This is actually something in a paper but the author claimed it without proof.
Let x be a positive elment of norm 1 in a $C^\*-$algebra A, and Her(x) is the hereditary subalegbra generated by x. Given $\epsilon>0$,let $f\_\epsilon$ be thecontinuous function on R defined as follow:


$f\_\epsilon \equiv 0 \quad on \quad [-\infty,\epsilon/2]$


$f\_\epsilon \quad is \quad linear \quad on\quad [\epsilon/2,\epsilon]$


$f\_\epsilon \equiv 1 \quad on\quad [\epsilon, +\infty]$


So $f\_\epsilon$ increase to the identiy function on [0,1] when $\epsilon$ decrease to 0, and $Her(x)=\overline{\cup\_{\epsilon>0} f\_\epsilon(x)Af\_\epsilon(x)}$. Let p be a projection in Her(x). then how do we know that there must exist a $\epsilon$ such that $p\in \overline{f\_\epsilon(x)Af\_\epsilon(x)}$? Or more generally, Let A be the inductive limit of {$A\_n$}, and p is a projection in A,does it follow that p is actually in some $A\_n$?



Projection in Hereditary C* subalgebra
No. Let $A$ be the C$^\*$-algebra of compact operators on $\ell\_2$ and $x$ is the diagonal operator $(1,1/2,1/3,\ldots)$. Then, $f\_\epsilon(x)Af\_\epsilon(x)$ is a matrix algebra in the left upper corner. The rank one projection corresponding to any vector of infinite support does not belong to $\bigcup f\_\epsilon(x)Af\_\epsilon(x)$.
However, it is a standard fact that if $a$ is a positive element such that $\| p - a \| < 1/2$, then the spectrum of $a$ has a gap around $1/2$ and $q=\chi\_{[1/2,3/2]}(a)$ is a projection in $C^\*(a)$ such that $\| p - q \| < 1$, which implies that $p$ and $q$ are unitarily equivalent.


