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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.
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 ...
The hyperbolic bound [7] is a tighter sufficient condition for schedulability than the one presented by Liu and Layland: = (+), where U i is the CPU utilization for each task. It is the tightest upper bound that can be found using only the individual task utilization factors.
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 ...
Thus, the infimum or meet of a collection of subsets is the greatest lower bound while the supremum or join is the least upper bound. In this context, the inner limit, lim inf X n, is the largest meeting of tails of the sequence, and the outer limit, lim sup X n, is the smallest joining of tails of the sequence. The following makes this precise.
On the other hand, / is a positive infinitesimal, since by the definition of least upper bound there must be an infinitesimal between / and , and if / < / then is not infinitesimal. But 1 / ( 4 n ) < c / 2 {\displaystyle 1/(4n)<c/2} , so c / 2 {\displaystyle c/2} is not infinitesimal, and this is a contradiction.
If T is the empty set, then {v} is an upper bound for T in P. Suppose then that T is non-empty. We need to show that T has an upper bound, that is, there exists a linearly independent subset B of V containing all the members of T. Take B to be the union of all the sets in T. We wish to show that B is an upper bound for T in P.
By the least-upper-bound property of real numbers, = {} exists and . Now, for every ε > 0 {\displaystyle \varepsilon >0} , there exists N {\displaystyle N} such that c ≥ a N > c − ε {\displaystyle c\geq a_{N}>c-\varepsilon } , since otherwise c − ε {\displaystyle c-\varepsilon } is a strictly smaller upper bound of { a n ...