$K_1(A)$ is countable when A is separable C*-algebra

Question

We know that when A is a separable C*-algebra then $K_0(A)$ is countable.

How can I show that $K_1(A)$ is also countable?

Answer 1

Here's an idea for how one can show that $K_1(A)$ is countable without taking suspensions: Since the unitization of a separable $C^*$-algebra is separable, we may assume without loss of generality that $A$ is unital. If $u,v$ are unitaries in $A$ and $\|u-v\|<2$, then $u$ is homotopic to $v$ in $\mathcal U(A)$ (the unitary group of $A$). Since $A$ is separable, there is a countable set $\{u_n\}$ of unitaries in $A$ such that $\mathcal U(A)\subset \cup_nB(u_n,2)$ (the ball of radius $2$ centered at $u_n$). Thus there are countably many homotopy classes of unitaries in $A$. The same applies to $M_n(A)$, and in the limit, we see that $K_1(A)$ is countable.