# Is $\Bbb R$ different from $\Bbb R^ⁿ$ where $n > 1$, in that $\Bbb R$ is not a set of $1$-tuples?

$\Bbb R^n$ where $n\gt1$ seems to be defined as a set of $n$-tuples, while $\Bbb R$ is defined a set of numbers. Some questions:

1. Does $\Bbb R^1$ equal $\Bbb R$, or is it the set $\{(a):a\in\Bbb R\}$?
2. Is e.g. $\Bbb R^2\times\Bbb R^2$ the set $\{((a,b),(c,d)):a,b,c,d\in\Bbb R\}$?

Apologies if it's been asked, couldn't find a matching question.

Edit:

Thanks for the comments and answers so far! There's been a few to the effect that distinctions of this kind don't have any practical consequences. I'm going to attempt to sketch out a scenario where it seems to me they do matter.

Suppose I'm presenting some mathematical argument that applies to the Cartesian product $A\times B$ of arbitrary sets $A$ and $B$. Then, it seems natural to think of the resulting entity as a set of ordered pairs, and e.g. "pick them apart" by saying e.g. "let $(a,b)\in A\times B$", where implicitly $a\in A$ and $b\in B$.

Now suppose you look at the special case where $A=B=\Bbb R^2$. It seems like I may still want $a$ and $b$ to still each denote elements of $\Bbb R^2$.

• We have that $\mathbb{R}^1=\{(a):a\in\mathbb{R}\}=\{a:a\in\mathbb{R}\}=\mathbb{R}.$ There is no difference between the $1$-tuple $(a)$ and the real number $a.$ When we have a $3$-tuple $(a,b,c)$ we have an ordered set of three numbers. So we can think of it as $(a,b,c)=((a,b),c)=(a,(b,c)).$ That is $\mathbb{R}\times \mathbb{R}^2=\mathbb{R}^3=\mathbb{R}^2\times \mathbb{R}.$
– mfl
Commented Jul 10, 2017 at 19:10
• On Wikipedia, a 1-tuple is defined as a singleton set, so wouldn't (a) = a imply that {a} = a? Commented Jul 10, 2017 at 19:17
• I'll wait for someone with more expertise comes along, but the way I understand it is that you are correct simplex... that $\Bbb R^1$ is technically not "equal" to $\Bbb R$, but we usually choose to ignore that because they "behave the same way and for all intents and purposes are identical" (with the exception of notation). It is the same as how $A\times B$ is different than $B\times A$, though there is an obvious homeomorphism between the two which preserves any and all structure but just changes the representation. Commented Jul 10, 2017 at 19:26
• TLDR: $\Bbb R^2\times \Bbb R^1$ is technically different than $\Bbb R^3$, but not essentially different. Commented Jul 10, 2017 at 19:28
• These kinds of questions depend on the underlying formalism for constructing ordered pairs, which may be different depending on who you ask. So mathematicians often sweep them under the rug a bit. Certainly if you write $\mathbb{R}^2 \times \mathbb{R}^2 = \mathbb{R}^4$ no one will complain, unless you're in a context where the difference is important (perhaps introductory set theory or category theory) Commented Jul 10, 2017 at 20:17

Formally speaking from a set theoretic perspective, yes, $\Bbb R^1$ and $\Bbb R$ are different. You are correct to observe that an element of $\Bbb R$ is not generally a $1$-tuple.

You are also correct that $\Bbb{R^2\times R^2}$ is a set of $2$-tuples that in each coordinate also host a $2$-tuple of elements from $\Bbb R$.

But here's the thing. There is a canonical identification map between $\Bbb{R^2\times R^2}$ and $\Bbb R^4$. And there is a canonical way to identify $\Bbb R^1$ and $\Bbb R$ itself. Since these identification are so canonical they preserve, it is often easier to abuse the notation and just omit them and replace $\Bbb{R^2\times R^2}$ with $\Bbb R^4$ at our convenience (and sometimes replace $\Bbb R^4$ with some decomposition into a product of some kind).

These things can be somewhat more expressed when you consider infinite dimensional spaces. Like $\Bbb R^\infty$, the space of eventually $0$ sequences, which also can be thought of as $\Bbb R[x]$. As a vector space, $\Bbb R\times\Bbb R^\infty$ is just $\Bbb R^\infty$ again. But what about $\Bbb R^\infty\times\Bbb R$? Now these are sequences which are eventually $0$, and have "an extra coordinate at the end". For most people these wouldn't make much sense, and they would go about their way arguing that the product is commutative so we can change the order and it's fine. And again, in a lot of contexts, this is very much the case. But from a formal perspective, this is not the case, since $1+\omega\neq\omega+1$ as ordinal addition (which is what we have here, really).

A $1$-tuple is a function $\{1\} \to \mathbb R$. Seen as such, a $1$-tuple is not a real number. Hence, if one defines $\mathbb R^1$ as the set of all $1$-tuples of real numbers, then $\mathbb R^1 \neq \mathbb R$. However, of course $\mathbb R^1 \cong \mathbb R$, so they may be regarded as the same for all purposes.

$\mathbb R^2 \times \mathbb R^2$ is what you mentioned, and this way it's not the same thing as $\mathbb R^4$, though as above it is $\cong$ to it, so they may be regarded as the same for all purposes.

Note, however, that usually, for any set $X$, $X^1$ is defined to be $X$ itself.

• Thanks for your answer! I've updated my question with a scenario where it seems to me the distinction might still matter. Would you mind having a look? Commented Jul 11, 2017 at 9:46
• Forgetting the distinction is one tactic, remembering it is another. It's up to you to decide which one to use in each situation. One part of mathematical maturity is knowing when it's safe to forget it, another is knowing how to combine clarity and precision when necessary. Commented Jul 11, 2017 at 10:44
• No, $A^1$ is defined to be all the functions from $\{0\}$ into $A$. This is not $A$ itself, unless $A$ is empty (at least when working with ZF-derived axiomatic set theories as a foundation). However, there is a canonical way to identify $A$ and $A^1$ and we simply abuse that notation. Commented Jul 11, 2017 at 10:54
• @AsafKaragila I said "usually". Commented Jul 11, 2017 at 19:42
• Usually sets are not empty.... Commented Jul 11, 2017 at 19:58