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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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What have you tried? –  Thomas Andrews May 29 '12 at 12:32
    
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. –  Thomas Andrews May 29 '12 at 14:01
    
It was wrongly edited. Have fixed it. –  Bilash Majumder May 29 '12 at 14:24
    
What is a "positive unit of $C(X)$?" A little confused by that term. –  Thomas Andrews May 29 '12 at 16:16
    
What is a $z$-ideal? –  Dylan Moreland Jul 23 '12 at 22:59

1 Answer 1

Define a continuous $g$ such that $g(x)=0$ for $|x|\leq\frac{1}{2}$ and $g(x)=1$ for $|x|\geq 1$ and $0\leq g(x)\leq 1$ for all $x$.

Define for any $u>0$, $$f_u(x) = xg(\frac{2x}{u})$$ Show that this has the property that $f_u(x)\in O'$ and, for all $x$, $|f_u(x)-x| < u$.

If you assume $u(x)\in C(X)$ is a strictly positive function, you can do much the same by defining:

$f_u(x) = xg\left(\frac{2x}{u(x)}\right)$.

It's only slightly trickier to show that this $f_u\in O'$.

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