Any II$_1$-factor contains the hyperfinite II$_1$-factor I want to show that any II$_1$-factor $N$ contains the hyperfinite II$_1$-factor $R$. The hyperfinite factor is constructed to be the weak closure of $A:= \bigotimes_{n = 1}^\infty \text{Mat}_{2 \times 2}(\mathbb{C})$ in the GNS-space $L^2(A)$.
My idea was to construct a sequence $A_1 \subset A_2 \subset \cdots$ of subalgebras of $N$ such that $A_i \cong \bigotimes_{j = 1}^i \text{Mat}_{2 \times 2}(\mathbf{C})$ as algebras, and then use that the trace is unique on a factor. However, I don't really know how to construct this sequence of subalgebras... 
So my question is: Am I on the right track? And if yes, how can I construct those subalgebras?
Thanks a lot!
 A: Yes, you are on the right track. 
For any $t\in[0,1]$, there exists a projection $p\in N$ with $\tau(p)=t$. Start with $p_1^0=I$. Then $\text{span}\{p_1^0\}=\mathbb C$.
Now assume for induction you have pairwise orthogonal projections $p_1^n,\ldots,p_{2^n}^n$, with $\tau(p_j^n)=2^{-n}$ and corresponding matrix units $e_{kj}^n$, i.e. $$e_{kj}^ne_{gh}^n=\delta_{jg} e_{kh}^n,\ \ e_{kk}^n=p_k^n. $$
Now choose a projection $p_1^{n+1}\leq p_1^n$ with $\tau(p_1^{n+1})=2^{-(n+1)}$. Let $p_2^{n+1}=p_1^n-p_1^{n+1}$, and let $e_{12}^{n+1}$ be a partial isometry with $e_{12}^{(n+1)*}e_{12}^{n+1}=p_2^{n+1}$, $e_{12}^{n+1}e_{12}^{(n+1)*}=p_1^{n+1}$. Next define 
$$
e_{1,2k-1}^{n+1}=p_{1}^{n+1}e_{1k}^{n},
\ \ \ \ 
e_{1,2k}^{n+1}=p_{2}^{n+1}e_{1k}^{n}
$$
and extend to matrix units by $$ e_{kj}^{n+1}=e_{1k}^{(n+1)*}e_{1j}^{n+1}$$ (one has to check that these are actually matrix units). Now we can define $p_{k}^{n+1}=e_{kk}^{n+1}$, which completes the induction. The linear map induced by $e_{kj}^n\longmapsto e^{n+1}_{2k-1,2j-1}+e_{2k,2j}^{n+1}$ is a $*$-monomorphism, and so we have unital, tracial, embeddings
$$ M_1(\mathbb C)\hookrightarrow M_2(\mathbb C)\hookrightarrow M_4(\mathbb C)\hookrightarrow \cdots\hookrightarrow M_{2^n}(\mathbb C)\hookrightarrow \cdots \hookrightarrow N.
$$
