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.)
How to show that the identity is in closure of $O' \subset C(\mathbb R)$?
2
$\begingroup$
general-topology
-
0I'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
-
0It was wrongly edited. Have fixed it. – 2012-05-29
-
0What is a "positive unit of $C(X)$?" A little confused by that term. – 2012-05-29
-
0What is a $z$-ideal? – 2012-07-23