A filter concentrates on a set Given a filter $\mathcal F$ on some nonempty set $X$ and some $Y \subseteq X$, people often say that "$\mathcal F$ concentrates on $Y$".
Questions: Does this simply mean 
$$\forall Z \subseteq X \colon Z \in \mathcal F \leftrightarrow Z \cap Y \in \mathcal F$$
If it is of any help: I encountered this non-defined term in Jech's beautiful 3rd millennium set theory edition when he introduced extenders.
I am aware that similar question have already been asked and answered on MSE, but in very different contexts. It therefore seems justified to ask this question again - just to make sure that I don't get this wrong.

Yes. This is indeed the case. Which is really equivalent to saying that $Y\in\cal F$.
For the case of extenders, we usually assume that we have an ultrafilter on $[\lambda]^{<\omega}$ which is $\sigma$-complete. In that case, consider the partition $\{[\lambda]^n\mid n<\omega\}$, exactly one of these parts lies in the ultrafilter.
