2
$\begingroup$

I am reading Rordam's paper "On the Structure of Simple C$^{*}$-Algebras Tensored with a UHF-Algebra." I came across the following passage:

enter image description here

I am having trouble understanding the $(\impliedby)$ direction in (2.3) and also the statement (2.4).

The $(\implies)$ direction in (2.3) is easy: if $a<\alpha$, then, given $e\in GL(A)$, we have $$ \|x-e\|\leq\|x-x_{a}\|+\|x_{a}-e\|\leq a+\|x_{a}-e\|, $$ which implies that $$ \|x_{a}-e\|\geq \|x-e\|-a\geq\alpha-a>0. $$ Thus, $x_{a}\notin\overline{GL(A)}$.

I would really appreciate some help with the other direction and (2.4). It is probably quite simple and I must have forgotten something, but I have been stumped for a while.

EDIT:

As André S. points out in the comments, the reverse implication in (2.3) follows from (2.4), so

all that remains is to prove (2.4).

Thank you.

$\endgroup$
  • $\begingroup$ I don't understand why $x_a\in A$, would you like to explain for me? $\endgroup$ – C.Ding Sep 30 '18 at 8:19
  • 1
    $\begingroup$ @C.Ding $f_a(0) = 0$ ensures that $x_a \in A$. Indeed, you approximate $f_a$ by polynomials with vanishing constant term. Back to the question: (2.4) implies (2.3) because if $a \geq \alpha$ you approximate $x_a$ by $x_{a+\delta}$ for some $\delta > 0$ and by (2.4) $x_{a+\delta} = u \lvert x_{a+\delta} \rvert$ for a unitary $u$. Now, approximate by $u (\lvert x_{a+\delta} \rvert + \delta')$ which is invertible. $\endgroup$ – user42761 Sep 30 '18 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.