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## Month: August, 2014

This is an article I wanted to write for a long time.

What exactly is topological continuity? A mapping $f:X\to Y$ between two topological spaces is said to be continuous if for any open set $U\subset Y$, $f^{-1}(U)$ is open in $X$.

After reading this definition, very few students understand the motivation behind this definition. By bringing these concepts to Euclidean space, a  picture slowly begins to form. This definition is valid in $\Bbb{R^n}$. However, what was the need to unnecessarily generalize continuity to non-Euclidean spaces? What does continuity really mean in a general setting?

Given below is my impression of continuity, and has not been influenced by any mathematical work that I have come across.

If there is a continuous mapping from $X$ to $Y$, this means that $Y$ is at least as “solid and smooth”, or more “solid and smooth” than $X$. I have no intention of making the phrase “solid and smooth” precise, and it is an arbitrary phrase concocted by me to give readers a “feel” for the nature of continuity. As an intuitive explanation, a circle is more “solid and smooth” then a line, which has two sharp points. The graph of $y=x$ is more “solid and smooth” than $y=\lfloor x\rfloor$. Developing more examples of this kind would be a useful exercise. The less the number of “sharp points”, the more an object is “solid and smooth”.

Now we swoop back to my original claim: that if $f:X\to Y$ is continuous, then $Y$ is at least as or more “solid and smooth” than $X$. As an example, consider the continuous mapping of $[0,1]$ to $\Bbb{S}^1$. The continuous map for this would be $t\to (\cos 2\pi t,\sin 2\pi t)$. We know that $[0,1]$ has two “sharp points” (at 0 and 1), while $\Bbb{S^1}$ has none. Hence, $\Bbb{S}^1$ is at least as or more solid and smooth than $[0,1]$. However, there is no continuous mapping from $\Bbb{S}^1$ to $[0,1]$, as the latter is less solid and smooth than the former.

It is a useful exercise to use this concept in more complicated topological cases.

Now we come to path homotopy. When we say there is a continuous mapping from $[0,1]\times [0,1]$ to a topological space such that $f(I\times \{0\})=C_1$ and $f(I\times \{1\})=C_2$, then we are in essence saying that the space encompassed and bordered by $C_1$ and $C_2$ is as or more solid and smooth then $I_0^2$. Hence, there are no breaks in between. It can be pictured as a solid block of ink. No spaces.

This is the reason why $[0,1]$ is generally used as the domain for a continuous map. We want the image to be solid. No gaps.

### Valuations, Absolute Values and Ostrowski’s theorem

This is a guest post by Abhishek Gupta.
We know that there is an absolute value on the set of real numbers which assigns to any real number $x$ the non negative real number $|x|$, its multiplicative and satisfies the triangle inequality. In this post we will abstract the notion absolute value for an arbitrary field and also see the related notion of valuations on a field.

Definition. A valuation on a field $F$ is a map $v:F\rightarrow \mathbb{R}\cup\{\infty\}$ which satisfies the following conditions:

1. $v(xy)=v(x)+v(y)$
2. $v(x)=\infty\Leftrightarrow x=0$
3. $v(x+y)\geq \inf(v(x),v(y))$

In all the examples below we set the valuation of $0$ as $\infty$.

Examples of valuations:

1. Fix a prime say $p$ for any rational number $\frac{m}{n}$ extract all powers of $p$ from the numerator and denominator so that $\frac{m}{n}=p^k\frac{a}{b}$ where $p\nmid a,b$, then the function $v_p(\frac{m}{n})=k$ defines a valuation called the $p-$adic valuation on $\mathbb{Q}$.
2. We define a valuation on the field $F(T)$ of all rational functions(quotients of polynomials) of $T$ as follows. For a rational function $\frac{f(T)}{g(T)}$ define $v_\infty(\frac{f(T)}{g(T)})=deg(g)-deg(f)$. This defines a valuation $v_\infty$ on the function field.
3. We define another valuation on the function field. Fix an irreducible polynomial say $p(T)\in F[T]$, then for rational function $\frac{f(T)}{g(T)}$ extract all the powers of $p(T)$ from the numerator and the denominator so that $\frac{f(T)}{g(T)}=p(T)^k\frac{r(T)}{s(T)}$ where $r(T),s(T)$ are not divisible by $p(T)$. Define $v_p(\frac{f(T)}{g(T)})=k$. This defines another valuation on the function field.

