# How to show multiplication and inversion are continuous in this topology on $C(X)$?

The sets$$\big\{ \{f\in C(X) : |g-f| \le u \} \;\big\vert\; g\in C(x) \text{ and } u \text{ is a positive unit of } C(X)\big\}$$ form a base for some topology on $C(X)$. Corresponding to this topology, how to show that multiplication is continuous on $C(X)^2$? Moreover if $U$ denotes the set of all units in $C(X)$, how to show the function $f \mapsto 1/f$ is continuous on U?

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Sorry, but what is $C(X)$? I think I could do the question if I knew what you mean't. –  simplicity May 18 '12 at 16:43
C(X) is the set of all real-valued continuous functions on X. Here X is an arbitrary topological space. –  Bilash Majumder May 19 '12 at 1:15

I assume $C(X)$ is the space of continuous real-valued (or perhaps complex-valued) functions on some topological space $X$. Any (strictly) positive continuous function is a positive unit of $C(X)$.
Note that $$|f_1 f_2 - g_1 g_2| \le |f_1| |f_2 - g_2| + |g_2| |f_1 - g_1| \le (|g_1| + |f_1 - g_1|) |f_2 - g_2| + |g_2| |f_1 - g_1|$$ Given $g_1, g_2 \in C(X)$ and positive unit $u$, if $|f_1 - g_1| < \min(1, u/(2 |g_2|))$ and $|f_2 - g_2| < u/(2 |g_1| + 2)$ then $|f_1 f_2 - g_1 g_2| < u$. This shows that multiplication is continuous.
For $f \mapsto 1/f$, note that $\dfrac{1}{f} - \dfrac{1}{g} = \dfrac{g-f}{fg}$ If $g$ is a unit and $u$ a positive unit, then for any $f$ with $|f - g| < \min(u|g|^2/2, |g|/2)$ we have $|f| \ge |g| - |f - g| > |g|/2 > 0$ and so $$\left|\dfrac{1}{f} - \dfrac{1}{g}\right| \le \dfrac{|f-g|}{|f||g|} < \dfrac{u |g|^2/2}{|g|^2/2} = u$$
@BilashMajumder: if $g_2=0$, then $\min(1, u/(2 |g_2|))=1$. –  robjohn May 19 '12 at 13:16
@BilashMajumder: $\min(x,y)$ is the minimum of $x$ and $y$, so if $y\to\infty$, $\min(x,y)\to x$. –  robjohn May 20 '12 at 3:30
Rather than arguing over what $\min(1, u/(2 |g_2|))$ means, just not that if $g_2 = 0$, $|f_1 - g_2| < 1$ and $|f_2 - g_2| < u/(2(|g_1|+2)$ will be enough. –  Robert Israel May 20 '12 at 5:46