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We write $\mathbb{R}^n$ to be the set $\{(x_1,\dots,x_n): x_i \in \mathbb{R} \mbox{ for } i=1,\dots,n\}$, so $\mathbb{R}^n$ is the set of $n$-tuples of real numbers. Is then $\mathbb{R}^1$ a set of 1-tuples but still "somehow the same" as $\mathbb{R}$? Is there any subtle difference between the two?

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    $\begingroup$ Usually $\Bbb R$ is meant to denote the field of real numbers, while $\Bbb R^1$ is meant to denote the one dimensional vector space over $\Bbb R$. This could be a subtle difference. In this sense, there is a priori no "product" of two elements of $\Bbb R^1$. $\endgroup$
    – WhatsUp
    May 2, 2021 at 20:32
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    $\begingroup$ They aren't the same but since there's a very obvious identification between the sets that respects essentially all properties of the sets respectively, we often treat them as the same. $\endgroup$ May 2, 2021 at 20:34
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    $\begingroup$ @AlohaSine That depends entirely on how you set theoretically build your tuples. With the construction I'm used to they aren't the same. $\endgroup$
    – Arthur
    May 2, 2021 at 21:08
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    $\begingroup$ This discussion parallels the discussion of whether integers are rational numbers, and whether rational numbers are real numbers. $\endgroup$
    – Arthur
    May 2, 2021 at 21:09
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    $\begingroup$ Like all mathematicians I'll ask "what does 'same' mean?"... I'm more inclined to think like AlohaSine that they are the same because there is a natural and obvious equivalence between them even if we interpret them as different types of objects. But I can see Don Thousand and Arthurs point. ... If things are equivalent is there any point in noting differences in construction? Is the integer $5$ we use to count five apples, the same or different than the $5$ we use to measure geometric distance? $\endgroup$
    – fleablood
    May 3, 2021 at 17:59

2 Answers 2

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Here is a somewhat different perspective.

It is quite common when we talk about $\mathbb{R}$ to think of the complete ordered field of $\left(\mathbb{R},+,\cdot,<\right)$. But we also have to consider the context in which the symbol $\mathbb{R}$ is used and this can make a big difference.

Here are some examples of text blocks we can sometimes read:

  • (1) Let's consider the $\color{blue}{\text{additive group}}$ of real numbers $\mathbb{R}$ ...
  • (2) Let's consider the $\color{blue}{\text{topological vector space}}$ $\mathbb{R}$ together with the usual euclidean topology ...
  • (3) Let's consider the $\color{blue}{\text{ring}}$ $(\mathbb{R},+,\cdot)$ of real numbers where we use the usual addition and multiplication of real numbers ...
  • (4) Let's consider the real numbers $\mathbb{R}$ ...
  • (5) Let's consider $\color{blue}{\text{the set}}$ of real numbers $\mathbb{R}$.

In the first three examples we have additionally to the set $\mathbb{R}$ also some kind of mathematical structure given. Each of these three examples provides us with some algebraic, or topological, or whatsoever mathematical context in which the set $\mathbb{R}$ should be considered.

Usually we do not have any problems to immediately switch into the right context (given we are already familiar with it). We then consider the real numbers precisely within this specific and restricted environment.

The example (4) is somewhat special, since here we do not have any additional context given. In this case it is common to consider $\mathbb{R}$ as complete ordered field $\left(\mathbb{R},+,\cdot,<\right)$ and often this is the correct view.

The example (5) is different to (4) since it explicitly puts the focus on the plain set structure of the real numbers: Let's consider the set of real numbers ... In this case the indicated context is rather the plain set structure and nothing else.

We have to keep in mind that a set per se is just a set without any additional mathematical structures.

When looking at OP's question, the OP is addressing explicitly the set structure of $\mathbb{R}$ and $\mathbb{R}^1$.

