Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $e$ be an idempotent element of $C(X)$ (where $X$ is a tychonoff space) such that if $f \in C(X)$ and $ 0 \leq f \leq e$ then $f = ce$ for some constant value $c$. I want to show that $e$ is the characteristic function of a singleton.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

For the completeness' sake I'll work with the following definition of Tychonoff -- completely regular and Hausdorff.

First of all, $e(x)$ is either $0$ or $1$ for all $x$ because $e(x)^2 = e(x)$. For the sake of contradiction, suppose now that there are two distinct points $x, y$ such that $e(x) = e(y) = 1$. Now by Tychonoff property, there exists a function $f \in C(X)$ such that $f(x) = 0$ and $f(y) = 1$. The function $g$ defined by $g(z) := \min\{e(z), f(z) \}$ is continuous (this is a nice little exercise itself) and such that $0 \leq g \leq e$. Therefore $g = ce$ which implies both $c=1$ and $c=0$, which is absurd, so we are done.

share|improve this answer
    
To whoever down-voted: please explain what is wrong with my answer. If there is a mistake I'd love to hear about it. –  Marek Sep 26 '13 at 15:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.