Proof of sentence - 2014-12-03

evinda

18:15

Hey @DanielFischer!!! Are you there? :)

Daniel Fischer

A little. Still not completely fit.

evinda

Have you been ill for such a long time? :o

Daniel Fischer

A week ain't a long time.

evinda

Yes, you are right...

Could we coninue now with the proof or better an other time?

Daniel Fischer

Might as well try to get it done. Where were we?

evinda

18:32

I hadn't understood how we concluded that there is an $a_k \in \mathbb{Z}/p^{k+1}\mathbb{Z}$ such that $x^{(i)}_k = a_k$ for all $i \geqslant N_k$..

Daniel Fischer

We chose $N_k$ such that $\lvert x^{(i)} - x^{(j)}\rvert_p < p^{-k}$ for all $i,j \geqslant N_k$.

Next, $\lvert u-v\rvert_p < p^{-k}$ means $u_k = v_k$ (and $u_m = v_m$ for all $m \leqslant k$).

And so we have $x^{(i)}_k = x^{(j)}_k$ for all $i,j \geqslant N_k$.

And we call that value $a_k$.

evinda

A ok.. I will think about it... And how do we use this?

Daniel Fischer

That gives us a candidate for the limit of $(x^{(i)})$. Then we show first that $(a_0,a_1,\dotsc)$ is an element of $\mathbb{Z}_p$, and second, that it is indeed the limit of $(x^{(i)})$.

Then we will have proved that $\mathbb{Z}_p$ is complete.

evinda

Do we look at the limit of $(x^{(i)})$ as for $i$ ?

Daniel Fischer

Yes, of course, $i$ is the sequence index.

evinda

18:43

So, do we conclude that $a_k$ is a candidate for the limit of $(x^{(i)})$, because $a_k=x^{(i)}_k, \forall i \geq N_k$?

Daniel Fischer

@evinda $a_k$ is not the right type of thing to be the limit of $(x^{(i)})$. Remember, $x^{(i)} \in \mathbb{Z}_p$, and $a_k \in \mathbb{Z}/p^{k+1}\mathbb{Z}$.

The collection of all the $a_k$ gives us an element of $\mathbb{Z}_p$.

evinda

So, does this: $a_k=x^{(i)}_k$ mean that $a_0=x^{(i)}_0, a_1=x^{(i)}_1, a_2=x^{(i)}_2$ and so on?

@DanielFischer Also, how could we show that $(a_0,a_1,\dotsc)$ is an element of $\mathbb{Z}_p$, and that it is indeed the limit of $(x^{(i)})$?

Daniel Fischer

@evinda Not for all $i$, only for $i \geqslant N_k$, so $a_0 = x^{(i)}_0$ for $i \geqslant N_0$, $a_1 = x^{(i)}_1$ for $i \geqslant N_1$, $a_2 = x^{(i)}_2$ for $i \geqslant N_2$ etc.

@evinda To show that $(a_0,a_1,a_2,\dotsc)\in \mathbb{Z}_p$, we must check the compatibility relations, namely $$a_{k+1} \equiv a_k \pmod{p^{k+1}}$$ for all $k$.

evinda

19:00

@DanielFischer And can we use this:

$a_0 = x^{(i)}_0$ for $i \geqslant N_0$, $a_1 = x^{(i)}_1$ for $i \geqslant N_1$, $a_2 = x^{(i)}_2$ for $i \geqslant N_2$

to show this?

Daniel Fischer

But that follows immediately if we look at an $i \geqslant N_{k+1}$, for then $a_{k+1} = x^{(i)}_{k+1}$ and $a_k = x^{(i)}_k$, and we have the compatibility since $x^{(i)}\in \mathbb{Z}_p$.

evinda

@DanielFischer Why do we take an $i \geqslant N_{k+1}$?

Daniel Fischer

Because that tells us something about $a_k$ and $a_{k+1}$. An $i < N_{k+1}$ wouldn't tell us anything about $a_{k+1}$.

evinda

@DanielFischer So, does this mean that we have $a_k=x_k^{(i)}, \forall i \geq N_k$ and $a_{k+1}=x_{k+1}^{(i)}, \forall i \geq N_{k+1}$ and because of the fact that we want that both of the above relations stand, we take the greatest $i$, which is $N_{k+1}$ ?

Daniel Fischer

Not "the greatest $i$". We have arbitrarily large $i$. But we must take a "large enough" $i$, and that means we take the greater of the two lower bounds.

evinda

19:12

@DanielFischer I see.. And how could we prove that $a_k$ is indeed the limit of $(x^{(i)})$ ?

