Show that $K_0(A)$ is a countable group if $A$ is a unital, separable C* algebra

Question

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.

Answer 1

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.

Answer 2

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 $$ \mathcal D(A) = \lim_{n \to \infty} \frac{\mathrm{Proj(M_n(A))}} {\sim_{\mathrm {MvN}}}. $$