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edited so that answer gives credit where it is due
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Yemon Choi
Yemon Choi
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The question is in subject.

I suppose theUpdate: See Andreas Thom's answer to be "yes", because every separable $C^*$-algebra could be represented as a subalgebra of the algebra $B(H)$ of bounded operators on (the) countably generated Hilbert space $H$, and the subalgebras of $B(H)$ should form a set.

Is there any flaw in this reasoning?

The question is in subject.

I suppose the answer to be "yes", because every separable $C^*$-algebra could be represented as a subalgebra of the algebra $B(H)$ of bounded operators on (the) countably generated Hilbert space $H$, and the subalgebras of $B(H)$ should form a set.

Is there any flaw in this reasoning?

The question is in subject.

Update: See Andreas Thom's answer.

Probable answer added
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Kolya Ivankov
Kolya Ivankov
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The question is in subject. Although 

I assumesuppose the answer to be "no", it would have been very nice if it was "yes", because every separable $C^*$-algebra could be represented as a subalgebra of the algebra $B(H)$ of bounded operators on (the) countably generated Hilbert space $H$, and the subalgebras of $B(H)$ should form a set.

Is there any flaw in this reasoning?

The question is in subject. Although I assume the answer to be "no", it would have been very nice if it was "yes".

The question is in subject. 

I suppose the answer to be "yes", because every separable $C^*$-algebra could be represented as a subalgebra of the algebra $B(H)$ of bounded operators on (the) countably generated Hilbert space $H$, and the subalgebras of $B(H)$ should form a set.

Is there any flaw in this reasoning?

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Kolya Ivankov
Kolya Ivankov
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Do separable $C^*$-algebras form a set?

The question is in subject. Although I assume the answer to be "no", it would have been very nice if it was "yes".