What is the definition of $\mathbb R^n$?

Question

What is the definition of $\mathbb R^n$ space ? The set of the $n\times 1$ matrix, where coefficients are in $\mathbb R$ ?
Or, the set of mappings from $\{1,2,...,n\}$ to $\mathbb R$ ?

It's the cartesian product $\mathbb{R}\times\mathbb{R}\times\cdots\times\mathbb{R}$, where there are $n$ $\mathbb{R}$'s. So it depends on how you define the cartesian product. — Yuxiao Xie, Jul 30 '17 at 11:54
@JoséCarlosSantos Well, I thought it was too short to be an answer. :) — Yuxiao Xie, Jul 30 '17 at 11:56
$\Bbb R^n\cong \Bbb R^{n\times 1}\cong\Bbb R^{1\times n}$, so you can see it as $n\times 1$ matrices, $1\times n$ matrices or just vectors of $n$ coordinates (without an explicit representation). — Masacroso, Jul 30 '17 at 11:57
@Phoenix: If $X$, $Y$, and $Z$ are sets, there is a canonical bijection from $(X \times Y) \times Z$ to $X \times (Y \times Z)$. Is that not associative enough for you? :) — Andrew D. Hwang, Jul 30 '17 at 12:02
@Phoenix Intuitively, it's not the elements themselves that we care about. We care about the defining property stating that $(x_1,x_2,...,x_n)=(y_1,y_2,...,y_n)$ if and only if $x_i=y_i$ for all $1\leq i\leq n$. So long as that is satisfied, whether the ordered tuple $(x_1,x_2,...,x_n)$ is defined to be a matrix or vector or any other thing does not matter. — Yuxiao Xie, Jul 30 '17 at 12:07
@Yuxiao Xie I agree with you, but, since these objects are "ill-defined" it occurs some abuses... espacially in linear algebra where linear mappings are "identified" to be matrices :/ — Célestin, Jul 30 '17 at 12:12
I don't think there is anything to worry about. Identification does not mean they are the same. Rather, it actually means that there is some kind of bijection between the objects that preserves our desired properties. In your case this is actually a (somewhat natural) isomorphism between the rings $\mathcal{L}(V,W)$ of all linear maps from $V$ to $W$ and $\mathcal{M}_{m,n}(F)$ of all $m\times n$ matrices in a field $F$. — Yuxiao Xie, Jul 30 '17 at 12:17
In this case, we have to work with isomorphisms using the bijections. But, for example write that $\mathbb R \ni x = [x] \in \mathcal M_{1,1}(\mathbb R)$ is absurd... the same things are reproduced in linear algebra. — Célestin, Jul 30 '17 at 12:23
Why? Clearly $\mathbb{R}\cong\mathcal{M}_{1,1}(\mathbb{R})$ as rings and so their elements can be safely identified. In mathematics (not only linear algebra) this kind of situation happens all the time... — Yuxiao Xie, Jul 30 '17 at 12:29
Yes $\mathbb R\cong \mathcal M_{1,1}(\mathbb R)$ but strictly speaking, the elements of $\mathbb R$ are distinct of those in $\mathcal M_{1,1}(\mathbb R)$. Adding the bijection $\phi : \mathcal M_{1,1}(\mathbb R) \to \mathbb R$, writting $x = \phi([x])$ is correct. — Célestin, Jul 30 '17 at 12:35
You're right, strictly speaking $\mathbb{R}$ and $\mathcal{M}_{1,1}(\mathbb{R})$ are disjoint sets. But actually there is no point in distinguishing the two. Let me take an example. The fields $\mathbb{R}$ and $\mathbb{C}$ are disjoint by construction, but would you say $\mathbb{R}\ni1\neq1+0i\in\mathbb{C}$? Actually all the number systems $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$ are disjoint, yet we have a natural imbedding from all of the sets to $\mathbb{C}$, which allows us to identify all these to be complex numbers. — Yuxiao Xie, Jul 30 '17 at 12:42
From axiom of infinity, we can construct a set $\hat{\mathbb N}$ (as the smallest inductive set) which provide a "fake natural numbers set". Then... $\hat{\mathbb N} \hookrightarrow \hat{\mathbb Z} \hookrightarrow \hat{\mathbb Q} \hookrightarrow \hat{\mathbb R} \hookrightarrow \hat{\mathbb C}$ (fake numbers set). Let define $\mathbb C$ to be equal to $\hat{\mathbb C}$, then through images of isomorphisms, we're able to define our final numbers sets, and the true inclusion $\mathbb N \subset \mathbb Z\subset \mathbb Q \subset \mathbb R \subset \mathbb C$. — Célestin, Jul 30 '17 at 12:57
Because it's amusing and at least tangentially relevant: From a MO thread on mathematical epigrams: Mathematics is the art of giving the same name to different things. (Henri Poincaré) — Andrew D. Hwang, Jul 30 '17 at 13:00
Part of mathematical training is being able to move forward without getting stuck putting your toe over the first line. A mathematician needs to understand many basic ways to identify different models of a mathematical object, and to use those models productively without getting stuck on expressing those models with absolute precisition whenever they are used. One of those isomorphisms, mentioned in the comment of @AndrewD.Hwang, is $(X \times Y) \times Z = X \times (Y \times Z)$. — Lee Mosher, Jul 30 '17 at 13:11
Not $=$, only $\cong$... otherwise, it condraticts the axiom of extensionnality in ZF set theory... — Célestin, Jul 30 '17 at 13:14
Keep in mind, humans are not computers, and they can understand and express things with remarkable precision even in the face of ambiguity between the choice of models. For instance, the precision of coordinate 3-dimensional geometry was understood by Descartes and many since then, despite the ambiguity between three common models for $\mathbb{R}^3$, namely $(\mathbb{R} \times \mathbb{R}) \times \mathbb{R}$, or $\mathbb{R} \times (\mathbb{R} \times \mathbb{R})$, or ${f : {1,2,3} \to \mathbb{R}}$. — Lee Mosher, Jul 30 '17 at 13:15
So for a mathematician $\text{cat} = \text{dog}$ because they've the same numbers of paws, etc. however human being is able to distinguish them :/ — Célestin, Jul 30 '17 at 13:20

