I was reading the proof of the Proposition (3.1, IV) in Hartshorne's Algebraic Geometry book.
Proposition: Let $D$ be a divisor on a curve $X$. Then:
(a) the complete linear system $\vert D \vert$ has no base points if and only if for every point $P\in X$, $$ \dim \vert D-P \vert = \dim \vert D \vert - 1;$$ (b) $D$ is very ample if and only if for every points $P, Q\in X$(including the case $P=Q$), $$ \dim \vert D-P-Q \vert = \dim \vert D \vert - 2.$$
In the proof, I could not understand the line in the proof which says that if $D$ satisfies the condition (b), then we have $$ \dim \vert D-P \vert = \dim \vert D \vert - 1$$ for every $P\in X.$
(It may be silly question.)
I would be thankful if someone could elaborate this sentence.
Let's get this off the unanswered list while it's active.
As Mohan points out in the comments, for any divisor $D$ on a curve $C$ and any point $P\in C$, we have the inequality $$\dim |D|-1\leq \dim |D-P|\leq \dim |D|.$$ This comes from taking global sections of the short exact sequence of sheaves $$0\to\mathcal{O}_C(D-P)\to\mathcal{O}_C(D)\to k(P)\to 0,$$ which gives the half-exact sequence of vector spaces $$0\to H^0(\mathcal{O}_C(D-P)) \to H^0(\mathcal{O}_C(D)) \to k.$$ The second map has image either $0$ or $k$, and therefore either $\dim H^0(\mathcal{O}_C(D-P)) = \dim H^0(\mathcal{O}_C(D))$ or $\dim H^0(\mathcal{O}_C(D-P)) = \dim H^0(\mathcal{O}_C(D))-1$, respectively. After projectivizing, we get that $\dim |D-P|$ is either $\dim |D|$ or $\dim |D|-1$ and the claim is proven.
To use this to attack the statement you had trouble with, note that if $\dim |D-P-Q| = \dim |D|-2$ for all $P,Q$, then we must have $\dim |D-P|=\dim |D|-1$ for all $P$.