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Until earlier today, when I thought of "real numbers" I thought of only the pure set of real numbers. It seems I was mistaken to do so, as this seems to be the set-theoretical Baire space. Instead, real numbers seem to be the set and what you can do with it, namely their field operations (addition and multiplication, their algebraic inverses, namely subtraction and division, and all the rules regarding associativity, commutativity and so on) and ordering. So the real numbers can be written as $(R, +, \cdot, <)$. Usually we write them as $\mathbb R$ though and know that we can do these things with the real numbers.

Next, we can build a product space (not inner product space!), which

is the Cartesian product of a family of topological spaces equipped with a natural topology called the product space,

from a finite number ($n\in\mathbb N$) of real numbers, which would be $\mathbb R^n$. A couple of people in How do I formally write down a Euclidean space with symbols? state that $\mathbb R^n$ is already the Euclidean space. In other words, every finite dimensional product space of the real numbers is a Euclidean space.

But I do not see what that should be the case. For a Euclidean space I would also need the

Now it seems to me that

  • completeness is inherited by every product space over the real numbers from the completeness of the real numbers
  • linear combination can be derived using the direct product. Even though I am not sure if the direct product is always implied for product topologies!
  • the Euclidean distance is induced by the Euclidean norm
  • the Euclidean norm is induced by the "normal" dot product, or according to J.W.Tanner in How do I formally write down a Euclidean space with symbols?, the dot product is induced by the Euclidean norm. However, one of them must be derived from somewhere else.

So, it seems to me that many of the properties of the Euclidean space are properties every product space over the real values has. However, I do not see why every product space over the real values should necessarily have the dot product defined as the Euclidean space has. For example I could choose to use a different inner product, which is not the Euclidean dot product, without having the dot product induced

So, are all product spaces over real numbers Euclidean spaces? If so, how is the dot product necessarily induced?

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  • $\begingroup$ I am not sure I understand the question. Are you talking about infinite products of $\Bbb R$? If yes, how are they defined? $\endgroup$
    – supinf
    Nov 27, 2020 at 19:51
  • $\begingroup$ Also the first paragraph does not seem to be relevant to the question (and the stuff about the Baire space seems wrong) $\endgroup$
    – supinf
    Nov 27, 2020 at 19:52
  • $\begingroup$ @supinf: Not the infinite product. I am asking about the finite Cartesian product of the real numbers, which I would denote with $\mathbb R^n = \mathbb R \times \mathbb R \times ... \times \mathbb R = (\mathbb R, \mathbb R, ... , \mathbb R)$. It seems to me that for this to be a Euclidean space, we are missing the added algebraic element of the dot product. But maybe the dot product is induced - if so I would like to know how. $\endgroup$
    – Make42
    Nov 27, 2020 at 19:57
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    $\begingroup$ What is your definition of a Euclidean space? Is it, for example, the one in the Wikipedia article? $\endgroup$
    – Hew Wolff
    Nov 27, 2020 at 19:58
  • $\begingroup$ @Make42 Then the answer is yes, as you wrote already: "In other words, every finite dimensional product space of the real numbers is a Euclidean space." $\endgroup$
    – supinf
    Nov 27, 2020 at 19:59

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I don't understand your comment about the Baire space.

For example I could choose to use a different inner product, which is not the Euclidean dot product

Yes, that's true. What you are learning is that "$\mathbb{R}^n$" is actually extremely ambiguous notation: depending on context it could refer to any of

This is part of a standard convention in mathematics, which is to refer to a structured set using the underlying set (also known as the carrier set) only, without naming explicitly the rest of the structure. This is for convenience; mostly it would be too annoying to do this and mostly people understand what you mean from context anyway.

However, I do not see why every product space over the real values should necessarily have the dot product defined as the Euclidean space has.

It's not necessary; it's a convention that if someone says "the inner product space $\mathbb{R}^n$" with no further elaboration they're referring specifically to the diagonal inner product defined above. This convention is relatively harmless because, among other things, all inner products on $\mathbb{R}^n$ (here I mean $\mathbb{R}^n$ the real vector space!) are related by a linear change of coordinates, so it doesn't really matter which one you pick and the diagonal one is maximally easy to compute with.

