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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.
Then has an upper bound (, for example, or ) but no least upper bound in : If we suppose is the least upper bound, a contradiction is immediately deduced because between any two reals and (including and ) there exists some rational , which itself would have to be the least upper bound (if >) or a member of greater than (if <).
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.
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 ...
For example, if the domain is the set of all real numbers, one can assert in first-order logic the existence of an additive inverse of each real number by writing ∀x ∃y (x + y = 0) but one needs second-order logic to assert the least-upper-bound property for sets of real numbers, which states that every bounded, nonempty set of real numbers ...
Proof of the Extreme Value Theorem. By the boundedness theorem, f is bounded from above, hence, by the Dedekind-completeness of the real numbers, the least upper bound (supremum) M of f exists. It is necessary to find a point d in [a, b] such that M = f(d). Let n be a natural number. As M is the least upper bound, M – 1/n is not an upper ...
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.
13934 and other numbers x such that x ≥ 13934 would be an upper bound for S. The set S = {42} has 42 as both an upper bound and a lower bound; all other numbers are either an upper bound or a lower bound for that S. Every subset of the natural numbers has a lower bound since the natural numbers have a least element (0 or 1, depending on ...