Why are every structures I study based on Real number?

Question

I've been studying basic concepts of inner product vector space, normed vector space and metric space. And all the inner products, norms and metrics are defined to be real-valued functions in my textbook. I doubted about it: why do I have to restrict the measurement standard to real number? Why not any ordered field? Why not… anything else?

I do know that there can be a lot of different ways to define 'distance': I'm studying topology, too. I'm just wondering if there is any more general definition to the inner product space, normed space, or metric space(my definition is just the standard one being taught at undergraduate level: only real-valued ones as I've mentioned).

For example, let $V$ be an $F$-vector space and $F_s$ be an ordered field embedded into $F$. Then I defined like this:

1. $\langle v, v\rangle=0$ if and only if $v$ is the $0$ vector
2. $\langle v, v\rangle>0$ (so the value is in $F_s$)
3. $\langle av, u\rangle=a\langle v, u\rangle$ where $a\in F$
4. $\langle v+u, z\rangle=\langle v, z\rangle+\langle u, z\rangle$
5. $\langle v, u\rangle=\langle u, v\rangle$

We can define the normed vector space in this fashion as well. I think this is a more general definition: I couldn't find anything wrong about it. I mean, if we define $\|v\|=\langle v, v\rangle^{1/2}$ then indeed $(V, \|\,\|)$ becomes a normed vector space according to my definition, as it should in the ordinary definition (I think the Cauchy-Schwarz inequality which connects the two spaces also hold in this general definition: I've checked the proof in my textbook and it does not use any property of the real number).

Is my definition not right, or not useful? If so, then why? I know that the real number is the only (up to isomorphism) ordered field with the least upper bound property and the property that every increasing bounded sequence converges. But is that enough to justify that every metric spaces use real number? I want to be more convinced, then.

A little bit disorganized, but I just wanted to hear some other people's opinion about this. Thanks as always.

P.S. I also know about the complex inner product vector space. The inner product there is complex-valued, so I think it was the start of my questioning.

Answer 1

why do I have to restrict the measurement standard to real number? Why not any ordered field?

You can use any ordered field, and the axioms will still make sense. The thing is, though, that most of our geometric intuition is built on the Archimedian property. But if we include this in our field, then it becomes a subfield of $\Bbb R$. In this case, we are really not restricting ourselves to $\Bbb R$ at all: it's the largest Archimdean ordered field!

Thinking about ordered fields that aren't Archimedian gets a little more weird. Are you prepared, for example, to have vectors $v,w$ such that $v$ and $w$ point in the same direction, and yet $vn$ is shorter than $w$ for every natural number $n$?

I mean, if we define $\|v\|=\langle v, v\rangle^{1/2}$ then indeed $(V, \|\, \|)$ becomes a normed vector space

No, not quite! Normally we want the norms to be in the base field, but the vector $(1,1)\in \Bbb Q^2$ would have a norm outside of $\Bbb Q$, with that definition. To fix things, you'd need something called a Pythagorean field, and that's enough to guarantee this definition of norm works.

If you don't ask for the norm to be in the base field, then you may not be able to carry out normalization, because dividing by the norm won't give you a vector in the field.

I know that the real number is the only (up to isomorphism) ordered field with the least upper bound property and the property that every increasing bounded sequence converges. But is that enough to justify that every metric spaces use real number? I want to be more convinced, then.

Yes, your intuition is along the right lines. The fact that $\Bbb R$ is "maximal" among Archimedian ordered fields makes it special. It's a smooth connected piece with no holes. In geometry this is important since it ensures that lines and circles cross where you expect them to. For example in $\Bbb Q^2$, you can find an example of a line and a circle that would intersect in $\Bbb R^2$, but they pass through each other in $\Bbb Q^2$ without touching.

Why not... anything else!!

Actually, geometers study generalizations of the norm in the form of bilinear forms over any field. The idea is that rather than focus on the norm, you instead focus on a (generalized) inner product. They can be quite different from what you're used to with run-of-the-mill normed real spaces. They can have, for example, nonzero vectors with length $0$ or even negative length.

Even more than that, "length" loses meaning totally when you're working in a field that isn't ordered: there's no such thing as positive or negative, there. Still, there is a huge theory for these types of spaces with (generalized) inner products.

Answer 2

You can define inner products for vector spaces and normed vector spaces over any ordered field.

You can define metrics over any ordered field: they are called generalised metrics.

Answer 3

Not any ordered field possesses the square root operation defined wherever $x ≥ 0$; rational numbers form an obvious counter-example known from ancient times. Yes, it is essentially the same problem as famous one faced by Pythagoreans, but in modern formulation: how to define number field to make things going well with geometry?

There can be numerous ways to tweak definition of an inner-product space, at expense of some nice properties (metric completeness, separability, symmetry of the inner product… ). We are even not obliged to compute the norm in the same field as the ground field of our vector space; the concept of valued field provides a possibility to do otherwise.

Real numbers are so popular not because they form “very correct” field, but because their arise from requirements deemed very natural in analysis: order, metric completeness, separability. Wherever you can sacrifice any of it, you are welcome to work over alternative fields.

Answer 4

Just to add some information about $\mathbb{R}$ as an ordered set.

As an ordered set, $\mathbb{R}$ is the minimal dense, complete, endless, non-empty ordered set up to order-isomorphism.

Similarly $\mathbb{R}_{\geq0}$ is the minimal dense, complete, no maximal, non-empty ordered set with a minimal element up to order-isomorphism.

The reason to define things like metric/distance function/absolute value/norm is to assign a topology on the algebraic object so we can do analysis/taking limit and so on. To take limit of a sequence, we expect to have $\forall\varepsilon>0,\exists N,\forall n\geq N,|x_n-x|<\varepsilon$. Hence an ordered set with a minimal element ${0}$ is required to be the targeted set $A$.

Now we have $A$ need to be an ordered set with a minimal element. The image in $A$ is another ordered set with a minimal element, so it suffices to pick a big enough $A$ that contains enough ordered subsets that interests us. Unfortuanately, there is no the largest ordered set with a minimal element, but there is a category of them that is in some sense the largest:

Let $B$ be a non-trivial ordered set (resp. with a minimal element), then we can enlarge it to an endless ordered set (resp. a no-max ordered set with a minimal element) $B_0$, then a dense ordered set (resp. with a minimal element) $B_1$, then we can complete it using Dedekind cuts and we have an endless (resp. no-max), dense and complete ordered set (resp. with a minimal element) $B_2$.

It can be shown that every ordered set that is contained in every such ordered set is order-isomorphic to $\mathbb{R}$ (resp. $\mathbb{R}_{\geq0}$). (Clearly $\mathbb{R}$ is contained in every such ordered set, for the reverse direction, see Is every dense complete endless linearly ordered subset of real order-isomorphic to real?)

So even if we discard all the algebraic property on $\mathbb{R}$, we can see that $\mathbb{R}$ (resp. $\mathbb{R}_{\geq 0}$) is the most natural choice for ordered sets (resp. with a minimal element) that contains enough ordered sets (resp. with a minimal element), i.e. every ordered set is contained in an endless, dense, complete ordered set, and $\mathbb{R}$ is the unique ordered set that is contained in every endless, dense, complete, non-empty ordered set, up to order-isomorphism.