Proof of sentence - 2014-11-20

evinda

14:53

Hey!!!
I am looking at the proof of the following sentence:
The p-adic numbers are complete with respect to the p-norm, ie every Cauchy sequence converges.

PROOF:

Let $(x_i)_{i \in \mathbb{N}}$ a Cauchy-sequence in $\mathbb{Q}_p$.

We want to show that, without loss of generality, we can suppose that $x_i \in \mathbb{Z}_p$.Let the set $\{ |x_i|_p | i \in \mathbb{N}\} \subset \mathbb{R}$. We suppose that it is upper bounded. If not, there are $\forall m \in \mathbb{N}$ and $N \in \mathbb{N}$ two indices $i,j \geq N$ with $|x_i|_p> |x_j|_p \geq p^m$.

Daniel Fischer

Let's start with something non-mathematical. It seems your native language is like German in that it uses the same word for "sentence" and for "theorem" resp. "proposition". In English, the word sentence is not what is meant there. It should be either theorem or proposition.

evinda

A ok!!! I am sorry...

Daniel Fischer

No need to be.

So let's see the proof. It starts with showing that the valuations of the terms are bounded, hence the powers of $p$ in the $p$-adic expansions $$x_i = \sum_{k = m_i}^\infty b_k\cdot p^k$$ are bounded below, there is an $m$ such that $m_i \geqslant m$ for all $i$.

Well, hum, do we already know that every $x\in \mathbb{Q}_p$ has such an expansion?

What definition of $\mathbb{Q}_p$ is used?

evinda

15:09

The definition of $\mathbb{Q}_p$ is:

$$\mathbb{Q}_p=\frac{r}{s},r,s \in \mathbb{Z}_p, s \neq 0$$

$$\mathbb{Z}_p:=\{ (\overline{x_n})_{n \in \mathbb{N}_0} \in \Pi_{n=0}^{\infty}\mathbb{Z}/ p^{n+1} \mathbb{Z}| x_{n+1} \equiv x_n \pmod{p^{n+1}}\}$$

Daniel Fischer

Sigh. Inconvenient. Let's see how to translate.

Okay, the important point is that every $x\in \mathbb{Z}_p$ with $x_0 \neq 0$ is invertible in $\mathbb{Z}_p$. Is that known?

evinda

According to my notes, it is known that the units of the ring $\mathbb{Z}_p$, $\mathbb{Z}_p^*=\mathbb{Z}\p\mathbb{Z}$ are equal to $\{ \sum_{n=0}^{\infty} a_np^n| a_0 \neq 0\}$.

Daniel Fischer

15:27

Okay, then we can see that every $x\in \mathbb{Q}_p$ has a unique representation of the form $\frac{r}{p^k}$ with $k \geqslant 0$.

Do you see how that follows?

evinda

At the beginning could you explain me why and how we have shown that $\{ |x_i|_p | i \in \mathbb{N} \} \subset \mathbb{R}$ is upper bounded? :/

Daniel Fischer

Why: to move the problem into $\mathbb{Z}_p$, because that is simpler to handle. How: using the ultrametric property of $\lvert\,\cdot\,\rvert_p$, we have $\lvert x_i\rvert_p = \lvert (x_i - x_j) + x_j\rvert^p \leqslant \max \{\lvert x_i-x_j\rvert_p,\, \lvert x_j\rvert_p\}$. Then choose $j$ so large that $\lvert x_i-x_j\rvert_p < 1$ for all $i \geqslant j$ - that is possibly since $(x_k)$ is a Cauchy sequence.

Then $\lvert x_i\rvert \leqslant \max (\{ 1, \lvert x_j\rvert_p\} \cup \{ \lvert x_k\rvert_p : 0 \leqslant k < j\})$ for all $i$.

It is, however, not important that $d_p(x,y) = \lvert x-y\rvert_p$ is an ultrametric, any Cauchy sequence in any metric space is bounded.

But well, having seen that $\lvert x_i\rvert_p \leqslant p^m$ for some $m\in\mathbb{N}$, one then considers the sequence $(p^m\cdot x_i)_{i\in\mathbb{N}}$.

That is a sequence in $\mathbb{Z}_p$.

A Cauchy sequence.

evinda

15:51

Why having shown that $\{ |x_i|_p | i \in \mathbb{N} \} \subset \mathbb{R}$ is upper bounded, do we move the problem into $\mathbb{Z}_p$?

Also, could you explain me how you found the union at the inequality $|x_i|_p \leq \max (\{1,|x_j|_p \} \cup \{ |x_k|_p:0 \leq k \leq j \})$ ?