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  • $\begingroup$ How about if I add the dot product to the "finite dimensional product space over real numbers" - do I have a Euclidean space then, since all other necessary parts for a Euclidean space are already there, as I argued in my question? $\endgroup$
    – Make42
    Nov 27, 2020 at 20:17
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    $\begingroup$ @Make42: that depends on what you mean by "a Euclidean space," which is part of the point of my answer. In terms of the list above, "a Euclidean space" could refer to either a topological space, a smooth manifold, or an inner product space. Depending on what structures you equip $\mathbb{R}$ with, the induced product structure on $\mathbb{R}^n$ could be a topology, a smooth structure, or even an inner product, if you adopt the convention that if $V, W$ are two inner product spaces then $V \times W$ should be given the inner product $(v_1, w_1) \cdot (v_2, w_2) = v_1 \cdot v_2 + w_1 \cdot w_2$. $\endgroup$ Nov 27, 2020 at 20:24
  • $\begingroup$ This is the "natural" choice but it's not so common a convention that I'd assume that a general audience knew what I meant without more elaboration. $\endgroup$ Nov 27, 2020 at 20:25
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Let's say we have an inner product $\langle \cdot, \cdot \rangle$ on $\mathbb{R}^n$. We can then use the Graham-Schmidt process to find an orthonormal basis $\{v_1, \dots, v_n\}$ of $\mathbb{R}^n$ with respect to this inner product. Then, if $v = \sum x_iv_i$ and $w = \sum y_i v_i$, then by the bilinear properties of the inner product we find that $$\langle v, w \rangle = \sum_{i = 1}^n \sum_{i = 1}^n x_i y_j \langle v_i, v_j \rangle = \sum_{i = 1}^n x_i y_i \langle v_i, v_i \rangle = \sum_{i = 1}^n x_i y_i = \begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \cdot \begin{bmatrix}y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} $$ since $\langle v_i , v_j \rangle$ is $1$ if $i = j$ and $0$ if $i \neq j$. This is why inner products give way to dot products in the usual sense.

For linear combinations, you have to be a bit careful. Usually things work pretty nicely with products of vector spaces but they get a bit weird when the dimension is infinite (ie. there is no basis).

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  • $\begingroup$ Ok, interesting, but I am not having an inner product. I just have the product space (#Properties). Please note, that this is not an inner product space, but related to a Cartesian product. (I am saying, just in case, there was a misunderstanding.) $\endgroup$
    – Make42
    Nov 27, 2020 at 20:13
  • $\begingroup$ @Make42 Perhaps I misunderstand what you mean by product space. Many product spaces in topology have no algebraic structure even though they are formed from the real numbers. For example, $(0,1) \times (0,1)$ is not even a group under the usual addition operation. However, it inherets its topological properties of distance as a subspace of $\mathbb{R}^2$ which does have the algebraic properties necessary to have an inner product. $\endgroup$
    – Daniel
    Nov 27, 2020 at 20:23
  • $\begingroup$ Sorry, my last comment got messed up. I meant en.wikipedia.org/wiki/Product_topology when I wrote "product space". Also, you last comment was "way over my head". For example, I do not know what the notation $(0,1)\times(0,1)$ means. $\endgroup$
    – Make42
    Nov 27, 2020 at 20:26
  • $\begingroup$ I see. So $\mathbb{R}^n$ can have the topology induced by the product topology and the topology on $\mathbb{R}$. It can also have the euclidean topology induced by the inner product. Is your question about whether or not these topologies are the same? The answer is yes but it takes a bit of work to show it. $\endgroup$
    – Daniel
    Nov 27, 2020 at 20:33
  • $\begingroup$ No, my question is about whether the dot product is induced by the "finite dimensional product space over the real numbers" and - secondly - if the answer is "no", then if a "finite dimensional product space over the real numbers" to which I add the dot product would be a Euclidean space or if I need more things. By Euclidean space I guess I mean the things explained in Wikipedia (Magma me, what I mean by "Euclidean space" - I am still trying to figure that out). $\endgroup$
    – Make42
    Nov 27, 2020 at 20:35
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$\mathbb R$ is a lot of things. At its most basic it is just a set, but there are a lot of additional structure that you can tack onto the base set to make $\mathbb R$ a mathematical object of a different category, such as an ordered set, a group, a field, a vector space, a metric space, a topological space, a smooth manifold or an algebraic variety, just to name a few. Since there's usually only a single sensible/common way to do this, and since it is usually clear from the context which kind of object one is referring to, the resulting objects are also just called $\mathbb R$. Take note that all these objects differ in the additional structure they are given.

