Proving Cauchy Convergence in Natural Nos: Metrics & Converse

[solakis1]

Using the fact that the Natural Nos are complete .then prove that every Cauchy sequence in Natural Nos converges in N and the converse.

I do not even know if we can have a Cauchy sequence in Natural Nos.

What would be the appropriate metric to use in our Cauchy sequence??

 

[zzephod]

Using the fact that the Natural Nos are complete .then prove that every Cauchy sequence in Natural Nos converges in N and the converse.

I do not even know if we can have a Cauchy sequence in Natural Nos.

What would be the appropriate metric to use in our Cauchy sequence??

You will need to specify what you mean by complete here since the usual definition is to a hand waving approximation that a metric space is complete iff every Cauchy sequence converges.

The obvious metric here would be the discrete metric:

\(\displaystyle \rho (x,y)=\begin{cases}1,& {\rm{if}}\ x\ne y \\ 0,& {\rm{if}}\ x = y \end{cases}\)

or even:

\(\displaystyle \rho(x,y)=|x-y|\)

for the purposes of convergence of sequences these should be equivalent (as would most natural choices of metric) so the choice would be whichever is most convenient.

.

 

[solakis1]

You will need to specify what you mean by complete here since the usual definition is to a hand waving approximation that a metric space is complete iff every Cauchy sequence converges.

The obvious metric here would be the discrete metric:

\(\displaystyle \rho (x,y)=\begin{cases}1,& {\rm{if}}\ x\ne y \\ 0,& {\rm{if}}\ x = y \end{cases}\)

or even:

\(\displaystyle \rho(x,y)=|x-y|\)

for the purposes of convergence of sequences these should be equivalent (as would most natural choices of metric) so the choice would be whichever is most convenient.

.

How do we define |x-y| in natural Nos??

Are the sequences of the type : 1,1,1,1,1,1,1...

5,5,5,5,5,5,5......

The only Cauchy sequences in Natural Nos??

 

[zzephod]

How do we define |x-y| in natural Nos??

Are the sequences of the type : 1,1,1,1,1,1,1...

5,5,5,5,5,5,5......

The only Cauchy sequences in Natural Nos??

A metric on a space is a real functional defined on the space, so it takes values in \(\displaystyle \mathbb{R}\). By \(\displaystyle |x-y|\) we mean the real corresponding to the natural \(\displaystyle x-y\) when \(\displaystyle x\ge y\) and \(\displaystyle y-x\) otherwise, where our definition of naturals includes \(\displaystyle 0\). What that means depends on how you have constructed (or defined) the reals.

Note, even for the discrete metric the \(\displaystyle 0\) and \(\displaystyle 1\) are to be considered reals not naturals.

The Cauchy sequences are those that beyond a certain point are constant.

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[solakis1]

A metric on a space is a real functional defined on the space, so it takes values in \(\displaystyle \mathbb{R}\)

.

Or is the function defined on a space and takes values in the positive real Nos??

 

[zzephod]

Or is the function defined on a space and takes values in the positive real Nos??

I am not defining a metric here so this is sufficient for my purpose. Which is to point out that the range of the metric is a subset of the reals.

If I were defining a metric, that it is a non-negative would appear in the definition usually in the form \(\displaystyle d(x,y)\ge 0\), and \(\displaystyle d(x,y)=0 \Leftrightarrow x=y\) and this would appear after stating that a metric is a function from the Cartesian product of the space with itself to the reals satisfying the conditions. Which you of course already know as you have the definition of a metric space in front of you!

(the non-negativity condition is redundant as it follows from the triangle inequality and symmetry conditions in the usual presentation of the definition)

Finally, why are we even discussing this, your original question is resolved and these subsequent questions can be resolved by looking at the definition which you presumably have in front of you. To summarise stop wasting peoples time on questions to which you know the answers. It make you look like a troll (and quack like a troll ...).

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[Deveno]

There are two slightly different "competing views" of the natural numbers in this thread. I will attempt to make clear what I mean:

We can view $\Bbb N$ as "a space unto itself", without reference to it belonging to some "larger structure".

We can view $\Bbb N$ as a "distinguished subset of the reals, $\Bbb R$".

In the first view the "natural metric" to impose upon $\Bbb N$ is the discrete metric:

$d(k,m) = 1$ for $k \neq m$
$d(k,k) = 0$.

We can ask, what is a Cauchy sequence, with this first metric? To do this, we have to rephrase the Cauchy condition in terms of an arbitrary metric. We say that a sequence $\{a_n\}$ is Cauchy with respect to the metric $d$, if for any REAL $\epsilon > 0$, there is some natural number $N$, such that, for all natural numbers $m,n > N$:

$d(a_m,a_n) < \epsilon$

It is clear that any sequence which is eventually constant is Cauchy under this revised definition. It should also be clear that if a sequence is Cauchy, then for $\epsilon = \frac{1}{2}$ the only way we can have:

$d(a_{n+k},a_n) < \frac{1}{2}$

for all $n > N$ (no matter what $N$ may be), is for $a_{n+k} = a_n$ for all $n > N$ and all $k$, which is to say the sequence is eventually constant.

The second view of the natural numbers is to view them as a subset of the real numbers, which has the "usual metric":

$d(x,y) = |y - x|$.

We then view $\Bbb N$ as a subspace with the relative metric topology. Note that for $k,m \in \Bbb N$, that $d(k,m) = 0$ or $d(k,m) \in \Bbb N$. In particular, if:

$d(a_m,a_n) = |a_n - a_m| < \frac{1}{2}$

for all $m,n > N$ then we must have $|a_n - a_m| = 0 \implies a_n = a_m$.

So, even though this is "a different metric", we get the same Cauchy sequences as before: the ones that are eventually constant (that is, constant except for a finite number of terms at the beginning).