Localization of ring of continuous functions at an element

Question

Let $X$ be a topological space. Let $D_f= \left\{ x\in X:f(x)\neq 0 \right\}$. Let $\mathcal O_X$ be the sheaf of continuous functions on $X$. Is it true that $\mathcal O_X(D_f)= \left\{ \frac g{f^n}:g\in \mathcal O_X(X) \right\}$? With polynomials it seems easier to prove because the zeros are isolated...

Answer 1

You want to see whether it works like for principal open subsets of affine varieties. Unfortunately no. The continuous ( even smooth) functions are not that rigid, in the sense that the rate of decrease around a zero of a continuous function cannot be measured only with natural numbers ( like the order of a zero...).

Let's take the simplest case, $X=\mathbb{R}$. Consider a continuous function $f$ that has the only zero at $x=0$. What would it mean the fact you wondered about? "Every continuous function defined on $\mathbb{R} \backslash \{0\}$, multiplied by a sufficient large power of $f$, can be extended to a continuous function on $\mathbb{R}$." Can we prevent this from happening? Yes. Take $f$ that has a zero at $0$ with a very "low" order. Instead of decreasing like a power of $x$, make it decrease like a power of $ \frac{1}{\log |x|}$. Concretely, take $f(x) = \frac{1}{\log |x|}$ for $x\ne 0$ and $0$ at $x=0$. Moreover, take a function $g$ on $\mathbb{R} \backslash \{0\}$ that increases towards $0$ like $\frac{1}{x}$. Concretely, take $g(x) = \frac{1}{x}$. Now you will get your counterexample.

Similar counterexamples hold in the $C^{\infty}$ category. Say take $f(x) = x$ but now take $g$ so that $\frac{1}{g}$ has a zero of infinite order at $0$, like $\frac{1}{g}= e^{-1/x^2}$.

For analytic functions things are better if $f$ say has finitely many zeroes.