To a slightly lesser extent, this also applies to $\mathbb R^n$: it's (usually) the n-fold product of $\mathbb R$ with itself in whichever category we're currently looking at, whether it be sets, vector spaces, topological spaces, or lots of other things.

So you're talking about euclidean space, which itself is a name shared by several different types of object: when you say "euclidean space" without context, you could mean for example a metric space, a metric incidence geometry, or a Riemannian manifold.

Let's see how $\mathbb R^n$ is a metric space. A metric space is just a set X with a function $d : X \times X \to [0,\infty)$ that satisfies the properties of a distance function (symmetry, positivity, triangle inequality). We know the set, it's the usual cartesian product, now let's define the distance function:

$$d \colon \mathbb R^n \times \mathbb R^n \to [0, \infty)\\$$ $$d(x, y) = \sqrt{(x_1 - y_1)^2 + \ldots + (x_n - y_n)^2}$$

Simple enough. Here we're doing a bunch of arithmetic on real numbers, which we borrow from the usual structure of $\mathbb R$ as an ordered field, and we can use the theory of $\mathbb R$ as an ordered field to prove that this definition does indeed satisfy the of a metric space.

The definitions of the other meanings of "Euclidean space" will be very different, since they require a different type of additional structure. It turns out that these structures are very similar though, in that you can reconstruct each type of structure given only another type of structure.

Now let's talk a bit about cartesian products. In various categories, it is possible define a process that takes two objects of that category and defines, in a canonical and sensible manner, a new object of that category whose base set is exactly the cartesian product of the base sets of the two input objects. The product of two sets is a set, the product of two topological spaces is a topological space, the product of two metric spaces is another metric space.

Concrete example in a different category, this time the category of inner product spaces: given two inner product spaces $X$ and $Y$, with inner products $g_x$ and $g_y$ respectively, we can define their cartesian product like so:

An inner product space is a vector space with a dot product. For the vector space we pick $X \times Y$ (this is just a product of vector spaces), and we can define an inner product $g$ on it like so:

$$g((x_1, y_1), (x_2, y_2)) = g_x(x_1, x_2) + g_y(y_1, y_2).$$

You can check that this new space satisfies the definition of an inner product space. Indeed, using this definition of product it is now actually a mathematically provable statement that $\mathbb R^n \times \mathbb R^m \cong \mathbb R^{n+m}$, where the $\cong$ means that these two objects might be differently defined but they are indistinguishable within the category. It is in fact the case that $\mathbb R^n \times \mathbb R^m \cong \mathbb R^{n+m}$ in almost every category where you have both $\mathbb R^n$ and $\times$, and in other categories they still satisfy a weaker form of equivalence.

To conclude: there are many things called $\mathbb R$, many things called $\mathbb R^n$, many things called $\times$, because they lie in different categories. To get a Riemannian manifold called $\mathbb R^n$, it is not enough to take the product of $n$ copies of the set called $\mathbb R$, or the ordered field called $\mathbb R$, but you actually need to take the product of $n$ copies of the Riemannian manifold called $\mathbb R$.

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  • $\begingroup$ I read the Wikipedia articles to Riemannian manifold, metric space, incidence geometry. According to the article, the Euclidean space is a special case of either of them. So, I guess, when I am talking about a Euclidean space, I mean a special case of all of the above. To quote: "A geometric structure such as the Euclidean plane is a complicated object that involves concepts such as length, angles, continuity, betweenness, and incidence." $\endgroup$
    – Make42
    Nov 27, 2020 at 21:16
  • $\begingroup$ Where is your confusion? $\endgroup$
    – Magma
    Nov 27, 2020 at 21:24
  • $\begingroup$ You wrote "... euclidean space, which itself is a name shared by several different types of object" and then you mention these objects like Riemannian manifold. But, as far as I understand, none of them are Euclidean spaces, but generalizations of the Euclidean space. That is what confuses me. Maybe I should ask what a Euclidean space is... $\endgroup$
    – Make42
    Nov 27, 2020 at 21:31
  • $\begingroup$ okay, maybe this might allay your confusion: there is not a category of Euclidean spaces like there is a category of metric spaces. "n-dimensional Euclidean space" is just the name of a certain metric space, the name of a certain Riemannian manifold, and the name of a certain metric incidence structure. $\endgroup$
    – Magma
    Nov 27, 2020 at 21:55
  • $\begingroup$ Ok, so when I say "Euclidean space", I do not mean a certain metric space, certain Riemannian manifold, or certain metric incidence structure, but an object that is all of those things and more. I guess I mean this "complicated object that involves concepts such as length, angles, continuity, betweenness, and incidence", which en.wikipedia.org/wiki/Euclidean_space#Technical_definition defines. I think this is what most people mean in the machine learning community. $\endgroup$
    – Make42
    Nov 27, 2020 at 22:04
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  1. Euclidean Affine Spaces

