Recall that in real analysis, we usually say that given the convergence of the series,
$latex\displaystyle 1 + \frac 12 + \frac 14 + \frac 18 + \cdots = 2,$
and
However, if we manipulate the second sum formally,
yielding the rather suspect claim that
We brush this result aside as “absurd” since this series diverges. But what do we mean by diverge? We mean that the partial sums do not converge. What does “converge” mean? It means that terms get “closer” to each other. Well, what does “closer” mean? It turns out that we superimposed our intuition of closeness, that is, given two numbers , we measure their “closeness” by
. But what happens if we use a different kind of closeness?
To be absolutely clear, in order to even begin discussing many of our results, we had to construct from
, which we have in fact already done, so that the notion of a metric
is even well-defined (and useful via the supremum property).
Yet, we already have a useful result that Cauchy sequences in converge in
. We’re going to use this property to “(Cauchy-)complete” any metric space
into the metric space
that contains
as a dense subspace, and setting
equipped with a special metric
, the equally following suspect equation actually makes sense:
Henceforth, let denote any metric space.
Definition 1. Let denote the subset of Cauchy sequences in
. We remark that
contains convergent
-sequences as a subset.
Lemma 1. The relation defined on
by
if and only if
is an equivalence relation. In particular, for any
,
if and only if
.
Corollary 1. Denote . The canonical inclusion
defined by
is a well-defined injection. Henceforth, for
, we denote
without ambiguity, so that
.
Lemma 2. For any ,
This is a consequence of the reverse triangle inequality for normed spaces.
Lemma 3. The map defined by
is a well-defined metric on
. Furthermore,
.
Proof. The hardest part of this result is to verify that is well-defined.
Firstly, fix Cauchy -sequences
. Fix
. Then for
, there exists sufficiently large
such that for
,
By Lemma 2,
Setting , we conclude that
is Cauchy in
and hence converges in
.
To prove uniqueness, suppose and
. Applying bookkeeping on the estimate
yields the result.
We leave it as an exercise to verify that is a metric on
.
Lemma 4. .
Proof. Fix . We leave it as an exercise to show that the sequence
converges to
. This means
.
Theorem 1. For any metric space , there exists a unique complete metric space
that contains
as a subspace. Furthermore,
.
Proof. Combine Lemmas 1, 3, and 4 to construct with the desired properties. All that remains is to prove that
is complete.
Let be any Cauchy
-sequence. For each
, use Lemma 4 to choose any
. We leave it as an exercise to check that
is Cauchy, and define
. By the triangle inequality, given
,
For sufficiently large and
,
and
, yielding
.
To resolve the seemingly absurd equation, we will need to define the -adic metric, which requires a brief discussion on
-adic numbers. Let
be any prime number.
Lemma 5. For any nonzero rational number , there exists a unique integer
and a unique rational number
such that
Proof. Write for positive integers
, and
if and only if
. By the fundamental theorem of arithmetic, find unique positive integers
such that
Defining and
yields the desired result.
Now we are ready to define the -adic norm on
, which induces a metric on
.
Lemma 6. The map defined by
where is defined according to Lemma 5 if
, and
, is a norm on
.
Proof. For non-degeneracy, if , then
, so that
. For the triangle inequality, we leave it as an exercise to verify that
, so that
Roughly speaking, then, a rational number is small in the -adic sense if it “contains” a higher power of
.
Example 1. For any power of ,
.
Definition 2. The set -adic numbers, denoted by
, is defined to be the completion of
under the metric
induced by the
–adic norm defined in Lemma 5:
.
Theorem 2. Under the -adic metric,
Proof. Without reservation, the partial sum is given by the geometric sum
so that . Hence,
Therefore, , and we get
We leave it as an exercise to verify that
when convergence is defined by the -adic metric. In fact the expression
, while meaningless in
, does have a meaning in
, and we can evaluate it to
.
—Joel Kindiak, 12 Apr 25, 1758H
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