score 7 · Accepted Answer · answered Jul 31 '17 at 00:12

$\newcommand{\Reals}{\mathbf{R}}$Caveat: This answer is based on my experiences with other mathematicians and is intended to fairly represent the views of the (large) majority, but is written as first-person opinion to avoid putting words in others' mouths (particularly because the second section below may be controversial to those concerned with foundations).

To answer the question at face value: If pressed for a specific Cartesian product, I'd define $\Reals^{n}$ inductively by $$ \Reals^{1} = \Reals,\qquad \Reals^{n+1} = \Reals^{n} \times \Reals^{1}\quad n \geq 1, $$ "tacking on the new coordinate at the end". Consequently, a typical element of $\Reals^{n}$ would be $((\cdots((x_{1}, x_{2}), x_{3}), \dots), x_{n})$.

As the comments and existing answers suggest, I take a less ontological, more phenomenological viewpoint in practice, focusing on how elements of $\Reals^{n}$ behave rather than on what elements of $\Reals^{n}$ are as sets.

To make an analogy with object-oriented programming, the "public interface" of the Cartesian vector space $(\Reals^{n}, \oplus, \odot)$ consists of precisely the following:

An ordered collection of $n$ real-valued functions $(\pi_{j})_{j=1}^{n}$ (the Cartesian coordinates or coordinate projections) such that if $x$ and $y$ are arbitrary elements of $\Reals^{n}$, then $x = y$ (as elements of $\Reals^{n}$) if and only if $\pi_{j}(x) = \pi_{j}(y)$ for all $j = 1, \dots, n$.
A mapping $\oplus:\Reals^{n} \times \Reals^{n} \to \Reals^{n}$ satisfying $$ \pi_{j}(x \oplus y) = \pi_{j}(x) + \pi_{j}(y)\quad\text{for all $x$, $y$ in $\Reals^{n}$.} $$
A mapping $\odot:\Reals \times \Reals^{n} \to \Reals^{n}$ satisfying $$ \pi_{j}(c \odot x) = c\pi_{j}(x)\quad\text{for all $x$ in $\Reals^{n}$.} $$

Using $\oplus$ and $\odot$ to denote vector addition and scalar multiplication distinguishes these operations from the field operations of the real numbers; in practice, they are universally denoted by $+$ and either $\cdot$ or simple juxtaposition.

The philosophical point about phenomenology comes down to:

Any question that cannot be phrased in terms of the public interface is not (to me) a question about the Cartesian vector space.

