Let $A$ be a separable C*-algebra and $\{\pi_n\}_{n\in\mathbb N}$ be a sequence of finite dimensional representations of $A$ such that for every $a\neq 0$ there is $n$ such that $\pi_n(a)\neq 0$. In other words, $\bigoplus_n \pi_n $ is faithful. (such $A$ is said to be residually finite dimensional.)
Let $B$ be unital separable and $\{\pi_n\}_{n\in\mathbb N}$ be such sequence. Assume every $\pi_n$ is irreducible and has dimension $k(n)$. Also assume each representation in $\{\pi_n\}_{n\in\mathbb N}$ repeats infinitely many times.
Suppose $l:\mathbb N\to \mathbb N$ is an increasing sequence.
For each $n$, define $\phi_n:M_{k(n)}\to M_{k(n)}(B)$ by viewing $M_{k(n)}$ as a subalgebra of $M_{k(n)}(B)$, since $B$ is unital. Then define $\psi_n^1=\phi_n\circ \pi_n:B\to M_{k(n)}(B)$.
By defining $h_1:b\mapsto \text{diag}(b,\psi_1^1(b),...,\psi_{l(1)}^1(b))$ , $h_1$ is a homomorphism from $B=M_1(B)=M_{I(1)}(B)$ to $M_{I(2)}(B)$, where $I(1)=1$ and $I(2)=1+k(1)+...+k(l(1))$.
Suppose $h_m:M_{I(1)}(B)\to M_{I(2)}(B)$ and $\psi_n^m:M_{I(m)}(B)\to M_{I(m)k(n)}(B)$ are defined. Define $\psi_n^{m+1}=\psi_n^1\otimes 1_{I(m+1)}:M_{I(m+1)}(B)\to M_{k(n)I(m+1)}(B)$ and define $h_{m+1}:b\mapsto \text{diag}(b,\psi_1^{m+1}(b),...,\psi_{l(m+1)}^{m+1}(b))$ as homomorphism from $M_{I(m+1)}(B)$ to $M_{I(m+2)}(B)$, where $I(m+2)=I(m+1)(1+k(1)+...+k(l(m+1)))$.
Then we have unital homomorphisms $h_m:M_{I(m)}(B)\to M_{I(m+1)}(B)$ and therefore we can define the inductive limit C*-algebra $A=\lim_{n\to\infty}M_{I(m)}(B)$.
Proposition. $A$ is simple.
This is from An introduction to the classification of amenable C*-algebras, p158. This definition is too complicated for me to understand.
Since $h_m$ maps $b$ to the matrix with $b$ on its diagonal, $h_m$ is a norm preserving injection. Therefore there are $M_{I(n)}(B)\simeq B_n\leq A$ with $\mathbf{closure}[\bigcup B_n]=A$. It only needs to show if ideal $I$ intersects with large enough $M_{I(n)}(B)$ then $M_{I(n)}(B)\subseteq I$. However, the definition is so complicated that I don't even know where to start.
As in the proof the book gives, it says,
For $0\neq y\in M_{l(n)}$, there is $\pi_m$ such that $\pi_m\otimes \text{id}_{M_{I(n)}}(y)\neq 0$. Let $n+l\geq m$, by considering $h_{n+l}\circ h_{n+l-1}\circ...\circ h_{n+1}$ we may write $h_{n,n+l}(y)=H(y)\oplus (\pi_m\otimes \text{id}_{M_{I(n)}}\otimes \text{id}_{M_J})$ for some positive integer $J$ and some homomorphism $H$. Since $\psi_m^1\otimes \text{id}_{M_{I(n)}}\otimes \text{id}_{M_J}(y)$ is a nonzero element in $M_{I(n+l)}$, the ideal generated by $\psi_m^1\otimes \text{id}_{M_{I(n)}}\otimes \text{id}_{M_J}(y)$ contains $M_{I(n+l)}$.
I do not understand it at all.
I get it now.
The point is, if $y\in M_{n}(B)$ contains nonzero constants among its elements, the ideal generated by $y$ is the whole $M_{n}(B)$.
Suppose $y\in M_{I(n)}(B)$ is nonzero, by the assumption, there is a sufficiently large $m$ such that $\pi_m\otimes 1_{I(n)}(y)\neq 0$, which means $\psi_m^n(y)$ has nonzero constant elements. Then $h_{m-1}\circ ...\circ h_{n}(y)$ therefore generates $M_{I(m)}(B)\supseteq M_n(B)$.
As simple as it is.