I had the following question when I am learning von Neumann algebras: Let p be a finite projection in a finite von Neumann algebra $M$, let $p>p_1>p_2>\cdots$ be a decreasing sequence of projections such that $p_i\neq p_{i+1}$. Is such a sequence necessarily of finite length? Any help or recommendations on books/papers are greatly appreciated.

The answer to your question is no. Abelian von Neumann algebras are finite, but it is easy to find examples of such algebras with infinite decreasing sequences of projections. For instance, one may take $M = L^\infty(\mathbb{R},\lambda)$ where $\lambda$ is the Lebesgue measure, and let $p_i \in M$ be the element corresponding to the characteristic function of $[0,\frac{1}{i}]$. Update: Now that you've seen some counterexamples, this might be a good followup "homework" problem for you (depending on how much you know about measure theory and abelian von Neumann algebras): prove that a (finite) von Neumann algebra $M$ satisfies your descending chain condition on projections if and only if it is a finite dimensional algebra. (Hint: think about a type decomposition.) 


I do not think that this true. Take for example the sequence of matrix algebras with $A_0 = \mathbb{C}$ and $A_n = M_2(\mathbb{C}) \otimes A_{n1}$ equipped with their normalized traces and the diagonal embeddings $A_{n} \to A_{n+1}$. The direct limit $A$ of this sequence is a $*$algebra with a trace and the GNSconstruction with respect to this trace state yields a (faithful) representation $H$, on which $A$ acts. The weak closure $M$ of $A$ is a von Neumann algebra, which is of type $II_1$ (in particular, it is finite). This is known as the hyperfinite type $II_1$factor. Now consider the sequence of projections starting with $p_0 = 1 \in A_0$, $$ p_1 = \begin{pmatrix} 1 & 0 \\\\ 0 & 0 \end{pmatrix} \in A_1 = M_2(\mathbb{C}) $$ $$ p_2 = \begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0\end{pmatrix} \in A_2 \cong M_2(\mathbb{C}) \otimes M_2(\mathbb{C}) $$ etc. (I hope the pattern is clear). We have $tr(p_n) = 2^{n}$. Therefore $p_i \neq p_j$ if $i \neq j$. But $p_0 > p_1 > p_2 > \dots$ In fact, constructions like this are the reason for many of the cool features of type $II$ von Neumann algebras. 

