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The sets $\{f \in C(X) : |g-f| \leq u \}$ where $g\in C(X)$, $u$ a positive unit of $C(X)$ form a base for some topology on $C(X)$. Let $X = \mathbb{R}$, the set of real numbers. With the above topology, how do I show that the identity map $\mathbb{R}\to\mathbb{R}$ is in the closure of $O'$, where $O'$ is the set of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$ vanishing on a neighborhood of $0$? (Evidently, $O'$ is a $z$-ideal.)

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    I'm assuming you mean $g$ is fixed - that us, the set $O_{u,g}$ is defined above. If that's what you mean, we usually don't include the condition on $g$ inside the set.2012-05-29
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    It was wrongly edited. Have fixed it.2012-05-29
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    What is a "positive unit of $C(X)$?" A little confused by that term.2012-05-29
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    What is a $z$-ideal?2012-07-23

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