Daniel Fischer

@evinda Have you understood that $a\in\mathbb{Z}_p$, where $a = (a_0,a_1,a_2,\dotsc)$?

evinda

@DanielFischer Is it because of the following or have I understood it wrong?

Since $(x^{(i)}) \in \mathbb{Z}_p$, we have that $x^{(i)}_{k+1} \equiv x^{(i)}_{k} \pmod{p^{k+1}}, i \geq N_k \Rightarrow a_1 \equiv a_0 \pmod{p}, a_2 \equiv a_1 \pmod{p^2}, a_3 \equiv a_2 \pmod{p^3}$

Daniel Fischer

Yes, that's right.

evinda

Great!!! :) And how could we prove that $a_k$ is the limit of $(x^{(i)})$ ?

Daniel Fischer

So, to see that $a = \lim\limits_{i\to\infty} x^{(i)}$, can you give a bound on $$\lvert x^{(i)} - a \rvert_p$$ when $i \geqslant N_k$?

evinda

19:23

@DanielFischer What do you mean? :/ How can we find a bound on $$\lvert x^{(i)} - a \rvert_p$$ ?

Daniel Fischer

When $i \geqslant N_k$, what do you know about $x^{(i)}$ and $a$?

evinda

@DanielFischer
It holds that $a_k=x_k^{(i)}, \forall i \geq N_k$.
But how can we find a bound on $|x^{(i)}-a|_p$, now that we don't have a single point?

Daniel Fischer

We know $x^{(i)}_k = a_k$ for $i\geqslant N_k$.

Now, that tells us something about $\lvert x^{(i)} - a \rvert_p$.

What was it that that tells us?

evinda

19:39

@DanielFischer That it is $\lvert x^{(i)} - a \rvert_p<p^{-k}$ ? Or am I wrong?

Daniel Fischer

@evinda And how does $\lvert x^{(i)} - a\rvert_p < p^{-k}$ follow?

evinda

@DanielFischer We have that $\lvert x^{(i)} - a\rvert_p < p^{-k}$, since $x^{(i)}$ is a Cauchy sequence, right?

Daniel Fischer

No, scroll up and check again what we had about $\lvert\,\cdot\,\rvert_p$ and what $\lvert u\rvert_p < p^{-k}$ means.

evinda

@DanielFischer Do you mean this?

$\lvert u\rvert_p < p^{-k}$ means means that $w_p(u)>k$ ?

Daniel Fischer

@evinda Not quite that. There was also something it says about the components of $u$.

Anyway, $\lvert u\rvert_p < p^{-k} \iff u_k = 0$, and here, $\lvert x^{(i)} - a\rvert_p < p^{-k} \iff x^{(i)}_k = a_k$.

So $i \geqslant N_k \implies \lvert x^{(i)} - a\rvert_p < p^{-k}$, which shows that $$a = \lim_{i\to\infty} x^{(i)}.$$

evinda

20:04

@DanielFischer Why do we conclude from this: $x^{(i)}_k = a_k$ that
$$a = \lim_{i\to\infty} x^{(i)}$$
?

Daniel Fischer

Because $i\geqslant N_k \implies \lvert x^{(i)} - a\rvert_p < p^{-k}$.

Now recall what the definition of the limit of a sequence was.

evinda

@DanielFischer The definition of $\lim_{n \to +\infty} a_n=a \text{ in } \mathbb{Q}_p$ means that $\forall \epsilon>0 \exists n_0 \text{ such that} \forall n \geq n_0:$
$$|a_n-a|_p< \epsilon$$

So, we conclude that $\lim_{i \to +\infty} x^{(i)}=a$, from this $\lvert x^{(i)} - a\rvert_p < p^{-k}$, right?

Daniel Fischer

Since, whatever $\epsilon > 0$ is, we have $p^{-k} \leqslant \epsilon$ for all large enough $k$.

evinda

@DanielFischer And what do we conclude from this?

Daniel Fischer

That $\mathbb{Z}_p$ is complete?

evinda

20:18

@DanielFischer Could you explain me further why we conclude from that that $\mathbb{Z}_p$ is complete?

Daniel Fischer

What does "$\mathbb{Z}_p$ is complete" mean?

evinda

@DanielFischer I don't really know... Could you explain it to me?

Daniel Fischer

If you don't really know what it means that a metric space is complete, you should recapitulate that part of your analysis lectures. It is an important concept, you ought to know the definitions and understand the ideas well. By definition, a metric space is complete if and only if every Cauchy sequence in that space is convergent.

So, to prove that a metric space is complete, you pick an arbitrary Cauchy sequence in that space, and show it converges.

That's what we did.

