3
$\begingroup$

When definining the completion of a field $k$ by a norm one typically uses Cauchy sequences. More specifically the completion of $k$ is defined as the set of equivalence classes of Cauchy sequences in $k$. It seems natural that one would want Cauchy sequences to be convergent in order to say that the field "does not have holes". However I have never seen a justification that would make Cauchy sequences a canonical choice of type of sequence that should be used in the definition of completion. Perhaps there are other equally valid choices to define a completion and they would yield a non-isomorphic field?

Are Cauchy sequences indeed a canonical choice for a definition of the completion? Is there a formulation of the completion by a norm as a universal object in some category or any way to fit Cauchy sequences into a natural setting?

$\endgroup$
13
  • 4
    $\begingroup$ Perhaps you would like to convince yourself of the following fact: any convergent sequence is a Cauchy sequence. So if there is a sequence that becomes convergent after completion it must be a Cauchy sequence. $\endgroup$
    – Zhen Lin
    Jun 5, 2021 at 14:57
  • $\begingroup$ @ZhenLin yes that is true but aren't there any other interesting facts that should be satisfied by all convergent sequences? Why isn't the Cauchy property just one among many possible ones? $\endgroup$
    – mathma
    Jun 5, 2021 at 14:58
  • $\begingroup$ Whatever other interesting facts there may be, they are either equivalent to being a Cauchy sequence, or they are implied by being a Cauchy sequence. It's just logic. $\endgroup$
    – Zhen Lin
    Jun 5, 2021 at 15:05
  • $\begingroup$ @ZhenLin why would it be logic? it seems like a pretty strong statement to say that there are no other weaker conditions implied by convergence in a general normed space. My question is precisely about this. $\endgroup$
    – mathma
    Jun 5, 2021 at 15:09
  • 1
    $\begingroup$ @MartinBrandenburg the question was not to motivate the construction but the definition (the former is trivial enough). perhaps it is not so clear as it is stated, but a careful read should give you a better idea of what was asked. $\endgroup$
    – mathma
    Jun 5, 2021 at 19:43

1 Answer 1

11
$\begingroup$

The purpose of defining a Cauchy sequence is that the property is (a) preserved under isometric maps, meaning that a sequence remains Cauchy if you extend or restrict the ambient metric space, and thus (b) is “intrinsic” to the sequence.

Note: By an “isometric map”, I mean a map $f$ from a metric space $(X, d)$ to a metric space $(Y, \rho)$ such that $\rho(f(x_1), f(x_2)) = d(x_1, x_2)$ for all $x_1, x_2 \in X$. An isometric map describes how $(X, d)$ exists as a metric space inside of $(Y, \rho)$.

Here’s a fact you can check, and I hope it answers your question.

Claim: Let $(X, d)$ be a metric space, and let $(x_n)$ be a sequence in $X$. Then the sequence $(x_n)$ is Cauchy in $(X, d)$ iff there exists a metric space $(Y, \rho)$ and an isometric embedding $\iota: X \hookrightarrow Y$ such that the sequence $(\iota(x_n))$ is convergent in $(Y, \rho)$.

The forward direction is just the completion of a metric space. The backward direction follows because if $(\iota(x_n))$ is convergent, then it’s Cauchy, and so $(x_n)$ is also Cauchy.

In other words, a sequence is Cauchy iff there’s some extension of the metric space that makes the sequence convergent. A Cauchy sequence can be made convergent, but a non-Cauchy sequence can never be made convergent by extending the metric space.

$\endgroup$
3
  • 1
    $\begingroup$ that's actually a really satisfactory answer. Thank you! $\endgroup$
    – mathma
    Jun 5, 2021 at 15:44
  • 1
    $\begingroup$ You mean "convergent" in the claim? $\endgroup$ Jun 5, 2021 at 16:24
  • $\begingroup$ @MartinBrandenburg yes, good catch $\endgroup$
    – AJY
    Jun 5, 2021 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.