# Projection in Hereditary C* subalgebra

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 $A_n$?

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Please check your definition of $f_{\varepsilon}$. – Andreas Thom Oct 29 '10 at 8:05
To Andreas Thom: Corrected. Thanks – Qingyun Nov 1 '10 at 21:12

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.