We picked an arbitrary Cauchy sequence in $\mathbb{Z}_p$, the sequence $(x^{(i)})$, and then showed that it has a limit.

And thus, we have shown that $\mathbb{Z}_p$ is complete.

evinda

@DanielFischer A ok!!!! And do we have to check also where the limit of the sequence belongs to?

Daniel Fischer

20:34

If the limit exists, that means it belongs to the space.

Here, we needed to check that $a\in\mathbb{Z}_p$ in order to have a candidate for the limit.

evinda

@DanielFischer A ok... So, now we have shown that $\mathbb{Z}_p$ is complete, right?
How do we continue? Do we have to show now that $\mathbb{Q}_p$ is complete? Or something else? :/

Daniel Fischer

@evinda Do you remember the beginning? We found that to show the completeness of $\mathbb{Q}_p$, it is sufficient to show the completeness of $\mathbb{Z}_p$. Do you remember the argument for that?

evinda

@DanielFischer At the beginning, we showed that for every Cauchy sequence $(x_i)$ in $\mathbb{Q}_p$, there is a $c\in \mathbb{Q}_p\setminus \{0\}$ such that $(c\cdot x_i)$ is a sequence in $\mathbb{Z}_p$ and that $(x_i)$ is Cauchy if and only if $(c\cdot x_i)$ is Cauchy, right? :)

Daniel Fischer

Right.

evinda

@DanielFischer And we use this to conclude that $\mathbb{Q}_p$ is also complete?

Daniel Fischer

20:44

And now, all that's missing is to note that for $c\neq 0$, we have $x^{(i)} \to a \iff c\cdot x^{(i)} \to c\cdot a$.

evinda

@DanielFischer Do we have to prove this or can we just say that it is like that?

Daniel Fischer

Well, nothing can be just "said to be like that". Everything must be proved.

But, $$\lvert c\cdot x^{(i)} - c\cdot a\rvert_p = \lvert c\rvert_p\cdot \lvert x^{(i)} - a\rvert_p.$$

evinda

@DanielFischer At the beginning we said that $(x_i)$ in $\mathbb{Q}_p$ and $(c\cdot x_i)$ is a sequence in $\mathbb{Z}_p$...

Now, we have that $(x^{(i)}) \in \mathbb{Z}_p$ and $cx^{(i)} \in \mathbb{Q}_p$, right?

Daniel Fischer

If it confuses you, call the sequence in $\mathbb{Q}_p$ $(y^{(i)})$, then let $x^{(i)} = c\cdot y^{(i)}$, and then use $$\lvert y^{(i)} - c^{-1}\cdot a\rvert_p = \lvert c^{-1}\cdot x^{(i)} - c^{-1}\cdot a\rvert_p = \lvert c^{-1}\rvert_p\cdot \lvert x^{(i)} - a\rvert_p.$$

evinda

@DanielFischer We showed that for every Cauchy sequence $(x_i)$ in $\mathbb{Q}_p$, there is a $c\in \mathbb{Q}_p\setminus \{0\}$ such that $(c\cdot x_i)$ is a sequence in $\mathbb{Z}_p$ .. Do we know also know that for every Cauchy sequence $(x_i)$ in $\mathbb{Z}_p$, there is a $c\in \mathbb{Q}_p\setminus \{0\}$ such that $(c\cdot x_i)$ is a sequence in $\mathbb{Q}_p$ ?

Daniel Fischer

21:02

@evinda That part is trivial. Since $\mathbb{Z}_p \subset \mathbb{Q}_p$, and $\mathbb{Q}_p$ is a field, for every $c\in \mathbb{Q}_p$ and every sequence $(x^{(i)})$ in $\mathbb{Z}_p$, the sequence $(c\cdot x^{(i)})$ is a sequence in $\mathbb{Q}_p$. If $c\neq 0$, then by the multiplicativity of the $p$-norm, the sequence $(c\cdot x^{(i)})$ is a Cauchy sequence if and only if $(x^{(i)})$ is a Cauchy seqeunce.

evinda

@DanielFischer Why do we take $c^{-1}$ at $\lvert y^{(i)} - c^{-1}\cdot a\rvert_p $ ?

Also, do we know that $|c^{-1}|_p=1$ ?

Daniel Fischer

@evinda If $x^{(i)} = c\cdot y^{(i)}$, then $y^{(i)} = c^{-1}\cdot x^{(i)}$. If we used $c$ to move into $\mathbb{Z}_p$, then we use the reciprocal, $c^{-1}$, to move back where we came from.

And generally, we won't have $\lvert c^{-1}\rvert_p = 1$.