A Euclidean space, $E$, refers to an affine space; think of points and coordinates. We have a Euclidean distance, determined by the Pythagorean Theorem. The Euclidean distance, $d$, with the space $E$ makes the metric space $(E,d)$

  1. Euclidean Vector Spaces $(+,-,*)$

These are different from Euclidean spaces in that we are no longer using points. These two concepts are often used in conjunction, and it's easy to forget that they are actually two different structures.

The n-dimensional Euclidean vector space, $\overrightarrow{E}$, is an inner product space. We have vector addition, an inner product and and the Euclidean norm is analogous to the distance between two points as defined for $E$. The Euclidean norm is clearly a metric for our vector space. $\overrightarrow{E}$ is nearly identical to $\mathbb{R^n}$, so why do we bother distinguishing the two?

  1. Coordinate Spaces of Real Numbers/Affine Real Spaces

When we mention the real vector space $\mathbb{R}^n$ we are actually referring to some n-dimensional Euclidean vector space with the properties of an affine space included. Once we assign a coordinate system to $\mathbb{R}^n$ we have a "coordinate space". The Cartesian coordinate system is certainly the most familiar to impress on such a space, but I'm sure you can think of some other interesting ones.

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  • $\begingroup$ Ok, I see the distinctions, also when reading en.wikipedia.org/wiki/Euclidean_space#Technical_definition in parallel. According that wikipedia passage, it seems to me that the Euclidean metric belongs to the Euclidean vector space (which "is a finite-dimensional inner product space over the real numbers" and I am guess the dot product is the inner product in this case), because we need the inner product and the "$-$". That is at least how wikipedia presents it, right? $\endgroup$
    – Make42
    Nov 27, 2020 at 21:41
  • $\begingroup$ So... from the wikipedia article "A Euclidean vector space is a finite-dimensional inner product space over the real numbers. A Euclidean space is an affine space over the reals such that the associated vector space is a Euclidean vector space." and from what you wrote and what Magma wrote about $\mathbb R$, I infer that if I just mean the set with $\mathbb R$ I still have quite a lot of work to get to Euclidean space. If I mean with $\mathbb R$ the field of real numbers and with $\mathbb R^n$ the product topology (Cartesian product) of that field, I already have many properties of the ... $\endgroup$
    – Make42
    Nov 27, 2020 at 21:51
  • $\begingroup$ Euclidean space. Now, I just need to add the dot product as the inner product. This will induce the Euclidean metric (so this does not need to be mentioned). With $n\in\mathbb N$ I have the finiteness of the dimension of the space. I guess I still need to define it to be affine (this is not), as it seems this is not implied by the "finite-dimensional inner product space". I do not see yet, why the "Coordinate Spaces of Real Numbers/Affine Real Spaces" is including both, but the "Euclidean Affine Space" is not - considering the wikidia description. But maybe wikipedia got it wrong. $\endgroup$
    – Make42
    Nov 27, 2020 at 21:57
  • $\begingroup$ @Make42, yes. I guess there are some people who call the dot product the Euclidean inner product. As for the terminology, "Euclidean space" can sort of mean anything. I think usually when someone says to consider a "n-dimensional Euclidean space", they almost always mean the real vector space. Don't get caught up on it too much, instead think of when you've been doing a problem and it involved line segments. That's the affine space you were working with. Okay, now if you've been introduced to vector functions you know you would actively use properties of both to solve problems... $\endgroup$
    – Algebraic
    Nov 27, 2020 at 22:54
  • $\begingroup$ When you say vector function, do you mean e.g. scalar fields and vector fields? I am using those a lot. For those I use the term "Euclidean space". Btw, I am working in machine learning, specifically cluster analysis. $\endgroup$
    – Make42
    Nov 27, 2020 at 23:02

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