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Jul 11, 2019 at 9:15 comment added erz I don't think assuming that $A$ is a C* algebra is essential actually, since a collection of functions generates the same topology as the collection of all possible C* combinations of them, and a collection generates the same topology as its closure in $C_b(X)$.
Jul 9, 2019 at 20:42 comment added Douglas Somerset If you assume that $A$ is a C$^*$-algebra then $X/R$ will be the Gelfand space of $A$, where $R$ is the equivalence relation on $X$ consisting of points not separated by $A$. So $X/R$ will be locally compact and Baire, and hence $X$ will be too in the weak topology.
Jul 9, 2019 at 19:03 history rollback Merry
Rollback to Revision 1
Jul 9, 2019 at 19:01 comment added Merry @DouglasSomerset It indeed is!
Jul 8, 2019 at 22:38 comment added Douglas Somerset I think that the edit to the title is erroneous.
Jun 7, 2019 at 5:55 history edited user64494 CC BY-SA 4.0
The title is improved.
Jun 6, 2019 at 23:22 history asked Merry CC BY-SA 4.0