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I want to show that if $A$ is a unital, separable C* algebra then $K_0(A)$ is a countable group. To do this it is enough to show that for every projection $p\in A$, there is a projection $ q\in D$ such that $||p-q||<1$ (here $D$ is the countable dense det in $A$). This will show that $p$ and $q$ are homotopy equivalent which will show that they are unitarily equivalent which wil again imly that these projections are Murray von Neumann equivalent.

Parish
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  • What is $D$? $ $ – Martin Argerami Apr 02 '17 at 02:37
  • @MartinArgerami, $D$ is the countable dense set in $A$. Sorry, I forgot to metion that. – Parish Apr 02 '17 at 03:36
  • I don't follow. If you get to choose $D$, you could just take $q=p$. If $D$ is given, what guarantee is there that it has projections at all? – Martin Argerami Apr 02 '17 at 06:12
  • @MartinArgerami, $A$ is a separable C* algebra, so it has some countable dense set. Now To show that $K_0(A)$ is countable we will have to show that there are only countably many equivalence classes of projections in the semigroup $\mathcal{D}(A)$. So my idea was that if for any projection $p\in A$ we can find a projection $q\in \textbf{the dense set}$ which is 'close' enough to $p$ then $q$ and $p$ will be Murray von Neumann equivalent. And so every projection $p\in A$ will lie in the equivalence class of some projection from a countable dense set. But as you say this does not seem to work. – Parish Apr 02 '17 at 06:22
  • @MartinArgerami, So how do you suggest should I show that $K_0(A)$ is countable for a unital, separable C* algebra $A$? – Parish Apr 02 '17 at 06:25
  • I wrote a short answer below. – Martin Argerami Apr 02 '17 at 17:24
  • Is it possible to make a similar statement if A is not assumed to be unital but still separable? – Miep May 04 '20 at 17:24

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Since projections which are at less than distance $1$ are unitarily equivalent, the Murray-von Neumann equivalence classes of projections lie in disjoint balls, all within the ball of radius two (we are thinking of balls of radius one around elements of the unit ball). Now the fact that $A$ is separable gives us only countably many balls of a given radius within a ball. So we only have countably many classes in $A$. The same reasoning applies to $M_n(A)$. As a countable union of countable sets is finite, the total of all clases in $K_0(A)$ is countable.

Martin Argerami
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First, there are countably many equivalence classes of projections in $A$. Let $\{a_n\}_{\mathbb N}$ be dense in $A$. Then, define an injection $$ \phi : \frac {\mathrm{Proj}(A)}{ \sim_{\mathrm{MvN}} }\ \to \ \mathbb N, $$ where $\phi([p]) = n$ such that $\lVert p-a_n \rVert < \frac 1 2$. Then $\phi$ is well-defined and injective.

Now, one notes that the semi-group of projections is $$ \mathcal D(A) = \lim_{n \to \infty} \frac{\mathrm{Proj(M_n(A))}} {\sim_{\mathrm {MvN}}}. $$

  • Yeah, thanks man. But I think you need to be slightly careful with your definition of $\phi$. It may well happen that for a projection $p\in A$, we have $\Vert p-a_1\Vert, \Vert p-a_2\Vert <1$. This will lead to an ambiguity in your definition of $\phi$. But nevertheless, we may let the set ${C_1,C_2,\cdots }$ denote the M-V equivalence classes of projections where $C_i$ is the equivalence class of projections $p\in A$ such that $\Vert p-a_i\Vert <\frac{1}{2}$. So there may be repetitions in the above set of equivalece classes. But this shows that $\mathcal{D}(A)$ is countable. – Parish Apr 02 '17 at 14:20
  • Yes, that $\phi$ is not well defined. –  Apr 02 '17 at 17:52
  • But anyway modulo that small caveat, the answer is fine. – Parish Apr 02 '17 at 17:54
  • Okay, so just for the sake of completeness: What you say is the same as making a surjection $\mathbb N \to \mathrm{Proj(A)} / \sim$. –  Apr 02 '17 at 18:02
  • No, I don't mean that. I mean to say that the set ${C_1,C_2\cdots }$ may have repetitions and may just be say, ${C_{n_1}, C_{n_2}\cdots}$ for some positive numbers $n_1<n_2<\cdots$ – Parish Apr 02 '17 at 18:05