Are these two mathematical objects the same from a practical standpoint, or literally identical mathematical objects? This question is derived from another question that I recently asked.
Take the two mathematical objects $\{ \mathbf{x} \in \mathbb{R}^n \mid x_1, x_2, \ldots, x_n \in \mathbb{Z} \}$ and $\{ \mathbf{x} \in \mathbb{Z}^n \mid x_1, x_2, \ldots, x_n \in \mathbb{Z} \}$. If I'm interpreting his response correctly, Travis said in the comments of the previous question that these two are the same iff we put the addition operator on the latter (since, unless $\mathbb{R}^n$, it doesn't have a natural vector space structure). Assuming we put the addition operator on $\{ \mathbf{x} \in \mathbb{Z}^n \mid x_1, x_2, \ldots, x_n \in \mathbb{Z} \}$, does this mean that the two mathematical objects $\{ \mathbf{x} \in \mathbb{R}^n \mid x_1, x_2, \ldots, x_n \in \mathbb{Z} \}$ and $\{ \mathbf{x} \in \mathbb{Z}^n \mid x_1, x_2, \ldots, x_n \in \mathbb{Z} \}$ are identical from a practical standpoint, or are they literally identical mathematical objects (from the perspective of mathematical rigour and precision)?
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?
I would greatly appreciate it if people could please take the time to clarify this. Also, please note that I have not studied abstract algebra yet, so more elementary language is appreciated. 
 A: So, if you want to be really exact, we should think carefully about the condition $x\in\Bbb Z$ for an element $x\in\Bbb R$. Does this mean that you have defined $\Bbb R$ in such a way that $\Bbb Z$ is actually a subset of $\Bbb R$? The way it is often done, $\Bbb Z$ only embeds into $\Bbb R$ by mapping a number $x\in\Bbb Z$ to the equivalence class of the constant Cauchy sequence $(\frac{x}1,\frac{x}1,\ldots)$. So if you have defined $\Bbb R$ as a set of equivalence classes of Cauchy sequences in $\Bbb Q$, then $\Bbb R$ does not truly contain the set $\Bbb Z$, it "only" contains something that is isomorphic to $\Bbb Z$ in some very strong ways. If Mathematicians were unable to get over this fact, then we'd have to conclude that $\{ \mathbf{x}\in\Bbb R^n \mid x_1,\ldots,x_n\in\Bbb Z \}=\emptyset$.
of course, this would be madness.
The fact of the matter is that there is always a strongly intuitive, vastly structure-preserving, injective map $i:\Bbb Z\to\Bbb R$, no matter how you construct $\Bbb R$. This map allows you to think of $i(\Bbb Z)$ as $\Bbb Z$ itself. In fact, it means that you could have constructed $\Bbb R$ in such a way that $\Bbb Z$ is truly a subset of it. And if you are willing to accept that $\Bbb Z$ is actually a subset of $\Bbb R$, then indeed
$$
\{ x\in\Bbb R^n \mid x_1,\ldots,x_n\in\Bbb Z \} = \Bbb Z^n.
$$
Proof. In set theory, we usually define the tuple $(x_1,\ldots,x_n)$ recursively by the following rules:
\begin{align*}
(x_1,x_2) &:= \{ \{x_1\}, \{x_1,x_2\} \}, \\
(x_1,\ldots,x_k,x_{k+1}) &:= ((x_1,\ldots,x_k),x_{k+1})
\end{align*}
The set $\Bbb Z^n$ simply denotes the set of all tuples $(x_1,\ldots,x_n)$ with $x_1,\ldots,x_n\in\Bbb Z$. Since $\Bbb Z\subseteq \Bbb R$, we have 
\begin{align*}
\{ (x_1,\ldots,x_n)\in\Bbb R^n \mid x_1,\ldots,x_n\in\Bbb Z \} 
&= \{ (x_1,\ldots,x_n) \mid x_1,\ldots,x_n\in(\Bbb Z\cap\Bbb R) \} 
\\&= \{ (x_1,\ldots,x_n) \mid x_1,\ldots,x_n\in\Bbb Z \} = \Bbb Z^n.
\end{align*}
