Proof We have 1 implies 3 implies 2 in a evident manner; thus it suffices to prove 2 implies 1. If we suppose that $|m|\leq M$ such that $m\in\Z$ and if $x,y\in K$ and $n\in \N$, then $$|x+y|^n=|(x+y)^n|=\left\vert\sum_{i=0}^n \binom{n}{i}x^iy^{n-i} \right\vert\leq (n+1)M\sup(|x|,|y|)^n$$ Passing to the $n$th root we obtain $|x+y|\leq (M(n+1))^{1/n}\sup(|x|,|y|)$, which gives the required inequality.

Proof Upto a permutation of $x$ and $y$, we can suppose $|x|>|y|$ . We have then $$|x+y|\leq\sup(|x|,|y|)= |x|=|(x+y)-x|\leq \sup(|x+y|,|y|)\leq|x+y| $$ providing the conclusion $|x|=|x+y|$.

Proof The first point is the immediate consequence of lemma 1.4. If $x_1\in B(x_0,r)$ (ball with center $x_0$ and radius $r$ open or closed), and if $y\in B(x_1,r)$, then $d(x_0,y)\leq \sup(d(x_0,x_1),d(x_1,y))\leq r$ (or $<r$ if we talk about open balls), and thus $B(x_1,r)\subset B(x_0,r)$. The inclusion in the other direction is obtained by interchanging the roles of $x_0$ and $x_1$, this permits to prove 2. After 2, if the two balls have a non empty intersection, all elements of intersection is the center of the ball proving 3. Now, 5. is an immediate consequence of 4, and if $B$ is open ball of radius $r$, the complement of $B$ contains the open ball of radius $r$ around each of his points according to 3, which shows that this complementary is open and that $B$ is closed. If $B$ is a closed ball or non zero radius, the $B$ is a neighborhood of each of the points, since, these are "the" center.

Proof If $\norm{.}$ is a norm on a field $K$, then $\norm{x}<1$ if and only if the sequence of general terms $x^n\ra 0$ when $n\ra +\infty$. We deduce the fact that if $\norm{.}_1$ and $\norm{.}_2$ are equivalent then $$\{x\in K:\norm{x}_1< 1\}= \{x\in K:\norm{x}_2< 1\}$$ If the latter is reduced to $\{0\}$, we have $\norm{x}_1=\norm{x}_2=1$ for any $x\in \tio{K}$. If not, let $x\in \tio{K}$ and verify $\norm{x}_1<1$. If $y\in\tio{K}, a\in\Z$ and $b\in\N$, then $$\norm{y^bx^{-a}}_1<1\iff \norm{y^bx^{-a}}_2<1. $$ We deduce the fact that $$\left\{r\in\Q:r<\dfrac{\log\norm{y}_1}{\log\norm{x}_1}\right\}=\left\{r\in\Q:r<\dfrac{\log\norm{y}_2}{\log\norm{x}_2}\right\}$$ and thus the reals $\dfrac{\log\norm{y}_1}{\log\norm{x}_1}$ and $\dfrac{\log\norm{y}_2}{\log\norm{x}_2}$ define the same Dedkind cut, and thus are equal, which shows that the function $y\mapsto \dfrac{\log\norm{y}_1}{\log\norm{y}_2}$ is constant on $\tio{K}$ and permits the conclusion.

1. The trivial norm: We can equip any field $K$ with the trivial norm defined by $\norm{x}=1$ if $x\neq 0$. The associated topology is discrete on $K$.
2. Induced Norm: If $L$ is a normed field, and $K$ is a sub field, we can equip $K$ with the norm obtained by restriction to that on $L$. For example, we can equip all sub fields of $\C$ with the norm induced by the map $\norm{x+\ci y}=\sqrt{x^2+y^2}$.
3. Norms on $\Q$: Since $\Q$ is a subfield of $\C$, we can equip with the usual norm denoted as $\norm{.}_\infty$. Furthermore, if $p$ is a prime number, we can equip $\Q$ with the $p$ adic norm defined by $\anorm{\dfrac{a}{b}}_p=p^{v_p(b)-v_p(a)}$,where $n\in\Z\bs\{0\},v_p(n)$ is the biggest integer $v$ such that $p^v$ divides $n$. The norm $\norm{.}_p$ is clearly multiplicative and verifies the ultrametric inequality because if $\norm{z}\leq 1$, we can write $z$ in the form $\dfrac{a}{b}$ with $p\nmid b$ implying $$1+z=\dfrac{a+b}{b}\quad\text{implies}\quad p^{-v_p(a+b)}\leq 1.$$

