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

Question

$\Bbb R^n$ 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 $\Bbb R^2$.

Answer 1

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).

Answer 2

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.