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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.

Michael Hardy
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The Pointer
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  • Why work with $\mathbb{R}^n$ and $\mathbb{Z}^n$? The question remains the same in one dimension – thedude Feb 05 '18 at 15:46
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    "practical" ? If they satisfies the same axioms/theorems they are the same. – Mauro ALLEGRANZA Feb 05 '18 at 15:47
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    These two objects are sets so they are identical (or rather, equal) iff they have the same elements. Here they clearly have the same elements (assuming as usual $\mathbb{Z}\subset \mathbb{R}$) and thus they are equal – Maxime Ramzi Feb 05 '18 at 15:48
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    @Max Ok, thinking about it from this perspective makes sense. But, just as an aside, aren't all mathematical objects built from set theory anyway? – The Pointer Feb 05 '18 at 15:50
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    Yes in the most common setting – Maxime Ramzi Feb 05 '18 at 15:51
  • Ok. Thank you all for the clarification. – The Pointer Feb 05 '18 at 15:51
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    This is basically about whether you think of the integers as really being a subset of the reals (with operations being a separate matter somehow), or merely having some nice embedding mapping them into the reals. In practice no one really cares, in part because mathematicians understand that both approaches are equivalent. – Ian Feb 05 '18 at 15:52
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    This question is not well-posed since "practical" doesn't seem to have any real meaning. I've offered an answer at the original post, though, to try to clarify things. I think perhaps it was a little to early to jump the gun and ask a more general question without wrestling with the meaning of the previous question and its answer. – rschwieb Feb 05 '18 at 16:07
  • @rschwieb Yes, I accept that that may be the case. My apologies. It seems that I got carried away with an idea that I was wrestling with. – The Pointer Feb 05 '18 at 16:12
  • I think the concept you're looking for here is "isomorphic", which these two structures are. – eyeballfrog Feb 05 '18 at 17:13

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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*}

Jesko Hüttenhain
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