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Oct 20, 2010 at 23:55 comment added Jonas Meyer Andreas: That is a good question, and I do not know the answer. Perhaps I should have mentioned in the first place that I did not understand that line. I included it as part of the quote for completeness, because I hoped it would add better context. But in any case, I'm glad it inspired your question: mathoverflow.net/questions/41597/…
Oct 9, 2010 at 12:28 comment added Andreas Thom What is the reasoning in "Note that by 2.8.8 a transfinite (but countable) application of the operation a will produce $A''$." ? I see that $\omega_1$ applications of the operation $a$ produce $A''$ and also that each element of $A''$ appears at the $\alpha$-th application for some $\alpha < \omega_1$; but why is $a^{\alpha}(A)=A''$ for some $\alpha < \omega_1$? A related question: Is there some $\alpha < \omega_1$ which works for all $A \subset B(H)$?
May 12, 2010 at 21:41 history asked Jonas Meyer CC BY-SA 2.5