Daniel Fischer

We move the problem into $\mathbb{Z}_p$ by multiplying the sequence with a suitable power of $p$. Showing that $\lvert x_i\rvert_p$ is bounded shows that that is possible, i.e. there is an $m\in \mathbb{N}$ such that $p^m\cdot x_i \in \mathbb{Z}_p$ for all $i$ (with the usual identification of $\mathbb{Z}_p$ with the set of fractions of the form $\frac{r}{1}$).

From the Cauchy property, we got a bound on $\lvert x_i\rvert_p$ for $i \geqslant j$. Then to get a bound for all $i$, we need to include the $\lvert x_k\rvert_p$ for $k < j$. Thus the union.

evinda

I haven't understood how we conclude that there is an $m\in \mathbb{N}$ such that $p^m\cdot x_i \in \mathbb{Z}_p$ for all $i$.. :/

Daniel Fischer

@evinda Because $\lvert p^m\cdot x\rvert_p = p^{-m}\cdot \lvert x\rvert_p$.

And $\mathbb{Z}_p = \{ x \in \mathbb{Q}_p : \lvert x\rvert_p \leqslant 1\}$.

evinda

16:06

So, don't we get it from the fact that $\{ |x_i|_p | i \in \mathbb{N} \} \subset \mathbb{R}$ is bounded above?

Daniel Fischer

So if we know that $\lvert x_i\rvert_p \leqslant p^m$ for all $i$, then we get $\lvert p^m\cdot x_i\rvert_p \leqslant 1$ for all $i$.

And we get $(\forall i)(\lvert x_i\rvert_p \leqslant p^m)$ (for some $m\in\mathbb{N}$) from the boundedness.

evinda

You said the following:
How: using the ultrametric property of $\lvert\,\cdot\,\rvert_p$, we have $\lvert x_i\rvert_p = \lvert (x_i - x_j) + x_j\rvert^p \leqslant \max \{\lvert x_i-x_j\rvert_p,\, \lvert x_j\rvert_p\}$. Then choose $j$ so large that $\lvert x_i-x_j\rvert_p < 1$ for all $i \geqslant j$ - that is possibly since $(x_k)$ is a Cauchy sequence.
Then $\lvert x_i\rvert \leqslant \max (\{ 1, \lvert x_j\rvert_p\} \cup \{ \lvert x_k\rvert_p : 0 \leqslant k < j\})$ for all $i$.

How do we get the last inequality? :/

Daniel Fischer

Do you see how we get $$i \geqslant j \implies \lvert x_i\rvert_p \leqslant \max \{ 1,\, \lvert x_j\rvert_p\}\,?$$

evinda

Yes, I understood this..

Daniel Fischer

Good. Then we only need to bound the finitely many terms with index $< j$ in addition to that.

Since there are only finitely many, there is a bound for that finite set.

Then we only need to take the larger of the two bounds to get a bound for all $i$.

evinda

16:34

So, could we also write it like that?

$|x_i|_p \leq \max \{ \max \{ 1, |x|_p\},\max \{|x_k|_p: 0 \leq k <j \}$

Daniel Fischer

Yes.

$\max \{ \max A,\max B\} = \max (A\cup B)$.

evinda

And in that way we have shown that the set $\{ |x_i|_p | i \in \mathbb{N} \} \subset \mathbb{R}$ is bounded above, right?

Daniel Fischer

Yes.

evinda

A ok :) And how do we continue?

Daniel Fischer

Then we pick an $m\in \mathbb{N}$ such that $p^m \geqslant \text{ bound}$.

And look at the sequence $p^m\cdot x_i$.

evinda

16:39

Why can we pick an $m$, such that $p^m \geqslant \text{ bound}$?

Daniel Fischer

Because $p^m$ grows without bound for $m\to\infty$, for example $p^m \geqslant 2^m > m$ for all $m\in\mathbb{N}$.

evinda

So, because of the fact that $p^m \to \infty$, when $m \to \infty$, and $\max \{ |x_i|_p | i \in \mathbb{N}_0 \}$ is a number, so it does not change when $m \to \infty$?

Daniel Fischer

There is no $m$ involved in $b := \max \{ \lvert x_i\rvert_p : i \in \mathbb{N}\}$. Since $b$ is some fixed real number, there are $m\in\mathbb{N}$ with $p^m \geqslant b$.

(All sufficiently large $m$ will do.)

evinda

I see!!! :-) And how can we continue?

Daniel Fischer

Then we look at $(p^m\cdot x_i)_{i\in\mathbb{N}}$, because that is a sequence in $\mathbb{Z}_p$.

evinda

16:53

Do we know that it is a sequence in $\mathbb{Z}_p$, because it is of the form $p^m \cdot u$, or is there an other reason?

