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Mar 13, 2016 at 18:20 vote accept mikis
Mar 13, 2016 at 16:48 history edited Petr Naryshkin CC BY-SA 3.0
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Mar 13, 2016 at 16:34 comment added Petr Naryshkin Wait! In that case $C(X)$ will not be a Banach algebra at all!
Mar 13, 2016 at 16:27 comment added Petr Naryshkin @NikWeaver But is that all such sets? What if $X$ is not locally compact?
Mar 13, 2016 at 15:58 comment added Nik Weaver (It is usual to acknowledge comments when editing a post.)
Mar 13, 2016 at 15:58 comment added Nik Weaver In response to the edit, compact subset of $\mathbb{R}^n$ if $C(X)$, closed subset of $\mathbb{R}^n$ if $C_0(X)$.
Mar 13, 2016 at 14:25 history edited Petr Naryshkin CC BY-SA 3.0
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Mar 13, 2016 at 14:23 comment added Nik Weaver To tighten this up, assuming $X$ is locally compact the condition is that it should be homeomorphic to a closed subset of $\mathbb{R}^n$ for some $n$. Its spectrum including $0$ is homeomorphic to a compact subset of $\mathbb{C}^n$, and after removing $0$ it has the form I stated.
Mar 13, 2016 at 13:28 history answered Petr Naryshkin CC BY-SA 3.0