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Every von Neumann algebra is a C$^*$-algebra. So the usual theorem that a C$^*$-algebra $A$ is (norm) separable iff its state space is first countable in the weak-* topology (i.e. the topology $\sigma …

The answer is no (to the main question, not the title). Consider the *-algebra $S$ of *-polynomials generated by one variable $z$ such that $zz^* = z^*z$, i.e. the free commutative *-algebra on one ge …

As expressed in my comment, the finite-dimensional CAR algebras do have a complete projection lattice. Here I outline the proof that the CAR algebra with countably many degrees of freedom (equivalentl …

You can find, in any decent textbook on von Neumann algebras, a proof that if $A \subseteq B(\mathcal{H})$ is a von Neumann algebra, and the Hilbert space $\mathcal{H}$ is separable, then the unit bal …

I agree with Dmitri Pavlov that separability is not so important in the modern theory of von Neumann algebras, and this answers the second question. However, an example answering the first question ha …

For any C$^*$-algebra $A$, we can define its opposite algebra $A^{\mathrm{op}}$, which is the algebra where $ab$ is defined to be $ba$, as calculated in $A$. Let's restrict to unital algebras for simp …

Since questions recirculate to the front page forever if left unanswered, I will amalgamate the comments into an answer, which I've made community wiki. Firstly, by the characterization of the Cantor …