The definition of first selection principle is well known: $S\_1(A,B)$.
 Let $A$ and $B$ be families of sets. The symbol $S\_1(A, B)$ denotes the statement:
for each sequence $(A\_n : n \in \omega)$ of elements of $A$ there is a sequence
$(x\_n : n\in \omega)$ such that each $x\_n$ is an element of $A\_n$, and $\{x\_n : n\in \omega\}$ is an element of $B$.


Can we assume that $x\_{n\_k}=\emptyset$ for $k\in \omega$ ?



Definition of $S_1(A,B)$
I understand your question to mean "must we chose an element $x\_n$ for all $n$?". (This is in accordance with the comments.)


**Answer.** Officially, the answer is "Yes", you must chose an element for each $n$. However, as pointed out in the comments, often you can prove this makes no difference if you are allowed to *not* chose elements. This is true for all classic selective covering properties of type $S\_1(A,B)$. 


For some properties this is not entirely trivial, but still provable. For example, $S\_1(\Gamma,\Gamma)=S\_{\le 1}(\Gamma,\Gamma)$ because WLOG the covers get finer with $n$. If you think about it, this is the essence of the now-classic proof that $S\_1(\Gamma,\Gamma)=S\_\text{fin}(\Gamma,\Gamma)$.


