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I do not think that this true. Take for example the sequence of matrix algebras with $A_0 = \mathbb{C}$ and $A_n = M_2(\mathbb{C}) \otimes A_{n-1}$ 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 GNS-construction 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.

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I do not think that this true. Take for example the sequence of matrix algebras with $A_0 = \mathbb{C}$ and $A_n = M_2(\mathbb{C}) \otimes A_{n-1}$ 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 GNS-construction 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 von Neumann algebras.