Proof Begin by supposing that there exists a $k\in\N$ such that $\dnorm{k}>1$. Since, $\dnorm{1}=1$, the triangle inequality implies that $\dnorm{k}\leq k$ and there exists $\alpha\in(0,1]$ such that we have $\dnorm{k}=k^\alpha$. Let $m\in\N$, we can write $m$ in base $k$ under the form $m=\sum_{i=0}^na_ik^i$ with $a_i\in\{0,1,\ldots, k-1\}$ and $a_n\neq 0$ so that we have $m\geq k^n$. Since $\dnorm{a_i}\leq a_i\leq k-1$ and $\dnorm{k^i}=\dnorm{k}^i$, we obtain the inequality $$ \begin{align*} \dnorm{m}&\leq (k-1)\sum_{i=1}^nk^{i\alpha}\\ &=\frac{k-1}{k^\alpha-1}(k^{(n+1)\alpha}-1)\\ &\leq \frac{k^\alpha(k-1)}{k^\alpha-1}k^{n\alpha}\leq Cm^\alpha\\ \text{where }C&=\frac{k^\alpha(k-1)}{k^\alpha-1}\\ &\text{independent of }m. \end{align*} $$ We can apply the above inequality to $m^n$ which gives $\dnorm{m}^n\leq Cm^{n\alpha}$ and, taking the $n$th roots and passing to limit gives $\dnorm{m}\leq{m}^\alpha$. We thus have $\dfrac{\log\dnorm{m}}{\log m}\leq \alpha=\dfrac{\log\dnorm{k}}{\log k}$ for any $m\in\N$. By symmetry, we can deduce the fact that if $\dnorm{m}>1,$ then this inequality is an equality. In the general case, there exists an $n\in\N$ such that we have $\dnorm{k^nm}>1$, which shows that we have an equality for some $m\in\N$, utilizing the multiplicativity of the norm and the fact that $\dnorm{-1}=1,$ and $\dnorm{x}=\anorm{x}_\infty^\alpha$ for some $x\in\Q$. We have this shown that there exists $k\in\N$ such that $\dnorm{k}>1$, then $\dnorm{~}$ is equivalent ot the usual norm.
  In the contrary case, we have $\dnorm{\ell}\leq 1$ for all prime number $\ell$. As we suppose $\dnorm{~}$ is non trivial, there exists a prime $p$ such that $\dnorm{p}<1$. If there exists another $q$, then there is some $n\in\N$, such that $u_np^n+v_nq^n=1,u_n,v_n\in\Z$ (after the Bezout's Theorem). We thus obtain $$ \begin{align*} 1&=\dnorm{1}=\dnorm{u_np^n+v_nq^n}\\ &\leq \dnorm{u_n}\cdot\dnorm{p}^n+\dnorm{v_n}\cdot\dnorm{q}^n\\ &\leq \dnorm{p}^n+\dnorm{q}^n \end{align*} $$ which is impossible for $n$ big enough (since $\dnorm{p}\leq 1$ and $\dnorm{q}<1$). There thus necessarily exists a prime number $p$ such that $\dnorm{p}\leq 1$ and $\dnorm{~}$ is equivalent to the $p$ adic norm. $\qquad \blacksquare$.

Let $K$ be a field and $a\in(0,1)$
1. $f\mapsto\anorm{f}_\infty=a^{-\deg f}$ defines an ultrametric norm on $K(X)$.
2. Let $\scr{P}$ be the set of irreducible monic polynomials in $K(X)$. If $P\in\scr{P}$ and $f\in k(X)$ is non zero there exists a unique $v_p(f)\in\Z$ such that $P^{-v_p(f)}f=Q_1/Q_2$, with $Q_1$ and $Q_2$ prime to $P$. Then $f\mapsto \anorm{f}_P=a^{\deg P\cdot v_p(f)}$ defines a ultrametric norm on $K(X)$ and that $\anorm{~}_p$ is not equivalent to $\anorm{~}_\infty$.
3. All non trivial norms on $K(X)$, that are trivial on $K$ are equivalent to $\anorm{~}_\infty$, or $\anorm{~}_P$ for a unique $P\in\scr{P}$.
4. We have a product formula for $f\in \tio{K(X)}$ $$\anorm{f}_\infty\cdot\prod_{f\in\scr{P}}\anorm{f}_P=1 .$$

If $K$ is an ultrametric field and if $P(X)=\sum_{i=1}^na_iX^i\in K(X)$, we define the Gauss Norm $\dnorm{P}_G$ of $P$ by $\anorm{P}_G=\sup_i(\anorm{a_i})$.

