Why is the rank of a locally free sheaf same everywhere if $X$ is connected?

Question

Let $(X,\mathcal{O}_X)$ be a connected scheme.Let $\mathcal F$ be a locally free sheaf on $X$. This means that $X$ can be covered by open sets $U$ for which $\mathcal F|_U$ is a free $\mathcal O_X|_U$ - module. The rank of $\mathcal F$ on such a $U$ is the number of copies of $\mathcal O_X|_U$ required.

I want to show that when $X$ is connected this rank is the same everywhere. (that is same for all open sets in the cover)

Define a map $f:X\rightarrow \mathbb N$ as follows -

For any $x\in X$ there is a $U$ containg $x$ for which $\mathcal F|_U$ is free $\mathcal O_X|_U$ of rank $n$. Let $f(x)=n$ If $f$ is continuous and well-defined then I am done because then the only connected subsets of $\mathbb N$ with discrete topology are singletons. So my questions are -

1. Why is $f$ continuous?

2. If $V$ is another open set containing $x$ then why should rank of $\mathcal F|_V$ equal to $n$?

3. would the same proof work if rank is infinite?

Thank you.

Answer 1

1. A map to a discrete space is continuous, iff all fibres are open. And the fact that the sets $M_n := \{x \in X | \operatorname{rank} \mathcal F_x = n \}$ are open, is built in the definition: If you have $x \in M_n$, you find some open set $U \ni x$, such that $\mathcal F_{|U} \cong \mathcal O_{X|U}^n$, which shows $U \subset M_n$. Hence $M_n$ is open.

2. Let $U,V \ni x$ with $\mathcal F_{|U} \cong \mathcal O_{X|U}^n$ and $\mathcal F_{|V} \cong \mathcal O_{X|V}^m$. From the first iso, we deduce $\operatorname{rank} \mathcal F_x=n$. From the second iso, we deduce $\operatorname{rank} \mathcal F_x=m$, hence $n=m$.

3. Yes. Instead of $\mathbb N$, you let $f$ map into the set of all cardinals, endowed with discrete topology.