}$

for the set of bounded finite-rank operators (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). An operator $T\in \mathcal{B}(X,Y)$ is approximable when $T$ lies in the closure of $\mathcal{F}(X,Y)$ in the operator-norm metric (Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$, 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).

The closure is an epsilon statement. The zero operator lies in $\mathcal{F}(X,Y)$, since its range $\{0\}$ admits the empty ordered basis, so $\mathcal{F}(X,Y)$ is nonempty and the metric-space description of the closure applies: $T$ is approximable if and only if for every real $\epsilon >0$ there is a bounded finite-rank operator $R$ with $\mathrm{\parallel }T-R\mathrm{\parallel }<\epsilon$. No choice principle is used for this equivalence.

Approximable operators are compact, under countable choice. Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)), let $Y$ be a Banach space (Banach space) and let $T$ be approximable. Choosing for every $n\in \mathbb{N}$ a bounded finite-rank $R_n$ with $\mathrm{\parallel }T-R_n\mathrm{\parallel }<1\mathrm{/}(n+1)$ is a countable selection from nonempty sets, so such operators exist; each $R_n$ is compact (Bounded finite rank operators are compact) and $\mathrm{\parallel }{R}_n-T\mathrm{\parallel }\to 0$, so $T$ is compact by the norm-limit theorem (Norm limit of compact operators is compact, Compact linear operator).

No converse is asserted here. The statement that every compact operator into $Y$ is approximable is the approximation-property question for the target $Y$; it is not a consequence of the definition and is not claimed. The companion page records the implication that holds when $Y$ has the approximation property, and the distinction between compact and approximable operators for a general Banach target is left open.