![]() ![]() ![]() If is a uniformly continuous map of metric spaces, then it is Cauchy-continuous. Next, the main result we’d like to prove is: To show that it’s not uniformly continuous, negating the definition means we need to find an ε>0 such that for any δ>0, there exist x and x’ such that | x – x’| 0, pick a positive integer n > 1/δ and let We now have: Take the function f : R + → R + given by f( x) = 1/ x as before. But here’s an example where the converse is not true. for any ε>0, there exists δ>0 such that whenever satisfies we haveĬlearly, a uniformly continuous function is also continuous (at every point of X).A map of metric spaces is said to be uniformly continuous if We had already seen it earlier in the case of R.ĭefinition. The answer to our problem is the following definition. To rectify that, we need a stronger form of continuity. Warning. Not all continuous functions are Cauchy-continuous. A function of metric spaces is said to be Cauchy-continuous if whenever is a Cauchy sequence in X, is a Cauchy sequence in Y. Put in another way, we can define two metrics on R + via and Then the sequence is Cauchy under the first metric but not the second.ĭefinition. The sequence is Cauchy, but the resulting sequence is not. In other words, it’s possible for two metrics on the same space to be topologically equivalent, but a sequence is Cauchy in one and not the other.įor example, consider the homeomorphism f : R + → R + of the space of positive reals given by f( x) = 1/ x. The concept of Cauchy sequences actually relies heavily on the metric and not just the underlying topology. One might ask if it’s necessary to consider all three metrics on X × Y since they all give rise to the same topology anyway. Thus, when m, n > max( M, N), we have – for any metric d on X × Y in the above list – there exists N such that when m, n > N, we have. ![]() there exists M such that when m, n > M, we have.If are sequences of metric spaces respectively, then is a Cauchy sequence of X × Y if and only if each of is Cauchy in the respective metric space.įor the metric of X × Y, we can pick any one of the following:įirst suppose and are Cauchy. For any ε>0, being Cauchy is a property of the sequence itself, regardless of the ambient space. There’s nothing to prove here since Y inherits the distance function from X. However, this serves to highlight the fact that while a sequence from Y which converges in X may not be convergent in Y, the same problem doesn’t hold for Cauchy sequences, i.e. Then a sequence in Y is Cauchy if and only if it’s Cauchy in X. For any ε>0, there exists N such that whenever n > N, we have Thus, whenever m, n > N, we have: for any ε>0, there exists N such that whenever m, n > N, we have.Recall that on an intuitive level, a Cauchy sequence is one where the elements get “closer and closer”.ĭefinition. For an example of the use of this test on a Banach space, see the article Fréchet derivative.We wish to generalise the concept of Cauchy sequences to metric spaces. | f n ( x ) | ≤ M n is the norm on the Banach space.Suppose that ( f n) is a sequence of real- or complex-valued functions defined on a set A, and that there is a sequence of non-negative numbers ( M n) satisfying the conditions It is named after the German mathematician Karl Weierstrass (1815-1897). It applies to series whose terms are bounded functions with real or complex values, and is analogous to the comparison test for determining the convergence of series of real or complex numbers. In mathematics, the Weierstrass M-test is a test for determining whether an infinite series of functions converges uniformly and absolutely.
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