Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $A_n$?

share|improve this question
2  
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
add comment

1 Answer

up vote 6 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.