Daniel Fischer

We know that $\lvert p^m\cdot x_i\rvert_p = p^{-m}\cdot \lvert x_i\rvert_p \leqslant p^{-m}\cdot b \leqslant 1$.

And $\lvert x\rvert_p \leqslant 1$ characterises the elements of $\mathbb{Z}_p$ in $\mathbb{Q}_p$.

evinda

So, we use the fact, that $\mathbb{Z}_p=\{x \in \mathbb{Q}_p: |x|_p \leq 1 \}$, right?

Daniel Fischer

Right.

And the next step is to show that $\mathbb{Z}_p$ is complete.

evinda

And how do we know that $p^{-m} \cdot x_i \i \mathbb{Q}_p$?

Daniel Fischer

Not $p^{-m}\cdot x_i$, $p^m\cdot x_i$.

You know that $\mathbb{Q}_p$ is the field of fractions of $\mathbb{Z}_p$. And $p^m \in \mathbb{Z} \subset \mathbb{Z}_p \subset \mathbb{Q}_p$ (with the usual tacit identifications).

Dinner time, bbl.

evinda

17:07

So,because $x_i \in \mathbb{Q}_p$ and $p^m \in \mathbb{Q}_p$, we conclude that $p^m \times x_i \in \mathbb{Q}_p$ ?

Enjoy your meal! :)

Daniel Fischer

18:37

Yes, every product of elements of $\mathbb{Q}_p$ is again in $\mathbb{Q}_p$. And since we've arranged $\lvert p^m x_i\rvert_p \leqslant 1$, we know we're now in $\mathbb{Z}_p$.

evinda

A ok.. And what have we proven, showing this? :/

Daniel Fischer

We have proven that up to multiplication with a constant factor, all Cauchy sequences in $\mathbb{Q}_p$ lie in $\mathbb{Z}_p$. Then one shows that every Cauchy sequence in $\mathbb{Z}_p$ converges, which is a bit simpler, since we need not occupy ourselves with the denominators. Then one notes that if $p^m\cdot x_i \to z\in \mathbb{Z}_p$, it follows that $x_i \to p^{-m}\cdot z \in \mathbb{Q}_p$, and then the proof of completeness of $\mathbb{Q}_p$ is complete.

evinda

19:10

What do we conclude from the fact that $p^m x_i \in \mathbb{Q}_p$ ?

Daniel Fischer

Nothing. We use the property $\lvert p^m\cdot x\rvert_p = p^{-m}\cdot \lvert x\rvert_p$ of the valuation to conclude that $(x_i)$ is a Cauchy sequence if and only if $(p^m\cdot x_i)$ is a Cauchy sequence.

So we know that our scaled sequence is a Cauchy sequence again.

evinda

Could you explain me further how we conclude the fact that $(x_i)$ is a Cauchy sequence if and only if $(p^m\cdot x_i)$ is a Cauchy sequence? :/

Daniel Fischer

If you have a real sequence $(\xi_k)$, and multiply that with a nonzero factor $c$, call $\eta_k := c\cdot\xi_k$, then you know that for all $k,n$, you have $\lvert \eta_k - \eta_n\rvert = \lvert c\xi_k - c\xi_n\rvert = \lvert c\rvert\cdot \lvert \xi_k - \xi_n\rvert$. And that tells you that $(\eta_k)$ is a Cauchy sequence if and only if $(\xi_k)$ is one, because $\lvert \eta_k - \eta_n\rvert < \varepsilon \iff \lvert \xi_k - \xi_n\rvert < \frac{\varepsilon}{\lvert c\rvert}$. The situation here

is the same, we just have $\lvert\,\cdot\,\rvert_p$ instead of the absolute value on $\mathbb{R}$.

evinda

19:32

And do we need to prove that it also stands for the p-norm?

Daniel Fischer

The $p$-norm is also multiplicative, so the only thing that changes is the subscript $p$.

evinda

So, to justify it, do I have just to say that the $p$-norm is multiplicative?

Daniel Fischer

19:48

If that fact has been proved somewhere, just referring to it is sufficient. If not, the special form of the constant here makes showing it for the case $c = p^m$ easy.

How exactly you'd show that depends on how exactly $\lvert\,\cdot\,\rvert_p$ was defined.

evinda

20:11

Do you mean the fact that the $p$-norm is multiplicative?

Daniel Fischer

Yes.

evinda

20:32

@DanielFischer I will look at my notes, if it has been proven.. And then, how do we continue?

Transcript for

$Mathematics$ Proof of sentence