I am trying to understand Lemma 2.1.18 in Jones paper
https://link.springer.com/content/pdf/10.1007/BF01389127.pdf
It seems to use that the commutant of a hyperfinite factor acting on a Hilbert space is again hyperfinite why is this the case? It further seems to assume that if $N\subset M$ and $M$ is hyperfinite $N$ is hyperfinite, why does this hold? It is not the case that $C^*$subalgebras of AF algebras are AF...
Let us clarify that we are talking II$_1$-factors here. The commutant of a II$_1$-factor is always type II but, depending on the representation, it can be II$_\infty$. This is easily seen if your factor is of the form $M\otimes I\subset B(H\otimes K)$; in such case, $(M\otimes I)'=M'\otimes B(K)$.
In Jones' case, with $N\subset M$, if we represent $M$ in the GNS for the trace, then $\dim_MH=1$. The fact that the index $[M:N]$ is finite then means that $\dim_N H=[M:N]<\infty$. Thus $N'$ is finite.
If $N$ is hyperfinite, we have $N=\overline{\bigcup_nA_n}$, an increasing union with $\dim A_n<\infty$ for all $n$. Then $N'=\bigcap_n A_n'$. Each $A_n'$ is type I, and so injective. Intersection of a decreasing family of injective algebras is injective, so $N'$ is injective. Since $N'$ is a II$_1$-factor, $M'\subset N'$ is injective. Then $M$ is the commutant of an injective algebra, so injective; and then, by Connes, $M$ is hyperfinite.