ag.algebraic geometry - Why the rank of a locally free sheaves is well defined?

Let $mathcal{F}$ be a locally free sheaf on $X$. For any $x$ in $X$ there exists $x in U subset_{open} X $ such that

$mathcal{F}|_U cong mathcal{O}_X|_U^{(I)}$ $ (star)$.

In particular, for each $y$ in this particular $U$, one has $mathcal{F}_y cong mathcal{O}_{X,y}^{(I)}$ (which is given by the isomorphism above!!!).

Suppose now $X$ is connected and $mathcal{F}$ is locally free (we need this). Fix an indexing set $I$ (and I think I need to take this $I$ to be one of the indexing sets from $(star)$ above). The properties of $mathcal{F}$ show that the set

$S_I = left(x in X : mathcal{F}_x cong mathcal{O}_{X,x}^{(I)}right)$

is both closed and open in $X$. We know that there exists

$x$ in $X$ with $mathcal{F}_x cong mathcal{O}_{X,x}^{(I)}$,

we have $S_I = X$.

In particular, $text{rank}_{mathcal{O}_{X,x}}(mathcal{F}_x)$ is constant as $x$ varies in $X$.