In the proof of the Castelnuovo theorem for curves in $\mathbb{P}^3$ (Hartshorne IV, 6.4.) the following is done: One considers a smooth, complete curve $C$ in the projective space $\mathbb{P}^3$ over an algebraically closed field. The degree of this curve is denoted by $d$. Then one takes a hyperplane section $D=P_1+...+P_d$ of the curve such that $P_1$,..., $P_d$ are different and no three of them are collinear. Then one wants to show that $P_i$ is not a basepoint of the linear system $|nD-P_1-P_2-...-P_{i-1}|$ if $i \leq \mbox{min}(d, 2n+1)$. At this point the following argument is given, which I do not understand: "To show that $P_i$ is not a basepoint it suffices to find a surface of degree $n$ in $\mathbb{P}^3$ that contains $P_1$, $P_2$,..., $P_{i-1}$ but not $P_i$". I do not get why this suffices. What is the reason that $P_i$ is not a basepoint, if it lies on there exists such a surface?
In the proof of the Castelnuovo theorem for curves in $\mathbb{P}^3$ (Hartshorne IV, 6.4.) the following is done: One considers a smooth, complete curve $C$ in the projective space $\mathbb{P}^3$ over an algebraically closed field. The degree of this curve is denoted by $d$. Then one takes a hyperplane section $D=P_1+...+P_d$ of the curve such that $P_1$,..., $P_d$ are different and no three of them are collinear. Then one wants to show that $P_i$ is not a basepoint of the linear system $|nD-P_1-P_2-...-P_{i-1}|$ if $i \leq \mbox{min}(d, 2n+1)$. At this point the following argument is given, which I do not understand: "To show that $P_i$ is not a basepoint it suffices to find a surface of degree $n$ in $\mathbb{P}^3$ that contains $P_1$, $P_2$,..., $P_{i-1}$ but not $P_i$". I do not get why this suffices. What is the reason that $P_i$ is not a basepoint, if it lies on such a surface?