I am working through Rordam's book on C$^{*}$-Algebras and I am having trouble with exercise 6.7 (iii). I have reduced the problem to the following.
We are given an inductive sequence
$$ A_{1}\overset{\varphi_{1}}{\longrightarrow}A_{2}\overset{\varphi_{2}}{\longrightarrow}A_{3}\overset{\varphi_{3}}{\longrightarrow}\cdots\longrightarrow (A,\{\mu_{n}\}_{n=1}^{\infty}), $$ where the $\mu_{n}\colon A_{n}\to A$ are the natural maps. I am trying to prove the following:
If all of the $\varphi_{n}$'s are injective and each of the $A_{n}$'s is not unital, then $A$ is not unital.
Assuming that the $\varphi_{n}$'s are injective, I've managed to deduce that the $\mu_{n}$'s are all injective. I also know that $A=\overline{\bigcup_{n=1}^{\infty}\mu_{n}(A_{n})}$. Aiming at a contradiction, I assumed that $A$ is unital with unit $1_{A}$. Then, if $1_{A}=\mu_{n}(a_{n})$ for some $n\in\mathbb{N}$ and $a_{n}\in A_{n}$, it follows that for all $c_{n}\in A_{n}$, $\mu_{n}(a_{n}c_{n})=\mu_{n}(a_{n})\mu_{n}(c_{n})=\mu_{n}(c_{n})=\mu_{n}(c_{n})\mu_{n}(a_{n})=\mu_{n}(c_{n}a_{n})$. Thus, by injectivity, $a_{n}c_{n}=c_{n}=c_{n}a_{n}$, which implies that $a_{n}=1_{A_{n}}$, contradicting that $A_{n}$ is not unital.
Therefore, I just have to show that we cannot have $1_{A}=\lim_{m}\mu_{n_{m}}(a_{n_{m}})$, where the $m\uparrow \infty$. However, I do not know how to rule out this case. Any help would be very much appreciated.
Thank you.
The invertible elements of a $C^*$ algebra form an open subset. This means that if $\mu_{n_m}(a_{m})\to\Bbb1$ you must have that eventually $\mu_{n_m}(a_m)$ are all invertible in $\mathcal A$. So lets look at one $a$ so that $\mu_n(a)$ is invertible.
The $C^*$ algebra generated by $\mu_n(a)$ consists of all convergent power series with $\mu_n(a)$ and $\mu_n(a)^*=\mu_n(a^*)$ terms. Since $\mu_n$ is an injective $*$-morphism it is an isometry and a power series of $\mu_n(a),\mu_n(a^*)$ converges iff the same power series in $a, a^*$ converges. Now $\Bbb1$ must lie in the $C^*$ algebra generated by $\mu_n(a)$ and thus must be the image of element in the $C^*$ algebra generated by $a$. Your argument from before then shows that $A_n$ had to be unital.