Let $S$ be a non-empty set of simple C$^*$-subbalgebras of a C$^*$-algebra $A$. Let us also suppose that $S$ is upwards-directed and that the union of all element of $S$ is dense in $A$. Then $A$ is simple.
I can't show this. Could you tell me how to show this?
Write $S=\{A_j\}$. Then, for any $a\in A$, we have $a=\lim a_j$ with $a_j\in A_j$.
Now fix $I\subset A$, a nonzero ideal. As $A_j\cap I$ is an ideal of $A_j$, we either have $A_j\cap I=0$ or $A_j\cap I=A_j$. We also have that $\{A_j\cap I\}$ is an increasing net of ideals of $A$.
For $a\in I$ and $a=\lim a_j$ with $a_j\in A_j$, we can use an approximate unit $\{e_k\}$ of $I$ to get $a=\lim_kae_k=\lim_k\lim_ja_je_k$. As $a_je_k\in A_jI=A_j\cap I$ (proof here), we get that $I=\overline{\bigcup_jA_j\cap I}$. This shows that, eventually, $A_j\cap I=A_j$. Thus $$ I=\overline{\bigcup_jA_j\cap I}=\overline{\bigcup_jA_j}=A. $$