Proof Since $\anorm{~}$ is ultrametric the inequality $\anorm{PQ}_G\leq\anorm{P}_G\anorm{Q}_G$ is immediate. Inversely, let $i_0$ (resp. $j_0$) be the smallest integer $i$ (resp. $j$) verfifying $\anorm{a_i}=\anorm{P}_G$ (resp. $\anorm{b_j}=\anorm{Q}_G$). Let $k_0=i_0+j_0$ and $P(X)Q(X)=\sum_{k=0}^{n+m}c_kX^k$. Then $$c_{k_0}=a_{i_0}b_{j_0}+\sum_{i\leq i_0}a_ib_{k_0-i}+\sum_{j<j_0}a_{k_0-j}b_j $$ and by definition of $i_0$ and $j_0$, the only term of this sum of norm $\anorm{P}_G\anorm{Q}_G$ is $a_{i_0}b_{j_0}$, the other being of norm strictly less, and using Lemma 1.4, this shows that we have $\anorm{c_{k_0}}=\anorm{P}_G\anorm{Q}_G$ which permits the conclusion. $\qquad \blacksquare$

The lemma permits to show that if $f=P/Q$ and if we set $\anorm{f}_G=\anorm{P}_G/\anorm{Q}_G$, this expression does not depend writing $f$ under the form $P/Q$ and thus defines a map $\anorm{~}_G$ of $K(X)$ in $\RR_+$ verifying the properties (i) and (ii) of the norms. On the other hand, if $f=P/Q$, we can, divide $P$ and $Q$ by an element of $K$ of norm $\anorm{Q}_G$, suppose that $\anorm{Q}_G=1$ and $\anorm{f}_G\leq 1$ is then equivalent to $\anorm{P}_G\leq 1,$ and implies $\anorm{P+Q}_G\leq 1$ in an evident manner. This permits to show that $\anorm{~}_G$ is an ultrametric norm on $K(X)$.

Remark Let $K$ be an ultrametric field, we denote by $K[[X]]^b$ as the set of series $f=\sum_{n\geq 0}a_nX^n$ such that the sequence ${(a_n)}_{n\in\N}$ is bounded in $K$. We equip $K[[X]]^b$ by the Gauss norm $\anorm{~}_G$ defined by $\anorm{\sum_{n\geq 0}a_nX^n}_G=\sup_{n\in \N}\anorm{a_n}$.
1. If $f,g\in K[[X]]^b$, then $\anorm{fg}_G=\anorm{f}_G\anorm{g}_G$.
2. If $f=\sum_{n\geq 0}a_nX^n$ is invertible in $K[[X]]^b$ if and only if $a_0\neq 0$ and $\anorm{a_0}\geq \anorm{a_n}$ for any $n\in\N$.

Proof The triangular inequality implies that we have $\anorm{\anorm{a_{n+p}}-\anorm{a_n}}\leq\anorm{a_{n+p}-a_n}$ for any $n$ and $p$ and thus the sequence with general term $\anorm{a_n}$ is Cauchy, from where we deduce (i).
On the other hand, if $a={(a_n)}_{n\in\N}\in \tilde{K}-I$ there exists $\delta>0$ such that we have $\anorm{a_n}\geq \delta$ for infinite $n$ and the limit of the sequence $\anorm{a_n}$ is greater or equal to $\delta$. Thus, there exists $N\in \N$ such that if $n\geq N$, then $\anorm{a_n}>\dfrac{2}{3}\delta$ and $\anorm{a_{n+p}-a_n}<\delta/2$ for any $p\in\N$. This implies $\anorm{a_{n+p}-a_n}<\anorm{a_n}$ and thus, since $\anorm{~}$ is supposedly ultrametric, $\anorm{a_{n+p}}=\anorm{a_n}$ for some $p\in\N$, this shows (ii).
We have $\anorm{\anorm{a_n}-\anorm{b_n}}\leq\anorm{a_n-b_n}$ and the hypothesis implies that this latter sequence tends to zero and hence (iii).$\qquad\blacksquare$

