The question is in subject.
Update: See Andreas Thom's answer.
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1$\begingroup$ I guess, this can also be formulated as: do there exist separable $C^*$-algebras of arbitrarily large cardinality? $\endgroup$– Péter KomjáthCommented Nov 20, 2010 at 13:35
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6$\begingroup$ Do you mean isomorphism classes of separable C*-algebras? $\endgroup$– Qiaochu YuanCommented Nov 20, 2010 at 13:43
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1$\begingroup$ I am surprised that you changed the question quite a bit after my answer. What is your intention? If you do not believe the correctness of the answer, the appropriate way would be to put a comment below the answer, ask for more details or point out some problem. $\endgroup$– Andreas ThomCommented Nov 20, 2010 at 18:23
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1$\begingroup$ I have edited the question back towards the original form, and added a note to give credit to Andreas' remarks rather than merely copying them $\endgroup$– Yemon ChoiCommented Nov 20, 2010 at 18:40
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1$\begingroup$ I'm sorry for this. I just have found that I know the answer myself, and so I've changed the question after that. By this time I still haven't seen Your answer - it was just because I haven't scrolled the page down. Perhaps, I should have given a reference as soon as I have seen the answer You have posted. I truly confess in this fault of mine. Also, I have accepted Your answer as soon as I have seen it. If You can somehow look in the log of the post, You will see that the last modification just precedes the acceptance of your answer. Once more, I'm sorry, and I didn't meant any dishonest. $\endgroup$– Kolya IvankovCommented Nov 20, 2010 at 19:14
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2 Answers
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It is not so clear what you mean.
However, every separable $C^\ast$-algebra embeds in $B(\ell^2 \mathbb N)$. Hence, the isomorphism classes of separable $C^\ast$-algebras form a set.
answered Nov 20, 2010 at 13:35
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5$\begingroup$ And meanwhile, each isomorphism class is a proper class, since we can find an isomorphic copy of any given $C^*$-algebra by replacing any element of it with any other object in the set-theoretic universe not already appearing in the algebra. $\endgroup$ Commented Nov 20, 2010 at 14:56
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$\begingroup$ Once more, I apologize for not giving a reference and changing the question. Also, the way I tried to answer my question was still \emph{wrong}, because there still was written \emph{$C^*$-algebras} instead of \emph{isomorphism classes}. Thus, Your answer was the FULL answer anyway. In order to recover my fault, I promise to include Your name in the list of acknowledgments to my thesis. In fact, I was determined to do so anyway, because You and others have answered my questions before, and these answers were truly helpful, as well as this one. Again, I'm thankful and beg your pardon. $\endgroup$ Commented Nov 20, 2010 at 19:38
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Long comment:
It should be pointed out that actually much less structure than what Andreas Thom uses in his answer is needed to show that the isomorphism classes of separable $C^*$-algebras have a set of representatives:
The crucial fact is that
there is a set of representatives of isometry classes of separable metric spaces.
This is essentially because separable metric spaces are of bounded size (see Komjath's comment), namely of size at most $2^{\aleph_0}$.
Each separable metric space carries only a set of vector space structures over $\mathbb C$.
Each metric vector space over $\mathbb C$ only carries a set of binary and unary operations.
So, we obtain a set of representatives of the isomorphism classes of separable
$C^\ast$-algebras
without ever using the structure of $C^\ast$-algebras. Just the fact that they are separable metric spaces with a vector space structure over $\mathbb C$ and a fixed number of binary and unary operations.
Note that I have never assumed that the metric, the vector space structure, and the additional operations interact in any way whatsoever.
answered Nov 20, 2010 at 19:20
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$\begingroup$ Interesting from a $C^{\star}$-point of view: Any two isometric $C^\star$-algebras are isomorphic or anti-isomorphic as $C^\star$-algebras. $\endgroup$ Commented Nov 20, 2010 at 19:43
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$\begingroup$ I'm sorry for taking away the answer flag. Although this approach is more general than the answer given by Andreas Thom, I'm just new on MO, and didn't know that there may be only one answer. So I give the flag back to Andreas, because he was the first one, but still I regard Your answer as very helpful. $\endgroup$ Commented Nov 20, 2010 at 20:08