A rational curve (function field $\cong k(t)$) has trivial $\mathrm{Pic}^0$ (every degree-$0$ divisor is principal) and infinitely many points.
A rational curve admits an injection from $\mathbb{N}$, witnessing it has infinitely many points.
Every degree-zero divisor on a rational curve is principal.
$\mathrm{Pic}^0(C) = 0$ for a rational curve: every divisor of degree $0$ is principal.
A nonzero rational function $f$ has prescribed poles given by an effective divisor $A$ if $\mathrm{div}(f) + A \geq 0$ and $\mathrm{ord}_P(f) = -A(P)$ for every $P$ with $A(P) > 0$.
Instances For
On a rational curve (hence infinite), any divisor $A$ admits a point $P_0$ outside its support.
On a rational curve, given any effective divisor $A$ of positive degree, there exists a rational function $f$ with prescribed poles equal to $A$.
A function with prescribed poles bounded by $A$ lies in the Riemann-Roch space $L(A)$.
If $A \leq B$ as divisors then $\deg B - \deg A \geq 0$.
Inductive step lemma for Riemann-Roch on $\mathbb{P}^1$: when $A \leq B$ are effective with $\deg B - \deg A = n$, we have $\dim_k L(B) = \dim_k L(A) + (\deg B - \deg A)$, proved by induction on $n$.
On a rational curve, for effective divisors $A \leq B$, $\dim_k L(B) = \dim_k L(A) + (\deg B - \deg A)$.
On a rational curve, for any effective divisor $D$, the Riemann-Roch space satisfies $\dim_k L(D) \geq \deg D + 1$.
Riemann-Roch on $\mathbb{P}^1$: for any effective divisor $D$ on a rational curve, $\ell(D) = \deg D + 1$.