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  "body": "\n## Definition\n\nLet $A$ be a C*-algebra.\nThen $A$ is *approximately finite-dimensional*, or *AF*, if it has finite-dimensional C\\*-subalgebras $A_1 \\subseteq A_2 \\subseteq \\cdots$ whose union is dense in $A$.\nEquivalently, $A$ is isomorphic to the C\\*-algebraic direct limit of a sequence $A_1 \\to A_2 \\to \\cdots$ of finite-dimensional C\\*-algebras and $*$-homomorphisms.\n\n## Remarks\n\nEvery finite-dimensional C\\*-algebra is isomorphic to a direct sum $M_{n_1}(\\C) \\oplus \\cdots \\oplus M_{n_k}(\\C)$ of [[matrix rings|0FHVPRAE60]].\nAn AF algebra in this sense is separable.\nSome authors call a nonseparable C\\*-algebra AF when every finite subset can be approximated by elements of a finite-dimensional C\\*-subalgebra.\n\nExample: the compact operators $K(H)$ on a separable infinite-dimensional Hilbert space $H$ with orthonormal basis $(e_k)$.\nLet $A_n$ be the operators that map $\\operatorname{span}(e_1, \\ldots, e_n)$ into itself and vanish on its orthogonal complement; then $A_n \\cong M_n(\\C)$ and $A_n \\subseteq A_{n+1}$.\nThe union is dense because the finite-rank operators are dense in $K(H)$, and $P_n T P_n \\to T$ for compact $T$, where $P_n$ is the projection onto the span.\n\nNon-example: $C([0, 1])$ is not AF.\nA finite-dimensional commutative C\\*-algebra is spanned by projections, and the only projections in $C([0,1])$ are $0$ and $1$, since $[0,1]$ is connected.\nSo every finite-dimensional C\\*-subalgebra lies in the constants $\\C 1$, which are not dense.\n",
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