So, what is the difference when restricting (naively, without formal considerations) on set structures? Let's exemplarily consider a finite set $X=\{a,b,c\}$. We have for $n\geq 1$: \begin{align*} X^n=\{(x_1,\ldots,x_n)|x_j\in X, 1\leq j\leq n\} \end{align*} and as special case $n=1$ we have \begin{align*} X^1=\{(a),(b),(c)\} \end{align*} clearly showing that $X^1$ and $X$ are the same set with only some minor notational differences.

Conclusion: Considering the set-structure of $\mathbb{R}$ and $\mathbb{R}^1$ the difference is solely notational.

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EDIT: I decided to edit my answer to make it more explicit.

To straight up answer the question: is there a difference between ${\mathbb{R}}$ and ${\mathbb{R}^1}$? Yes, technically there is a difference.

When we speak about ${\mathbb{R}^n}$, we are talking about a vector space of dimension $n$ made of ${n}$-tuples over the complete ordered field ${\mathbb{R}}$. If you plug in ${n=1}$, you end up with a $1$-dimensional vector space made of $1$-tuples over ${\mathbb{R}}$.

You mentioned in the comments "isn't ${\mathbb{R}}$ a vector space over ${\mathbb{R}}$?" - you have to be careful with what you mean by this. The notation ${\mathbb{R}}$ is just referencing a complete ordered field, the notation ${\mathbb{R}^1}$ is when we are building a vector space out of ${\mathbb{R}}$ over itself, and we call it ${\mathbb{R}^1}$. There are structural differences between ${\mathbb{R}^1}$ and ${\mathbb{R}}$. We have no order defined on ${\mathbb{R}^1}$ by default, and also no multiplication defined. Well, we have the dot product, but this doesn't take us from ${\mathbb{R}^1\times \mathbb{R}^1\to \mathbb{R}^1}$, it takes us from ${\mathbb{R}^1\times \mathbb{R}^1\to \mathbb{R}}$: $$ (x)\cdot (y) = xy \in \mathbb{R} $$ so we need to first fix this: $$ (x)\times (y) := (xy) \in \mathbb{R}^1 $$ and now we need to define an order, since by default ${\mathbb{R}^n}$ has no usual "order" attached to it. Define $$ (x) < (y) \Leftrightarrow x < y $$ and now if we define the map $$ \phi : \mathbb{R} \to \mathbb{R}^1 $$ by $$ \phi(x) = (x) \in \mathbb{R}^1 $$ it's an easy exercise now to verify that ${\phi}$ is an isomorphism with respect to addition, our fixed multiplication, and the defined order. So now for our "fixed" ${\mathbb{R}^1}$, ${\mathbb{R}^1\cong \mathbb{R}}$.

In many texts, however, you will see ${\mathbb{R}^1}$ and ${\mathbb{R}}$ mostly considered as the same thing. Implicitly in the background, the fixing of multiplication and creation of an order is done, but we don't ever explicitly do it since it's rather trivial to do. Ultimately, it's just a matter of semantics when it comes to looking at the difference between ${\mathbb{R}^1}$ and ${\mathbb{R}}$.

Hope that helps clear things up.

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  • $\begingroup$ Thank you @Rieamann'sPointyNose for your detailed response! However, I am still puzzled. Isn't $\mathbb{R}$ itself also a vector space (over $\mathbb{R}$)? There we already have multiplication and order defined. $\endgroup$ May 3, 2021 at 13:07
  • $\begingroup$ My question here is why do we need to consider $\mathbb{R}^1$ as a different vector space from $\mathbb{R}$? What are the elements of $\mathbb{R}^1$ exactly? $\endgroup$ May 3, 2021 at 13:17
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    $\begingroup$ @MathCurious I have edited my answer. I hope that helps clear confusion. Note that ${\mathbb{R}}$ is not referencing a vector space, it's referencing a complete ordered field - ${\mathbb{R}^1}$ is referencing "taking the complete ordered field and building a vector space out of it over itself" - it's a very pedantic difference, but it is technically a difference $\endgroup$ May 3, 2021 at 17:31

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