Search results
Results from the WOW.Com Content Network
In any metric space, a Cauchy sequence is bounded (since for some N, all terms of the sequence from the N-th onwards are within distance 1 of each other, and if M is the largest distance between and any terms up to the N-th, then no term of the sequence has distance greater than + from ).
A metric space is said to be totally bounded if every sequence admits a Cauchy subsequence; in complete metric spaces, a set is compact if and only if it is closed and totally bounded. [2] Each totally bounded space is bounded (as the union of finitely many bounded sets is bounded).
The theorem states that each infinite bounded sequence in has a convergent subsequence. [1] An equivalent formulation is that a subset of R n {\displaystyle \mathbb {R} ^{n}} is sequentially compact if and only if it is closed and bounded . [ 2 ]
(This limit exists because the real numbers are complete.) This is only a pseudometric, not yet a metric, since two different Cauchy sequences may have the distance 0. But "having distance 0" is an equivalence relation on the set of all Cauchy sequences, and the set of equivalence classes is a metric space, the completion of M.
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.
This defines two Cauchy sequences of rationals, and so the real numbers l = (l n) and u = (u n). It is easy to prove, by induction on n that u n is an upper bound for S for all n and l n is never an upper bound for S for any n. Thus u is an upper bound for S. To see that it is a least upper bound, notice that the limit of (u n − l n) is 0 ...
Cauchy continuity makes sense in situations more general than metric spaces, but then one must move from sequences to nets (or equivalently filters). The definition above applies, as long as the Cauchy sequence (,, …) is replaced with an arbitrary Cauchy net.
In a general metric space, however, a Cauchy sequence need not converge. In addition, for real-valued sequences that are monotonic, it can be shown that the sequence is bounded if and only if it is convergent.