Similarly, any question that cannot be phrased in terms of the first item alone is not (to me, in the sense I work with $\Reals^{n}$ in daily practice) a question about $\Reals^{n}$, but a meta-question.

By contrast, you're asking about ontology, namely, about the "private data" or "implementation".

To implement $\Reals^{n}$ in set theory, the inductive construction in the preceding section suffices. Happily, the order of grouping doesn't matter, because (as noted in the comments) the Cartesian product of sets is associative up to a canonical bijection. In other words, we can, if we wish, change the implementation details (regroup the Cartesian product) without touching the public interface (without altering "how elements of $\Reals^{n}$ behave" as mathematical objects or as vectors).

It's alternatively possible to implement $\Reals^{n}$ as the set of real-valued functions on the set $\{1, 2, \dots, n\}$, with $\pi_{j}$ denoting evaluation at $j$. Some mathematicians do this. One pedagogical advantage is that function spaces are immediately seen to be natural generalizations of $\Reals^{n}$.

Coda: These are good technical issues to iron out (+1). That said, in practice, I expect most non-foundational mathematicians view this type of distinction in the same light as the following quip of Woody Allen, on the fundamental material constituent of the universe, which I offer as a good-natured jest:

Democritus called it atoms. Leibniz called it monads. Fortunately the two men never met, or there would have been a very dull argument.

score 4 · Answer 2 · answered Jul 30 '17 at 12:39

Let $n \in \mathbb{N}$. Then the simplest definition of $\mathbb{R}^{n}$ may be $$ \mathbb{R}^{n} := \{ (x_{1},\dots, x_{n}) \mid x_{1},\dots,x_{n} \in \mathbb{R} \},$$ i.e. the set of all the $n$-tuples of real numbers. If you know of Cartesian product, then you see that $\mathbb{R}^{n} = \mathbb{R} \times \cdots \times \mathbb{R}$, the $n$-fold of $\mathbb{R}$. Note that it is customary to call a set a space if it is endowed with some structures such as a topology or a metric.

score 0 · Answer 3 · answered Jul 30 '17 at 16:10

The following is how I conceive the idea of a Cartesian product.

Given a set $X$, we define an $m-tuple$ of elements of $X$ by the map, $$ x:\{1,...,m\}\to X$$ where the image $x(i)$ in $X$ is denoted by $x_i$. We usually represent the map by $(x_1,...,x_m)$.

So, the set of all $m-tuples $ of $X$ is the set of all maps $x$ as defined above.

Let $\{A_1,...,A_m\}$ be a family of sets and let $X=A_1\cup ...\cup A_m$.

We define the $\text{cartesian product}$ of this family, denoted by $$A_1\times ...\times A_m $$ to be the set of all $m-tuples$ $(x_1,...,x_m)$ of elements of $X$ such that $x_i\in A_i$ for each $i.$

In particular taking $A_i=$ $\Bbb R$ for $i=1,...,m$ we get a rigorous definition of the $\text {cartesian product}$ on $\Bbb R$ denoted by, $$\Bbb R \times...\times \Bbb R \text{[m-times]}=\Bbb R^m$$

This is the definiton of a generalisation of the cartesian product for any family of sets $(A_i){i\in I}$ : the product of sets defined by $\prod{i\in I} A_i = \left{x : I\to \bigcup_{i\in I} A_i :\big\vert: \forall i\in I, x(i)\in A_i\right}$. I think that is the most rigorous way to define $\mathbb R^n$ (vector space and topology are easy to setup by the way). — Célestin, Jul 30 '17 at 16:26
Yes, it is indeed a rigorous definition of the cartesian product and its exactly the same as what i've mentioned. — Naive, Jul 30 '17 at 16:33

score 0 · Answer 4 · answered Jul 31 '17 at 00:30

In math there is a tendency to refer to objects that are technically different, but which are closely related, by the same name. So depending on the context, $\mathbb R^n$ could denote the set of all functions from $\{1,2,\ldots, n\}$ to $\mathbb R$, or it could denote the set of all $n \times 1$ matrices with real entries. And technically, those are two different things. Hopefully the meaning will be clear from context.

By the way, if you dig into the definitions, a real $n \times 1$ matrix is often defined to be a function from $\{(i,1)\mid 1\leq i \leq n\}$ to $\mathbb R$.

What is the definition of $\mathbb R^n$?

4 Answers4

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