Unitary element in an AF $C^*$-algebra can be approximated by sequence of unitaries

Question

Let $A$ be a unital AF $C^*$-algebra.
Write $A=\overline{\bigcup_{k\in \mathbb{N}}A_k}$ where each $A_k$ is a unital (with the same unit of $A$) finite dimensional $C^*$ subalgebra. Suppose $u\in A$ is a unitary element.
I want to show that given $\epsilon>0$ there exists $k\in \mathbb{N}$ and $u_k \in A_k$ unitary such that $||u-u_k||<\epsilon$.

I'm not sure that it is true, but if I could show the above claim then it helps me to conclude $K_1(A)$ is zero for A unital AF $C^*$-algebra.

Any help will be appreciated!

Answer 1

Somehow, after I post here a question the solution comes to my mind...
So, I'll use the following argument:
almost unitaries are close to a unitary element
Now, it's enough to show that:
If $A$ is a unital $C^*$ algebra and $A=\overline{\bigcup_{k\in \mathbb{N}} A_k}$ ,where each $A_k$ is a unital (same unit of $A$) $C^*$ subalgebra, then for any unitary $u\in A$ and any $\epsilon>0$ there exist $n\in \mathbb{N}$ and unitary $v\in A_n$ such that $||u-v||<\epsilon$.
(I dropped the assumption that each $A_n$ is f.d., it's not relevant).
So, for small enough $\delta>0$ (which depends on $\epsilon$), find $n\in \mathbb{N}$ and $a\in A_n$ s.t. $||a-u||<\delta$ and $||a||\leq 1$ (it is possible because the unit ball of the union is dense in the unit ball of $A$).
Now, $||a^*a-1||=||a^*a-a^*u+a^*u-u^*u||\leq ||a^*|| \ ||a-u||+||a^*-u^*|| \ ||u||\leq 2\delta$.
Similarly, we can show $||aa^*-1||\leq 2\delta$.
So, if $\delta$ small enough then $\exists v\in A_n$ unitary s.t. $||a-v||< {\epsilon\over 2}$.
Consequently, $||v-u||<{\epsilon \over 2} +\delta<\epsilon$ (by requiring also $\delta<{\epsilon\over 2}$) , as desired.

Answer 2

Ignoring your original question: Note that $A = \lim_k A_k$ so $K_1(A) = \lim_k K_1(A_k)$. But $K_1(A_k) = 0$ since $K_1(M_n(\mathbb C)) = 0$ for each natural number $n$. This is a general argument that AF algebras have trivial $K_1$-group.