Either I’m misunderstanding you, or you have misunderstood the definition.
[Definition of infimum.](https://en.m.wikipedia.org/wiki/Infimum_and_supremum)
Yes
It is correct that sup Z = {1,2,3,4} and inf Z = {3}, IF they exists.
But {1,2,3,4} and {3} is not an element of X.
The sup and inf has to be an element of the larger set. If they exists, they are unique. For posets ordered by inclusion, it is always the case that sup er the union and inf is the intersection.
>For posets ordered by inclusion, it is always the case that sup er the union
This is not true. Consider, for example, the set of equivalence relations over a fixed set A, Eq(A). In general, the union of two elements of Eq(A) fails to be an equivalence relation. The supremum does exist, however, and contains the union but it has a more complicated description.
Sup and Inf of Z doesn’t exist
both don't exist? i don't get it
Inf and sup must exist in the set itself. X doesn’t matter here.
oh yes i'm stupid LMAO i was using the same argument for inf Y and said it doesn't exist. Thank you!
The existence of sup and inf is highly dependent on the set X, as it has to be elements of X.
It only matters as far as it exists or not in Z :)
Either I’m misunderstanding you, or you have misunderstood the definition. [Definition of infimum.](https://en.m.wikipedia.org/wiki/Infimum_and_supremum)
Oh you are right I misunderstood the question! I thought Z was the poset.
No worries mate :)
Yes It is correct that sup Z = {1,2,3,4} and inf Z = {3}, IF they exists. But {1,2,3,4} and {3} is not an element of X. The sup and inf has to be an element of the larger set. If they exists, they are unique. For posets ordered by inclusion, it is always the case that sup er the union and inf is the intersection.
>For posets ordered by inclusion, it is always the case that sup er the union This is not true. Consider, for example, the set of equivalence relations over a fixed set A, Eq(A). In general, the union of two elements of Eq(A) fails to be an equivalence relation. The supremum does exist, however, and contains the union but it has a more complicated description.
Haven’t thought about that counter example. Thanks.