By F.D or AF $C^{*}$ algebra,we mean finite dimensional or approximately finite dimensional $C^{*}$ algebra.

Let $A$ be a unital commutative $C^{*}$ algebra with the property that for every unital $C^{*}$ subalgebra $B\subset A$, $Spec(B)$ is embeddable in $Spec(A)$.

What can be said about $A$? Is $A$ F.D or at least $AF$? does every $AF$ algebra satisfy the above property?

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I don't know what are the motivation for the formulaton in terms of $C^*$ algebras, but you are essentially asking for hausdorff compact/locally compact spaces such all their compact/locally compact quotients can be embeded into them.

Finite topological space do have this property. But the Hilbert cube also have this property (and the corresponding $C^*$-algebra really don't look approximately finite dimensional ).

Edit: I realize there is a small gap in the previous argument: If you consider non unital $C^*$-algebras of $C(X)$ then they do not correspond to quotients of $X$, but to proper quotients of an open subspace of $X$, so the topological interpretation I gave does not work. This being said, the Hilbert cube is still a valid counter exemple: any separated quotient of an open subspace is metrizable and separable and hence can be embedded in the Hilbert cube.

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  • $\begingroup$ Thanks for the very interesting answer. Some more question: By quotient, do you mean an arbitrary quotient by an equivalent relation or a particular collapsing $X/A$? Moreover, what do you mean by "separated" quotient? The last question what about $X$= the cantor set or arbitrary totally disconnected compact metric space? I mean is every AF algebra satisfies in the statment under my question? $\endgroup$ – Ali Taghavi Oct 18 '14 at 15:30
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    $\begingroup$ By quotient I mean by an aribitrary equivalence relation, but such that the quotient is separated (i.e. by a closed equivalence relation). That is what a unital subalgebra of C(X) is: a compact topological space endowed with a surjection from X. By compactness this is automatically a quotient map. For your other question, I don't know. $\endgroup$ – Simon Henry Oct 18 '14 at 16:14

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