From Dummit and Foote (emphasis mine):
Let $G$ be a transitive permutation group on the finite set $A$. A block is a nonempty subset $B$ of $A$ s.t. $\forall \sigma\in G:$ either $\sigma(B)\cap B = B$ or $\sigma(B)\cap B = \emptyset$.
However from wikipedia I can infer no constraint on the cardinality of $A$.
This question stems from the following proof attempt:
Prove that a doubly transitive group is primitive.
Let $B$ be a block in $A$. Consider $a\in B$, and take $b\in B-\{a\}$, $c\not\in B$. Then by double transitivity there is $\sigma\in G_a$ s.t. $c=\sigma b$. Now if $A$, and thus also $B$, is finite, we can conclude that $\sigma(B)\cap B$ cannot equal $B$ by a pigeon-hole style argument. Hence $\sigma(B)\cap B=\emptyset$.
Since $\sigma a= a$, we must have $\sigma(B)\cap B = B$.
This would now be a contradiction, so either we cannot choose $b\in B-\{a\}$ or we cannot choose $c\not\in B$, thus either $B=\{a\}$ or $B=A.\blacksquare$
The problem with the proof is indicated by the italic text, and this problem inspired my question about the definition of a block. I know of another proof that does work (I believe regardless of the cardinalities) but I'm still curious about the definition of a block.
I'm also curious if my proof is salvageable even if $B$ is infinite, since in that case I don't see how we can proof $\sigma(B)\cap B$ cannot equal $B$.