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If $\mathbf{x} \in \mathbb{R}^n$, then we would have $x_1, x_2, ..., x_n \in \mathbb{R}$, right? This is commonly known as $n$-dimensional space.

My question is, could we also have such a thing as $\mathbf{x} \in \mathbb{Z}^n$ such that $x_1, x_2, ..., x_n \in \mathbb{Z}$? In other words, could we also have $n$-dimensional integer space? Or would this simply be equivalent to $\{ \mathbf{x} \in \mathbb{R}^n | x_1, x_2, ..., x_n \in \mathbb{Z} \}$?

I would greatly appreciate it if people could please take the time to clarify this.

The Pointer
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    Yes, yes, and yes, though this set is sometimes called an integer lattice. Unless $\Bbb R^n$ it does not have a natural vector space structure, but the entrywise addition operation $+$ realizes it as a (the) free abelian group on $n$ generators. – Travis Willse Feb 05 '18 at 14:16
  • @Travis Thanks for the response. Unfortunately, I don't have the abstract algebra knowledge to understand the latter part of your comment (I haven't studied abstract algebra yet), but I'm guessing what you're saying is that, although the two objects are identical from a practical standpoint, they are actually different mathematical objects (from the perspective of mathematical rigour/precision)? – The Pointer Feb 05 '18 at 14:22
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    You're welcome. There's a typo in my comment---it should read, "Unlike $\Bbb R^n$..." and not "Unless..." The latter comment in particular means that $\Bbb Z^n$ as a set doesn't intrinsically carry any other structure---it's just a countable collection of points---but there's a natural way to put an operation $+$ on that set (arising from the usual addition operation on $\Bbb Z$). Then, the object $(\Bbb Z^n,+)$, which is the set together with that operation, is a particular well-known structure in group theory. In short, not only is $\Bbb Z^n$ a thing, but (with $+$) it's a famous thing. – Travis Willse Feb 05 '18 at 14:33
  • @Travis Ok, I think I understand. So would I be correct in saying that ${ \mathbf{x} \in \mathbb{R}^n | x_1, x_2, ..., x_n \in \mathbb{Z} }$ and ${ \mathbf{x} \in \mathbb{Z}^n | x_1, x_2, ..., x_n \in \mathbb{Z} }$ are identical iff we put the addition operation on the latter? But would they then be identical only from a practical standpoint, or would they actually then be identical mathematical objects in general? – The Pointer Feb 05 '18 at 14:38
  • It seems to me that they would be identical from a practical standpoint, but would still be different mathematical objects, no? So from a rigorous/precise mathematical perspective, they wouldn't be considered identical? – The Pointer Feb 05 '18 at 14:44
  • @ThePointer that depends. Do you consider those members of $\mathbb{R}$ that look like integers, and the integers, to be identical or not? https://math.stackexchange.com/questions/1055501/why-are-integers-subset-of-reals – AakashM Feb 05 '18 at 15:19
  • @AakashM This isn't a matter of subjective considerations; it's a matter of mathematical rigour and precision. As such, there should be a definitive, objective answer to the question, right? – The Pointer Feb 05 '18 at 15:21
  • Would you accept 'it depends' as a definitive, objective answer? See the q I linked and also https://math.stackexchange.com/questions/1819718/is-it-an-abuse-of-language-to-say-the-integers-the-rational-numbers-or?noredirect=1&lq=1 – AakashM Feb 05 '18 at 15:25
  • @AakashM There's a lot of content in those posts, so I'm not sure how they precisely address this question? Care to clarify? – The Pointer Feb 05 '18 at 15:26
  • Since this discussion is slightly off-topic from the original question, I have devoted a new thread to it: https://math.stackexchange.com/questions/2637184/are-these-two-mathematical-objects-the-same-from-a-practical-standpoint-or-lite – The Pointer Feb 05 '18 at 15:38
  • @ThePointer To answer your point starting "Ok, I think I understand", I wasn't making reference to the two sets in your post but rather $\Bbb Z^n$ (the set) and the pair $(\Bbb Z^n, +)$. I don't I could compactly improve on rschweib's below answer to the other question. – Travis Willse Feb 05 '18 at 17:17

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My question is, could we also have such a thing as $\mathbf{x} \in \mathbb{Z}^n$ such that $x_1, x_2, ..., x_n \in \mathbb{Z}$?

Yes, we talk about such things all the time: its called the notation for the Cartesian power as you are just using the Cartesian product $n$ times with a single set.

In other words, could we also have $n$-dimensional integer space?

Yes, you can always form the set, but using "dimensional" may not always be appropriate. Outside of vector spaces, which are direct products of copies of fields, the number of factors in the Cartesian power may not be a relevant quantity. And sometimes it is important and just has a different name. For example, if you're considering $R^n$ as a module over itself, we usually say "$R^n$ is a free module of rank $n$" instead of "$R^n$ is $n$ dimensional."

Or would this simply be equivalent to $\{ \mathbf{x} \in \mathbb{R}^n | x_1, x_2, ..., x_n \in \mathbb{Z} \}$?

The question is "what do you mean by equivalent?" The answer should something like "I mean up to isomorphism in some category."

In that case, the answer to this would be "yes": just as we can identify $\mathbb Z$ as a subset of $\mathbb R$ (there is an inclusion map from $\mathbb Z\to \mathbb R$) we can identify $\mathbb Z^n\subset\mathbb R^n$ via an inclusion map. This inclusion works for all of the categories you probably have in mind, namely the category of sets, abelian groups, or rings.

rschwieb
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