Search results
Results From The WOW.Com Content Network
Every bounded-above monotonically nondecreasing sequence of real numbers is convergent in the real numbers because the supremum exists and is a real number. The proposition does not apply to rational numbers because the supremum of a sequence of rational numbers may be irrational.
It is possible to prove the least-upper-bound property using the assumption that every Cauchy sequence of real numbers converges. Let S be a nonempty set of real numbers. If S has exactly one element, then its only element is a least upper bound. So consider S with more than one element, and suppose that S has an upper bound B 1.
Each of the probabilities on the right-hand side converge to zero as n → ∞ by definition of the convergence of {X n} and {Y n} in probability to X and Y respectively. Taking the limit we conclude that the left-hand side also converges to zero, and therefore the sequence {(X n, Y n)} converges in probability to {(X, Y)}.
The monotone convergence theorem (described as the fundamental axiom of analysis by Körner [1]) states that every nondecreasing, bounded sequence of real numbers converges. This can be viewed as a special case of the least upper bound property, but it can also be used fairly directly to prove the Cauchy completeness of the real numbers.
In particular, every subset Y of X is bounded above by X and below by the empty set ∅ because ∅ ⊆ Y ⊆ X. Hence, it is possible (and sometimes useful) to consider superior and inferior limits of sequences in ℘(X) (i.e., sequences of subsets of X). There are two common ways to define the limit of sequences of sets. In both cases:
Convergence proof techniques are canonical patterns of mathematical proofs that sequences or functions converge to a finite limit when the argument tends to infinity. There are many types of sequences and modes of convergence , and different proof techniques may be more appropriate than others for proving each type of convergence of each type ...
Proof: (sequential compactness implies closed and bounded) Suppose A {\displaystyle A} is a subset of R n {\displaystyle \mathbb {R} ^{n}} with the property that every sequence in A {\displaystyle A} has a subsequence converging to an element of A {\displaystyle A} .
[proof 1] The sequence a n is increasing and bounded above by b, so the limit A = lim n → ∞ a n exists. Similarly, the limit B = lim n → ∞ b n exists since the sequence b n is decreasing and bounded below by a. Also, a n < b n implies A ≤ B. If A < B, then for every n: x n ∉ (A, B) because x n is not in the larger interval (a n, b n).