In all the examples we have seen so far the image of the valuation map $v$, which is called as the valuation group is $\mathbb{Z}$. Any valuation with valuation group as a discrete subgroup of $\mathbb{R}$ is called as a discrete valuation. Thus all the examples above are discrete valuations. The valuation ring of a valuation is the subset of all elements of the field with nonnegative valuation. If the valuation is discrete then the valuation ring is called as a discrete valuation ring. These rings are important and occur often in algebraic number theory and algebraic geometry. We now see examples of valuations which are not discrete.

1. The field of formal Puisseux series $\cup_{n\geq1}k((x^{1/n}))$ with valuation $v:K^{\times}\rightarrow\mathbb{R}$ given by $v(x^q)=q$ has $\mathbb{Q}$ as its valuation group which is not a discrete subgroup of the reals.
2. This one is not as immediate as the previous one: the algebraic closure $\overline{\mathbb{Q}}_p$ of $\mathbb{Q}_p$ has a valuation extending the one on $\mathbb{Q}_p$ and the valuation group of this is also $\mathbb{Q}$.

Lets look at the more familiar concept of an absolute value on a field and its relation to valuations on a field.

Definition. An absolute value on a field $F$ is a map $|\cdot|:F\rightarrow \mathbb{R}$ satisfying the following conditions:

1. $|x|=0\Leftrightarrow x=0$
2. $|x y|=|x||y|$ (multiplicativity)
3. $|x|+|y|\geq \sup(|x|,|y|)$ (triangle inequality)

It is easy to see that the absolute value is independent of sign and that the absolute value of the unity of the field is one. Also if the field under consideration is that of the rational numbers then given the values of the absolute value on all primes we can tell the absolute value of any rational number because the numerator and the denominator can be factorised as products of primes and absolute values are multiplicative.
If $k\in \mathbb{N}$ then $|k|=|1+1+\dots+1|\leq 1+1+\dots+1=k$. ($\mathbb{N}$ denotes the set of positive integers).

Examples of absolute values:

1. For any field $F$ we can define an absolute value as $|x|=1$ for $x\neq 0$ and $|0|=0$. This is called as the trivial absolute value on $F$.
2. The ususal absolute value called the Euclidean absolute value on $\mathbb{R}$($|x|=\pm x$ depending on whether $x$ is positive or negative) and $\mathbb{C}$($|z|=\sqrt{x^2+y^2}$) are examples of absolute values in the sense defined above. Subfields of $\mathbb{R}$ and $\mathbb{C}$, like $\mathbb{Q}(\sqrt{i})$ or $\mathbb{Q}(\pi)$ inherit the absolute on these fields by considering them as subsets and restricting the absolute value to this subset. We shall denote the Euclidean absolute value on $\mathbb{Q}$ by $|\cdot|_\infty$.
3. Given a valuation on a field we can use it to define an absolute value on it. Let $v$ be a valuation on a field $F$ and $c\in \mathbb{R}, c>1$. Then for $x\in F$, define $|x|=c^{-v(x)}$. This defines an absolute value on $F$. In particular for the $p-$adic valuation on $\mathbb{Q}$, the corresponding absolute value will be denoted as $|\cdot|_p$ and will be referred to as the $p$-adic absolute value in what follows.
4. Its can be easily shown that an absolute value on a finite field is necessarily trivial.

Its easy to check that absolute values coming from a valuation satisfy a stronger inequality than just the triangle inequality: $|x+y|\leq\sup{(|x|,|y|)}$. Any absolute value satisfying this inequality is called as a ultrametric or a non-archimedean absolute value, it can be shown that for a given non-archimedean absolute value there is a valuation that gives rise to this absolute value(just take the $-\log$ of the absolute value to get the valuation and check that it satisfies the required properties). Hence we see that there is a one-one correspondence between valuations and non-archimedean absolute values.

Two absolute values $|\cdot|_1$ and $|\cdot|_2$ on a field $F$ are said to be equivalent if there is a positive real number $c$ such that $|x|_1=|x|_2^c,\;\;\forall x\in F$. One can get topology on the field by using the absolute value to get a metric, then equivalence of two absolute values in the sense just defined is equivalent to saying that the two absolute values give the same topology to the field.

Having introduced absolute values we now classify absolute values on $\mathbb{Q}$ up to equivalence. Fortunately the proof is completely elementary. This proof was given by Emil Artin and this is the one found in most textbooks.

Theorem(Ostrowski). Any nontrivial absolute values on the field of rational numbers is equivalent to either the Euclidean absolute value $|\cdot|_\infty$ or the p-adic absolute values $|\cdot|_p$.