Proof The fact that $\tilde{K}$ is a ring and $I$ an ideal is immediate. The unit element of $\tilde{K}$ is the sequence of constants $1$ all of whose terms are equal to $1$. If $a={(a_n)}_{n\in\N}\in\tilde{K}\bs I,$ after the preceding there exists $\delta>0$ and $N\in\N$ such that we have $\anorm{a_n}\geq\delta$ if $n\geq N$. The sequence $b={(b_n)}_{n\in N}$ defined by $b_n=0$ if $n<N$ and $b_n=a_n^{-1}$ if $n\geq N$ is Cauchy and $ab-1$ is element of $I$, this shows that $a$ is inversible in $\tilde{K}/I$ and permits the conclusion that $I$ is maximal.$\qquad\blacksquare$

Proof The multiplicativity of the norm and the triangle inequality (resp ultrametric) pass to the limit. On the other hand, $\anorm{a}=0$ if and only if $\lim_{n\ra\infty}\anorm{a_n}=0$ if and only if $a\in I$ and thus $\anorm{~}$ is the norm on $\hat{K}$ that is ultrametric if $\anorm{~}$ is ultrametric on $K$.
Now if $a={(a_n)}_{n\in\N}\in\tilde{K},$ then $\anorm{a-a_n}\leq \sup_{p\geq 1}\anorm{a_{n+p}-a_n}$ tends to zero as $n$ tends to $\infty$ since the sequence ${(a_n)}_{n\in \N}$ is Cauchy. We thus that $a=\lim_{n\ra\infty}a_n$ in $\hat{K}$ and thus $K$ is dense in $\tilde{K}$.
Finally, if ${(a_n)}_{n\in\N}$ is a Cauchy sequence in $\hat{K}$, as $K$ is dense in $\hat{K}$, we can find for each $n$ an element $b_n$ of K such that we have $\anorm{a_n-b_n}\leq 2^{-n}$ and the sequence $b_n$ is Cauchy in $K$ thus converges in $\hat{K}$ to a limit that is also the limit of the sequence ${(a_n)}_{n\in\N}$; which proves that $\hat{K}$ is complete. $\qquad\blacksquare$

1. $\RR$ is completion of $\Q$ with respect to absolute value.
2. $\C$ is the completion of $\Q(\ci)$ for the norm $\anorm{a+\ci b}=\sqrt{a^2+b^2}$.
3. More generally, if $K$ is a complete normed field, and $L$ is a dense subfield of $K$, then $K$ is completion of $L$ for the induced norm.
4. The completion of $K(X)$ for the norm $\anorm{~}_X$ is $K((X))$.

Proof It suffices to prove that that they are all equivalent to the $\sup$ norm, this is shown by induction on dimension of $V$. If the dimension is $1$, there is nothing to do. If not let $e_1,\ldots, e_n$ be a base of $V$, we have $$\dnorm{x_1e_1+\cdots+x_ne_n}\leq (\dnorm{e_1}+\cdots+\dnorm{e_n})\sup(\anorm{x_1}, \ldots, \anorm{x_n}), $$ from where one has to verify two inequalities. For demonstrating the other, we reason by contradiction. Suppose there exists a sequence $x_1^{(k)}e_1+\cdots+x_n^{(k)}e_n$ which tends to zero for the norm $\dnorm{~}$ but not for the $\sup$ norm. There exists $C>0, i\in \{1,\ldots, n\}$ and a infinite subsequence such that we have $\anorm{x_i^{(k)}}\geq C$ and thus the sequence with general term $v_k=\dfrac{x_1^{(k)}}{x_i^{(k)}}e_1+\cdots+\dfrac{x_n^{(k)}}{x_n^{(k)}}e_n$ tends again to zero for $\dnorm{~}$. We can deduce the fact that $e_i$ is in the closure of $W=\text{Vect}(e_1,\ldots, e_{i-1}, e_{i+1},\ldots e_n)$ which complete after the induction hypothesis, which implies that $e_i\in W$ and this is a contradiction since $e_i$ forms a base of $V$ and thus cannot be generated by other elements. $\qquad \blacksquare$