The Stacks project

"if each neighbourhood of $x$ has infinitely many points, and each $A_{x'}$ has exactly two elements, then $\mathcal{F}_x$ has infinitely many elements"

I think this statement is false.

Take for ease of notation $A_x=\{0,1\}$. Take $X$ an infinite set and take $\tau$ a non-principal ultrafilter of $X$, and take the open sets of $X$ to be exactly those sets which are in $\tau$. Then for any point $x\in X$, any neighbourhood of $x$ contains infinitely many points.

But $\mathcal{F}_x$ has four points, one is given by the section $s\in\mathcal{F}(X)$ constant with $s_y=1$ for all $y\in X$, another is given by $t\in\mathcal{F}(X)$ such that $t_y=1$ iff $y=x$. The other two are by interchanging $1$ and $0$ for this two.

This are the only four because for any $r\in\mathcal{F}(U)$ with $U\in\tau$ we can define $A=\{y\in U: r_y=0\}$ and $B=\{y\in U: r_y=1\}$ so that $A\cup B=U$, and as $\tau$ is an ultrafilter then either $A\in\tau$ or $B\in\tau$, this with $r_x=0$ or $r_x=1$ correspond to the four elements.

(There seems to be a problem in my computer with the subscripts in the preview)

The preview should now be fine, there was a silly mistake in the code. I will let Johan handle the math :).

Yes, very good! How bizarre to get 4 elements! I replaced it with the statement: $x = \lim x_n$ with $x_n \not = x$ and all sets have two elements then the stalks are infinite. The fix is here. Stop by my office if you are in NY and I'll give you a Stacks project mug, if I have't run out by then.