Every C*-algebra admits an approximate unit. I'm working my way through Murphy's, C-algebras and Operator Theory and I have a question concernig the proof that every C-algebra admits an approximate identity.
Let A be an arbitrary C*-algebra. We denote by $\Lambda$ the set of all positive elements a in A such that $||a||<1$. It can be shown that $\Lambda$ is a poset under the partial order of $A_{sa}$ (set of all hermitian elements in A). Partial order is defined by $a\leq b \iff b-a\geq 0$, where $a\geq 0$ means a is positive, in other words hermitian element such that $\sigma(a)\subseteq [0,\infty \rangle$. It is also shown $\Lambda$ is a upwards-directe set, so we can defined a net with $\Lambda$ as it's enumeration set. The proof goes on as follows:
How does $||f-\delta gf||\leq \epsilon$ follow ?. Also the choice od $\lambda_0$ is confusing to me, basically the rest of the proof I cant wrap my head around.
 A: $f\in C_{0}(\Omega)$ with $||f||_{\sup}=||a||<1$. Let $\omega\in\Omega$
be arbitrary. If $\omega\in K$, then $g(\omega)=1$. Therefore
\begin{eqnarray*}
\left|f(\omega)-\delta g(\omega)f(\omega)\right| & = & (1-\delta)|f(\omega)|\\
 & \leq & 1-\delta\\
 & < & \varepsilon.
\end{eqnarray*}
If $\omega\notin K$, then $|f(\omega)|<\varepsilon$ and $0\leq g(\omega)\leq1$.
Therefore
\begin{eqnarray*}
\left|f(\omega)-\delta g(\omega)f(\omega)\right| & \leq & |f(\omega)|\cdot|1-\delta g(\omega)|\\
 & \leq & |f(\omega)|\\
 & < & \varepsilon.
\end{eqnarray*}
It follows that $||f-\delta gf||_{\sup}\leq\varepsilon$.
A: By construction $\|f\|\leq1$.
If $t\in K$,
$$
|f(t)-\delta g(t)f(t)|=|f(t)|\,|1-\delta |\leq \varepsilon. 
$$
If $t\not\in K$ then $|f(t)|\leq\varepsilon$. As $0\leq g(t)\leq 1$,
$$
|f(t)-\delta g(t)f(t)|=|f(t)|\,|1-\delta g(t)|\leq |f(t)|\leq\varepsilon.
$$
The choice of $\lambda_0$ gives you,
$$
\|a-\lambda_0 a\|=\|\varphi^{-1}(f)-\varphi^{-1}(\delta g)\varphi^{-1}(f)\|=\|\varphi^{-1}(f-\delta g f)\|
=\|f-\delta g f\|\leq\varepsilon.
$$
A: for $w\in K$, $g(w) =1\implies \lvert f-\delta g f\rvert =(1-\delta) \lvert f\rvert <\epsilon $ (since Gelfand transformation is an isometry), for $$ w\in K^{c}, \lvert f(w) <\rvert \epsilon, \lvert g(w) \rvert \leq 1$$.
