Skip to main content
added 200 characters in body
Source Link
Yemon Choi
  • 25.8k
  • 9
  • 69
  • 156

EDIT: the following answer was based on the assumption that $W$ is Banach and the ring morphisms $R\to W\to {\mathbb C}$ were continuous. My apologies for misunderstanding the question.


I think the answer is yes, because $W$ must be isomorphic to $C(K)$ for some closed subset $K\subset [0,1]$. The condition about idempotents implies $K$ is connected. The only connected subsets of $[0,1]$ are closed sub-intervals, and these are either homeo to $[0,1]$ itself or are degenerate singletons.

I think the answer is yes, because $W$ must be isomorphic to $C(K)$ for some closed subset $K\subset [0,1]$. The condition about idempotents implies $K$ is connected. The only connected subsets of $[0,1]$ are closed sub-intervals, and these are either homeo to $[0,1]$ itself or are degenerate singletons.

EDIT: the following answer was based on the assumption that $W$ is Banach and the ring morphisms $R\to W\to {\mathbb C}$ were continuous. My apologies for misunderstanding the question.


I think the answer is yes, because $W$ must be isomorphic to $C(K)$ for some closed subset $K\subset [0,1]$. The condition about idempotents implies $K$ is connected. The only connected subsets of $[0,1]$ are closed sub-intervals, and these are either homeo to $[0,1]$ itself or are degenerate singletons.

Source Link
Yemon Choi
  • 25.8k
  • 9
  • 69
  • 156

I think the answer is yes, because $W$ must be isomorphic to $C(K)$ for some closed subset $K\subset [0,1]$. The condition about idempotents implies $K$ is connected. The only connected subsets of $[0,1]$ are closed sub-intervals, and these are either homeo to $[0,1]$ itself or are degenerate singletons.