Is the following filter known in set theory, and does it have a name ? For $k=1$ it is the filter of cofinite subsets.

Fix a natural number $k$ and a linear order $I$. Define a filter on the set of strictly increasing k-tuples $I^{(k)}:=\{(i_1,...,i_k): i_1<...<i_k \}$ as follows:

A subset $U$ of the set of strictly increasing k-tuples is big iff $U$ contains a tuple of strictly increasing $k$ elements from every infinite subset of I. In notation, call $U\subset I^{(k)}$ big iff for each infinite subset $J\subset I$ there is a $k$-tuple $(j_1,...,j_k)\in U$, $j_1<...<j_k$, $j_1,...,j_k\in J$.

For $k=1$ it means that $U$ intersects each infinite subset, i.e. is cofinite. For $k>1$ the proof that it is a filter uses the Ramsey theorem.

The motivation for the question is that this filter is implicit in the definition of dividing in model theory: a formula $\phi(-,b)$ does not k-divide in a model $M$ iff the subset $\{(b_1,...,b_k)\in M^k: tp(b_i)=tp(b), M\models\exists x \wedge_i \phi(x,b_i)\}$ is big in $p(M)^k=\{(b_1,...,b_k)\in M^k: tp(b_i)=tp(b)\}$,
for any (eqv.,each) linear order on $p(M)$, where $p(M)$ denotes the set of realisations of $tp(b)$ in $M$. Indeed, if it is not big, by definition there is an infinite subset $(b_J)_J$ of realisations of $tp(b)$ such that each $k$-tuple in $\phi$-inconsistent, i.e. $\phi(-,b)$ $k$-divides.

Is the following filter known in set theory, and does it have a name ? For $k=1$ it is the filter of cofinite subsets.

Fix a natural number $k$ and a linear order $I$. Define a filter on the set of strictly increasing k-tuples $I^{(k)}:=\{(i_1,...,i_k): i_1<...<i_k \}$ as follows:

A subset $U$ of the set of strictly increasing k-tuples is big iff $U$ contains a tuple of strictly increasing $k$ elements from every infinite subset of I. In notation, call $U\subset I^{(k)}$ big iff for each infinite subset $J\subset I$ there is a $k$-tuple $(j_1,...,j_k)\in U$, $j_1<...<j_k$, $j_1,...,j_k\in J$.

For $k=1$ it means that $U$ intersects each infinite subset, i.e. is cofinite. For $k>1$ the proof that it is a filter uses the Ramsey theorem. The notion is not quite trivial: for $k=2$ and $I=\Bbb Z$ the subset of pairs with even sum is big, even though the subset of pairs with odd sum is not (take $J$ to be the subset of even numbers).

Perhaps an equivalent description in terms of subsets is clearer. A subset $U$ of the set $P^k(I)$ of subsets of size $k$ is big iff each infinite subset has a subset of size $k$ in $U$, i.e. in notation, $P^k(J)\cap U \neq \emptyset$ for each infinite $J\subset U$.

The motivation for the question is that this filter is implicit in the definition of dividing in model theory: a formula $\phi(-,b)$ does not k-divide in a model $M$ iff the subset $\{(b_1,...,b_k)\in M^k: tp(b_i)=tp(b), M\models\exists x \wedge_i \phi(x,b_i)\}$ is big in $p(M)^k=\{(b_1,...,b_k)\in M^k: tp(b_i)=tp(b)\}$,
for any (eqv.,each) linear order on $p(M)$, where $p(M)$ denotes the set of realisations of $tp(b)$ in $M$. Indeed, if it is not big, by definition there is an infinite subset $(b_J)_J$ of realisations of $tp(b)$ such that each $k$-tuple in $\phi$-inconsistent, i.e. $\phi(-,b)$ $k$-divides.