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More generally, one may define upper bound and least upper bound for any subset of a partially ordered set X, with “real number” replaced by “element of X ”. In this case, we say that X has the least-upper-bound property if every non-empty subset of X with an upper bound has a least upper bound in X.
In the instance where for each task, its period is an exact multiple of every other task that has a shorter period, the task set can be thought of as being composed of n harmonic task subsets of size 1 and therefore =, which makes this generalization equivalent to Liu and Layland's least upper bound.
There is a corresponding greatest-lower-bound property; an ordered set possesses the greatest-lower-bound property if and only if it also possesses the least-upper-bound property; the least-upper-bound of the set of lower bounds of a set is the greatest-lower-bound, and the greatest-lower-bound of the set of upper bounds of a set is the least ...
If (,) is a partially ordered set, such that each pair of elements in has a meet, then indeed = if and only if , since in the latter case indeed is a lower bound of , and since is the greatest lower bound if and only if it is a lower bound. Thus, the partial order defined by the meet in the universal algebra approach coincides with the original ...
In fact, if this were false, then the integers would have a least upper bound N; then, N – 1 would not be an upper bound, and there would be an integer n such that n > N – 1, and thus n + 1 > N, which is a contradiction with the upper-bound property of N. The real numbers are uniquely specified by the above properties.
For any sets X and Y, X join Y, written X ⊕ Y, is defined to be the union of the sets {2n : n ∈ X} and {2m+1 : m ∈ Y}. The Turing degree of X ⊕ Y is the least upper bound of the degrees of X and Y. Thus is a join-semilattice. The least upper bound of degrees a and b is denoted a ∪ b.
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Now let a 0, a 1, … be pairwise disjoint infinite sets of naturals, and let A 0, A 1, … be their corresponding equivalence classes in P(ω)/Fin. Then given any upper bound X of A 0, A 1, … in P(ω)/Fin, we can find a lesser upper bound, by removing from a representative for X one element of each a n. Therefore the A n have no supremum.