How to show $X_f$ is open in $X$

Question

Let $X$ be a locally ringed space (not necessarily a scheme). Set $X_f$ to be the locally ringed space whose underlying topological space is the set of points $x$ in $X$ such that $f$ is not sent to $0$ under the map $\mathcal{O}_X (X) \rightarrow \kappa(x) = \mathcal{O}_X^x / \mathfrak{m}_x \mathcal{O}_X^x$. How do we make this into an open sub-locally ringed space of $X$? Most importantly to me, how do we show that this set is open? I know how to do this when $X$ is a scheme- we cover $X$ by open sets. But we can't do that here.

Answer 1

We'll show that if $f$ does not vanish at $x$, then there is some open neighbordhood of $x$ on which $f$ does not vanish.

Note that another way of saying the $f$ does not vanish at $x \in X$ is saying $f_x \in \mathcal{O}_{X,x}$ is invertible.

Suppose $f$ does not vanish at $x$. Then we may choose an inverse to it in $\mathcal{O}_{X,x}$. The inverse is represented by a pair $(U, g)$, where $U$ is some open subset of $x$ and $g \in \mathcal{O}_X(U)$. Thus, $f_x g_x = 1$ in $\mathcal{O}_{X,x}$. In other words, there is some open neighbordhood $V$ of $x$ contained in $U$ such that $f|_V \cdot g|_V = 1$ in $\mathcal{O}_X(V)$. In particular, $f_y \in \mathcal{O}_{X,y}$ is invertible for each $y \in V$.