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I am reading a accepted answer here. But I don't know why $\phi(C[0,1])$ generats $\ell^\infty({\mathbb{N}} )$ or what generates $\ell^\infty(\mathbb{N})$ as a von Neumann algebra. Can someone tell me?

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The mapping $\phi$ is given by an enumeration $(t_n)_{n\in\mathbb{N}}$ og $[0,1]\cap\mathbb{Q}$, and the formula $$\phi(f)=\bigl(f(t_n)\bigr)_{n\in\mathbb{N}}$$ Useful fact: Given any $x\in\ell^\infty(\mathbb{N})$ and a finite subset $F\subset\mathbb{N}$, there is some $f\in C([0,1])$ with $\lVert f\rVert_\infty\le\lVert x\rVert_\infty$ and $f(t_n)=x_n$ for all $x\in F$.

Now recall that a weak* neighbourhood in $\ell^\infty(\mathbb{N})$ is given by a finite set of sequences in $\ell^1(\mathbb{N})$. Given one of these sequences, say $y$, and any $\varepsilon>0$, there is a finite subset $F\subset\mathbb{N}$ so that $\sum_{n\notin F}|y_n|<\varepsilon$. The union of these sets is still a finite set, so you can apply the above “useful fact”.

I am taking the chance that you can see where to go with this now. If not, let me know, and I'll fill in some more detail.

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