Timeline for Is the strong operator topology metrizable?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 17, 2018 at 0:55 | comment | added | Nate Eldredge | The second paragraph is also a proof all by itself, since the weak-* topology on $X^*$ is not metrizable, so it cannot embed in any metrizable space. | |
| Apr 27, 2018 at 9:27 | history | edited | Jochen Wengenroth | CC BY-SA 3.0 |
added 410 characters in body
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| Apr 27, 2018 at 6:03 | comment | added | ABB | @ Tomek Kania Probably it works: the product on $B(X)$ is jointly norm continuous but this point is not valid concerning SOT. | |
| Apr 26, 2018 at 16:47 | comment | added | Tomasz Kania | Is there an easy argument to see that the SOT topology is different from the norm topology on a general Banach space without using, for example, the Josefson-Nissenzweig theorem? | |
| Apr 26, 2018 at 16:29 | vote | accept | ABB | ||
| Apr 26, 2018 at 14:33 | history | edited | Jochen Wengenroth | CC BY-SA 3.0 |
corrected spelling
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| Apr 26, 2018 at 13:44 | history | answered | Jochen Wengenroth | CC BY-SA 3.0 |