Proof. Let $|\cdot|$ be an absolute value on $\mathbb{Q}$. $\mathbb{N}$ denotes the set of positive integers. We divide into two possible cases:

Case 1. $|n|\leq 1,\;\;\;\forall n\in \mathbb{N}$
In this case there has to be a prime $p$ with $|p|<1$, if not then the absolute value for every prime is 1 and hence its 1 for all rationals because of the multiplicativity of the absolute value thus making it the trivial absolute value which we do not want. We now claim that $p$ is the only prime with absolute value less than 1. If not then let $q$ be another prime with $|q|<1$. Then by choosing $N$ large enough, we have $|p|^N,|q|^N<1/2$. Also $p^N,q^N$ are coprime, so there are integers $r,s$ such that $rp^N+sq^N=1$. This gives

$1 \leq|rp^N+sq^N|\leq |rp^N|+|sq^N|= |r||p^N|+|s||q^N|<\frac{|r|+|s|}{2}\leq \frac{1+1}{2}=1$

where we have used the fact that $|r|,|s|\leq 1$.Thus we have $1<1$ which is absurd. Hence the only prime with absolute value less than one is $p$. We now show that the absolute value in this case must be equivalent to the $p-$adic absolute value. Let us denote the absolute value of $p$ by $\lambda$, i.e. $|p|=\lambda$. Then for a positive integer $x\in \mathbb{N}$ with prime factorization $x=p^{v_p(x)}p_1^{\alpha_1}\dots p_t^{\alpha_t}$ the absolute value is $|x|=|p|^{v_p(x)}$ as the absolute value of other primes is one. Thus $|x|=\lambda^{v_p(x)}={(p^{-v_p(x)})}^{c}=|x|_p^c$ where $c=-\log_p \lambda$. Thus it is equivalent to the $p$-adic absolute value for any integer and therefore for any rational.

Case 2. $|n|> 1$ for some $n\in \mathbb{N}$
Let $a,b\in \mathbb{N}_{>1}$(a notation we shall follow throughout) and $n\in \mathbb{N}$. Expand $b^n$ in base $a$: $b^n=\sum_{i with $c_i\in \{0,1,\dots,a-1\}$. Thus $a^{m-1}\leq b^n$ in other words, $m\leq n\log_ab+1$. Then,

$|b^n|\leq\sum_{i

Now taking the $n$-th root of the two extremes of the inequality and using $m\leq n\log_ab+1$ we get:

$|b|\leq a^{\frac{1}{n}}(1+n\log_ab)^{\frac{1}{n}}\sup(1,|a|^{n\log_ab})$

Taking the limit $n\rightarrow \infty$ we get:

$|b|\leq \sup(1,|a|^{\log_ab})$

So far we have not made use of any assumptions on the absolute value under consideration so the inequality just obtained is always true for any absolute value, we shall make use of this later.

Having made this remark we now use the hypothesis of the case under consideration. Now choose $b=n$, that is $b$ has absolute value bigger than one. Then in the inequality $|b|\leq \sup(1,|a|^{\log_ab})$, the $\sup$ term must evaluate to $|a|^{\log_ab}$, as the other possibility that is 1, is forbidden.The inequality then becomes $|b|\leq |a|^{\log_ab}$. Moreover this forces that $|a|>1$, because if this does not happen then $|b|\leq |a|^{\log_ab}\leq 1^{\log_ab}$(${\log_ab}>0$), which is false by our choice of $b$. This implies that irrespective of what natural number $a$ is, $|a|\geq 1$!!

With the knowledge that the hypothesis for this case forces all natural numbers to have absolute value bigger than one we return to the general inequality: $|b|\leq \sup(1,|a|^{\log_ab})$. Now, we could choose $b$ to be anything and yet obtain $|b|\leq |a|^{\log_ab}$, the reason being that the absolute value of $b$ is always bigger than one. Hence, for any $a,b$ we have $|b|\leq |a|^{log_ab}$ which implies $\frac{|b|}{b}\leq \frac{|a|}{a}$ which is an equality by symmetry. Thus for any natural number $\frac{|b|}{b}=\lambda$ is a fixed constant. This gives $|b|=b^\lambda$ for all natural numbers. Thus for all integers and hence for all rational numbers $|x|=|x|_\infty^\lambda$.

This proves that in this case the absolute value is equivalent to the Euclidean absolute value. Hence the theorem is proved.

Q.E.D.

Analogues of this theorem exist for the fields $F_q(T)$ and finite extensions of $\mathbb{Q}$. A good place to find out about these besides other things is this.