Just in case anyone was wondering why we can’t have any finite sets in the ultrafilter:
If some finite set {n1, n2, …, n_k} is in an ultrafilter U, then either {n1, n2, …, n_(k-1)} is in U or I \ {n1, n2, …, n_(k-1)} is in U. In the latter case, the intersection with the original set is {n_k}, which must be in U. In the former case, you can keep repeating this until you are left with some other one-element set.
If any one-element set {n} is in U, then membership in U is just decided by whether a set contains n or not.
When you go through the equivalence construction, this means that two sequences are equivalent if and only if they agree at the n’th position, which means that all the operations are just the same as arithmetic on that position with the rest not mattering at all. So to get anything different, U really does have to be a non-principal ultrafilter.
Just in case anyone was wondering why we can’t have any finite sets in the ultrafilter:
If some finite set {n1, n2, …, n_k} is in an ultrafilter U, then either {n1, n2, …, n_(k-1)} is in U or I \ {n1, n2, …, n_(k-1)} is in U. In the latter case, the intersection with the original set is {n_k}, which must be in U. In the former case, you can keep repeating this until you are left with some other one-element set.
If any one-element set {n} is in U, then membership in U is just decided by whether a set contains n or not.
When you go through the equivalence construction, this means that two sequences are equivalent if and only if they agree at the n’th position, which means that all the operations are just the same as arithmetic on that position with the rest not mattering at all. So to get anything different, U really does have to be a non-principal ultrafilter.