2
$\begingroup$

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.)

  • 0
    What is a $z$-ideal?2012-07-23

1 Answers 1

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 $f_u\in O'$.