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Prove that $$\left(\mathbb{R}\setminus \mathbb{Z}\right) \times \mathbb{N}= \left(\mathbb{R} \times \mathbb{N}\right) \setminus \left(\mathbb{Z} \times \mathbb{N}\right)$$

I'm new to proofs with sets I justified to myself by doing a sketch of both graphs which showed me they should be equal. I am not sure how to go about proving they are equal though. I believe the general idea is I need to show each is a subset of the other but any help would be greatly appreciated.

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    $\begingroup$ When proving this kind of equations, it is usually helpful to use definitions. For example, $(\mathbb{R}-\mathbb{Z})\times\mathbb{N}=\{(x,y);x\in\mathbb{R}\land x\not\in\mathbb{Z}\land y\in\mathbb{N}\}$. $\endgroup$ – Colescu Sep 8 '16 at 7:23
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    $\begingroup$ I suppose you're using the minus sign to indicate set difference? In this case, in the future, I suggest using \setminus for that, as in $\mathbb{R}\setminus\mathbb{Z}$. This is to prevent confusion between: \begin{aligned}\mathbb{R}\setminus\mathbb{Z} &= \{x \in \mathbb{R}\,|\,x \notin \mathbb{Z}\}\\ \mathbb{R}-\mathbb{Z} &= \{ x\in \mathbb{R}\,|\,\exists r \in \mathbb{R}, \, \exists z \in \mathbb{Z}, \, x = r-z\}\end{aligned} In this particular case, $\mathbb{R}-\mathbb{Z}$ is simply $\mathbb{R}$, but generally these 'translations' can lead to different sets. $\endgroup$ – Fimpellizieri Sep 8 '16 at 7:35
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The usual route for proving that $A=B$ (set equality) is showing that $A \subset B$ and $B\subset A$, but in this case doing this is almost tautological. Instead, we can use Yuxiao Xie's suggestion.

By definition, we have that:

$$(\mathbb{R}\setminus\mathbb{Z})\times \mathbb{N}=\{(x,y)\,|\, x \in\mathbb{R}\setminus\mathbb{Z}\, \text{ and } y \in \mathbb{N} \}$$

On the other hand, we have that:

\begin{aligned} (\mathbb{R}\times\mathbb{N})\setminus(\mathbb{Z}\times \mathbb{N}) &=\{(x,y)\,|\, (x,y) \in\mathbb{R}\times\mathbb{N}\, \text{ and } (x,y) \notin \mathbb{Z}\times\mathbb{N} \}\\ &=\{(x,y)\,|\, [x \in\mathbb{R}\,\text{ and } y\in\mathbb{N}]\, \text{ and } [x \notin \mathbb{Z}\,\text{ or }y\notin\mathbb{N}] \}\\ &=\{(x,y)\,|\, x \in\mathbb{R}\,\text{ and } y\in\mathbb{N}\, \text{ and } x \notin \mathbb{Z} \}\\ &=\{(x,y)\,|\, x \in\mathbb{R}\setminus\mathbb{Z}\,\text{ and } y\in\mathbb{N}\}\\ &=(\mathbb{R}\setminus\mathbb{Z})\times \mathbb{N} \end{aligned}

It's still pretty tautological, but I guess that's because the exercise is perhaps a bit too simple; all one has to really do is check some definitions and apply basic logic (and, or, negation).

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  • $\begingroup$ Thank you it was a lot more clear with the definition approach. Yes I am just starting out with learning proofs with sets. $\endgroup$ – HighSchool15 Sep 8 '16 at 7:57

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