A linearly independent set about approximate units Let $B$ be a C*-algebra and $\{b_{i}\}_{i=1}^{n}\subset B$ be linearly independent. If we take $\{f_{k}\}\subset B$ which is approximate units, then can we find a large $k$, such that $\{b_{1}f_{k}, b_{2}f_{k},...,b_{n}f_{k}\}$ is a linearly independent set?
 A: I try to answer my own question, is there anything wrong in my answer? 
Since $\{f_{k}\}$ be the approximate units of $B$, then give arbitrary $\varepsilon>0$, we can find a $k$ such that 
$$||(b_{i}f_{k}-b_{i})-(\sum\mu_{i}b_{i}f_{k}-\sum\mu_{i}b_{i})||<\varepsilon.$$ 
Here, the $\mu_{i}$ denotes the arbitrary complex numbers that less than or equal to zero (this is necessary). This implies $$||b_{i}f_{k}-\sum\mu_{i}b_{i}f_{k}||>||b_{i}-\sum\mu_{i}b_{i}||-\varepsilon.$$
Assume that $\sum\limits_{i=1}^{n}\lambda_{i}b_{i}f_{k}=0$. If $\lambda_{i}\neq 0$, for all $i=1, 2,..., n$, then there exists  a $\lambda_{i}$ satisfies that $|\lambda_{i}|=$max$\{|\lambda_{1}|,...,|\lambda_{n}|\}$. Without loss of generality, we can set the $\lambda_{n}$ be the ``absolute maximum" (i.e., the $\lambda_{i}$ mentioned above). Hence, $b_{n}f_{k}=\sum\limits_{i=1}^{n-1}\frac{\lambda_{i}}{\lambda_{n}}b_{i}f_{k}$. Indeed,
$$||b_{n}f_{k}-\sum\limits_{i=1}^{n-1}\frac{\lambda_{i}}{\lambda_{n}}b_{i}f_{k}||>||b_{i}-\sum\limits_{i=1}^{n}\frac{\lambda_{i}}{\lambda_{n}}b_{i}||-\varepsilon.$$
As $b_{i}$'s are linearly independent, it follows the distance $d(b_{n},$ span$\{b_{1},...,b_{n-1}\})>0$. Then, take the $\varepsilon<d(b_{n},$ span$\{b_{1},...,b_{n-1}\}),$ we have $$||b_{n}f_{k}-\sum\limits_{i=1}^{n-1}\frac{\lambda_{i}}{\lambda_{n}}b_{i}f_{k}||>0$$
which leads a contradiction.
