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How do I show that the set of unitary equivalence classes of projections is countable in a unital separable $C^*$-algebra?

So I tried to show that the set of unitary elements in $C^*$-algebra is countable, but it was not successful.

Thanks.

Andrew39
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1 Answers1

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First, a comment: every unital C$^*$-algebra has uncountably many unitaries (simply because $\mathbb C $ does).

For your problem, if the projections $p $ and $q $ are not unitarily equivalent, then $\|p-q\|\geq1$; see this. So, uncountably many pairwise non-unitarily equivalent elements would all be at distance $1$ from each other, contradicting separability.

Martin Argerami
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  • So if we assume that $||p-q||<1$, then we have unitary elements $u$ such that $upu^=q$. Then how do we show such $u$ is countable in a separable C-algebra? Are you saying there's no such $u$? – Andrew39 May 02 '15 at 20:23
  • No one is saying that there are countably many unitaries; as I said, the only C$^*$-algebra with countably many unitaries is the zero algebra, which has none, What I'm saying that if you have uncountably many unitary equivalence classes of projections, then the algebra is non-separable. – Martin Argerami May 02 '15 at 20:35
  • oh I thought 'countably many unitaries' and 'countable unitary equivalence classes of projections' eventually meant the same thing. I guess it's not. And I will try to prove the contrapostion of the statement. Thanks. – Andrew39 May 02 '15 at 20:48
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    Yes, there is no relation. In $M_3 (\mathbb C) $, for instance, there are uncountably many unitaries, but four equivalence classes of projections. In $\mathbb C $ there are uncountably many uniunitaries but two equivalence classes of projections. – Martin Argerami May 02 '15 at 21:14
  • Thanks :) I will try to figure it out. Have a nice day. – Andrew39 May 02 '15 at 23:05
  • Martin Argerami, I think that the zero algebra has one unitary element, since $0^∗0=00^∗=0$, which is a unity in this algebra. – User Jun 01 '15 at 16:38
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    True, but it is standard to ignore that algebra; I shouldn't have mentioned it. It contributes nothing and only makes statements harder. For instance, in that algebra every element has empty spectrum, when it is considered an accepted result that every element in every C$^*$-algebra has nonempty spectrum. – Martin Argerami Jun 01 '15 at 17:47
  • Interesting, I had never thought about this fact and it's clearly true... – User Jun 03 '15 at 19:45
  • I don't understand the last statement: if we have uncountably many equivalence classes of projections,the algebra is not separable.Why is it true? – math112358 Jul 06 '18 at 04:12
  • Because you take a representative of each class, and they will all be at distance at least $1$ from each other. – Martin Argerami Jul 06 '18 at 04:44