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A C*-algebra $A$ is finite if $s^*s=1$ implies $ss^*=1$. A C*-algebra $A$ is AF if: for all $a_1,\ldots,a_n\in A$ and $\varepsilon>0$, there exists a finite-dimensional C*-subalgebra $B_n$ of $A$ and elements $b_1,\ldots,b_n\in B$ so that $\|a_i-b_i\|<\varepsilon$ for all $\varepsilon$.

How do you prove that every unital AF algebra is finite? I'm going through the following proof from Murphy's book, but I'm stuck at a detail. Here's how it goes so far.

Let $s\in A$. Since $A$ is AF we have $s=\lim s_n$, where $s_n$ belongs to some finite-dimensional subalgebra $A_n$ with $1\in A_n$. Now $1=s^*s = \lim s_n^*s_n$, so replacing $(s_n)$ by a subsequence we may assume $\|1-s_n^*s_n\|<1$ for all $n$, which implies that $s_n^*s_n$ is invertible in $A_n$. In particular $s_n^*$ is right-invertible, hence automatically left-invertible in $A_n$ because $A_n$ is finite-dimensional. Choosing $t_n\in A_n$ so that $t_ns_n^*=1$, Murphy concludes $s=\lim_{n\rightarrow\infty}t_n$ --- this is the part I don't understand. Perhaps the appropriate argument is $$ s = \lim s_n = \lim t_ns_n^*s_n = (\lim t_n)(\lim s_n^*s_n) = \lim t_n$$ but this assumes that $(t_n)$ converges, and I don't understand why. (A limit of units might not be a unit!) Once you manage to show that, it is easy to conclude $$ ss^* = (\lim t_n)(\lim s_n^*) = \lim t_ns_n^* = 1 $$ which is what we wanted to show.

Ehsaan
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1 Answers1

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As you wrote, you have that $s=\lim t_ns_n^*s_n$. You also have that $1=\lim s_n^*s_n$. Then $$ \|s-t_n\|\leq\|s-t_ns_n^*s_n\|+\|t_ns_n^*s_n-t_n\|=\|s-t_ns_n^*s_n\|+\|t_n(s_n^*s_n-1)\|. $$ The first term on the right goes to zero, and the second will if the sequence $\|t_n\|$ is bounded. The norms of the $t_n$ will stay bounded if the $s_n^*s_n$ are forced to be close enough to 1.

Martin Argerami
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  • Could you elaborate on the boundedness of $(||t_n||)_n$? I tried proving it using the Neumann series, i.e. if $||1-s_n^||<1$, then $||t_n||=||\left(s_n^\right)^{-1}||\leq\frac{1}{1-||1-s_n^*||}$, but did not succeed. – math-jl Nov 05 '23 at 12:46
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    There exists $n_1$ such that $|1-s_n^s_n|<\frac12$ when $n\geq n_1$. Then $|s_n^s_n|>\frac12$ and $|(s_n^s_n)^{-1}|<2$. Now you can use that there exists $n_2\geq n_1$ such that for $n\geq n_2$ $$ |s-t_ns_n^s_n|<\frac12. $$ Then, for $n\geq n_2$ $$ |t_n|=|t_ns_n^s_n(s_n^s_n)^{-1}|\leq2|t_ns_n^*s_n|\leq2\Big(|s|+\frac12\Big)=2|s|+1. $$ – Martin Argerami Nov 05 '23 at 20:13