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$I$ a closed ideal in the Banach algebra $C_0(X)$, $X$ locally compact Hausdorff space. Is the claim correct: For all $x\in X$ exists $f\in I$ such that $f(x)\neq 0$?

I need this for a proof. But I have no idea if it's correct or false.

taglap
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    The trivial ideal $I = {0}$ gives a counterexample. Are there further conditions on your ideal? – Daniel Fischer Sep 23 '15 at 08:53
  • no, nothing. Is the statement correct if $I\neq {0}$? Maybe the author has forgotten to assume $I\neq {0}$. – taglap Sep 23 '15 at 08:56
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    No, it's still false, we have the maximal ideals $\mathfrak{M}_p = { f \in C_0(X) : f(p) = 0 }$ for $p \in X$, which are defined so that the property doesn't hold. The claim holds for maximal ideals that are not of this form (if they exist). – Daniel Fischer Sep 23 '15 at 08:59
  • ok, this is really bad. thank you!! – taglap Sep 23 '15 at 09:01
  • What is the thing you want to prove that you would need the property for? – Daniel Fischer Sep 23 '15 at 09:05
  • If $C_0(X)$ is a $C^\ast$-algebra, $I$ a closed ideal in $C_0(X)$, then there is a closed set $Y\subseteq X$ such that $I=I_Y$. Someone asks the same yesterday here http://math.stackexchange.com/questions/1446809/prove-that-a-map-is-well-defined-and-closed-ideals-in-c-0x ,maybe he is in my course. I try to understand the proof of the unital case and meanwile I think, the author wants to apply urysohn. He claims: If $f\in I_Y$ and $x\in X\setminus Y$, then there is $g\in I$ such that $g(x)\neq 0$ and $g(x)=f(x)$. I think, by Urysohn you get $g(x)\neq 0$, but I don't know why g has to be in $I$. – taglap Sep 23 '15 at 09:17
  • but I found an other proof now, which is a littlebit different and seems correct for me. Thanks for your help=)! – taglap Sep 23 '15 at 09:26

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