Yes. We have 2=[(2,2,2,...)]. But we can compare 2 with (1,3,1,3,1,3,...) since (1,3,1,3,1,3,1,3,...)=1 (this happens when the set of all even natural numbers is in your ultrafilter) or (1,3,1,3,1,3,1,3,...)=3 (this happens when the set of all odd natural numbers is in your ultrafilter). Your partially ordered set is actually a linear ordering because whenever we have two sequences (an)n,(bn)n, one of the sets
{n:an>bn},{n:an<bn},{n:an=bn} is in your ultrafilter (you can think of an ultrafilter as a thing that selects one block out of every partition of the natural numbers into finitely many pieces), and if your ultrafilter contains
Yes. We have 2=[(2,2,2,...)]. But we can compare 2 with (1,3,1,3,1,3,...) since (1,3,1,3,1,3,1,3,...)=1 (this happens when the set of all even natural numbers is in your ultrafilter) or (1,3,1,3,1,3,1,3,...)=3 (this happens when the set of all odd natural numbers is in your ultrafilter). Your partially ordered set is actually a linear ordering because whenever we have two sequences (an)n,(bn)n, one of the sets
{n:an>bn},{n:an<bn},{n:an=bn} is in your ultrafilter (you can think of an ultrafilter as a thing that selects one block out of every partition of the natural numbers into finitely many pieces), and if your ultrafilter contains
{n:an>bn}, then [(an)n]>[(bn)n].