But it is some fixed nonzero constant. We don't really care whether it is $p^{-1234567890123456789}$, or $p^{9876543210987654321}$, or something in between, or something larger, or smaller.

evinda

21:25

@DanielFischer A ok.. And how can we continue, knowing that $|y^{(i)}-c^{-1}a|_p=|c^{-1}|_p |x^{(i)}-a|_p$ ?

Daniel Fischer

There really isn't anything to continue. Since $\lvert c^{-1}\rvert_p$ is a (non-zero) constant, it is directly clear that $y^{(i)} \to c^{-1}\cdot a \iff x^{(i)}\to a$.

evinda

@DanielFischer So can we always find an $\epsilon$ such that $|c^{-1}|_p \cdot |x^{(i)}-a|_p< \epsilon$ ?

Daniel Fischer

@evinda That's backwards.

You must, for every $\epsilon > 0$, find an $N_\epsilon\in\mathbb{N}$, such that for all $i \geqslant N_\epsilon$ ...

evinda

@DanielFischer We know that $\forall i \geq N_k , \exists \epsilon>0 \text{ such that } |x^{(i)}-a|_p<$, right?

Can we just multiply by $c$?

Daniel Fischer

@evinda You got it backwards again.

First you pick $\epsilon > 0$.

Then you find an $N$.

$$\bigl(\forall \epsilon > 0\bigr)\bigl(\exists N_\epsilon\bigr)\dotsc$$

evinda

21:43

@DanielFischer I am confused now..
We have that $\lvert y^{(i)} - c^{-1}\cdot a\rvert_p = \lvert c^{-1}\cdot x^{(i)} - c^{-1}\cdot a\rvert_p = \lvert c^{-1}\rvert_p\cdot \lvert x^{(i)} - a\rvert_p.$

Why do we write the definition of the limit?

Daniel Fischer

Sorry, what?

evinda

@DanielFischer What do we have to do, in order to show that $y^{(i)} \to c^{-1}\cdot a \iff x^{(i)}\to a$? :/

Daniel Fischer

Just note that the $p$-norm is multiplicative. And that $\lvert c^{-1}\rvert_p \neq 0$.

It's the same argument as one uses for $\lim (c\cdot x_n) = c\cdot \lim x_n$ for real sequences in the first semester.

evinda

@DanielFischer So, can we say it like that?

We have that $\forall \epsilon>0, \exists n_0=n_0(\epsilon)$ such that $\forall n \geq n_0$:

$$|x^{(i)}-x^{(j)}|_p< \epsilon$$

$$|y^{(i)}-c^{(-1)} \cdot a|_p=|c^{-1}|_p \cdot |x^{(i)}-a|_p< |c^{-1}|_p \cdot \epsilon$$

Can we conclude from that that $\lim_{i \to +\infty} y^{(i)}=c^{-1}a$?

Daniel Fischer

22:00

Yes, we can conclude. Given $\epsilon > 0$, we pick $n_0$ such that $\lvert x^{(i)} - a\rvert_p < \lvert c\rvert_p\cdot \epsilon$ for $i\geqslant n_0$. Then we have $$\lvert y^{(i)} - c^{-1}\cdot a\rvert_p = \lvert c^{-1}\rvert_p \lvert x^{(i)} - a\rvert_p < \lvert c^{-1}\rvert_p\cdot \lvert c\rvert_p\epsilon = \epsilon$$ for $i \geqslant n_0$.

evinda

22:29

@DanielFischer So, can we say it like that?

We have that $a=\lim_{n \to +\infty} x^{(i)}$.
That means that $\forall \epsilon'>0, \exists n_0$ such that $\forall i \geq n_0$:

$$|x^{(i)}-a|_p< \epsilon'$$

We pick $\espilon'=|c|_p \cdot \epsilon$.

$$|x^{(i)}-a|_p< |c|_p \cdot \epsilon$$
So:
$$|y^{(i)}-c^{-1}a|_p=|c^{-1}|_p |x^{(i)}-a|_p<|c^{-1}|_p \cdot |c|_p \epsilon= \epsilon$$

Therefore, $y^{(i)} \to c^{-1}a$.

Daniel Fischer

Yes.

evinda

@DanielFischer So have we shown now that each Cauchy sequence in $\mathbb{Q}_p$ converges, i.e that $\mathbb{Q}_p$ is complete?

Daniel Fischer

Yes.

evinda

And what does remain to show now?

Daniel Fischer

Nothing.

evinda

22:50

@DanielFischer Ok :) I will read tomorrow again all what we said and will tell you if I have more questions... Thank you very much for your help!!!!!!!!! :)

Transcript for

$Mathematics$ Proof of sentence