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I wish to prove there does not exist a bijection relation $$F:\mathbb{Z}\times\mathbb{Z}_{2}\rightarrow\mathbb{Z}$$ Im thinking that there is more elements in $\mathbb{Z}\times\mathbb{Z}_{2}$ but both sets have an infinite number of elements so Im not too sure how that works. But I think another way of saying this is that for each integer we can assign three elements in $\mathbb{Z}\times\mathbb{Z}_{2}$ and hence the relation cant be injective. I am unsure how to formalize this and if this is even enough to show there cannot be a bijection.

user3258845
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    Why do you wish to prove this? It's not true... – Eric Wofsey Mar 29 '17 at 03:17
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    Both sets are countable, so there exists a bijection of each with $\mathbb N$ and therefore with each other. That's not the same as saying there exists an isomorphism (which preserves group structure), however. Is that what you meant? – MPW Mar 29 '17 at 03:20
  • Most likely yes, sorry. I will have to read about group structure thank you. – user3258845 Mar 29 '17 at 03:22
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    Note that $\mathbb Z$ and $3\mathbb Z$ have the "same number" of elements, even though one has three times as many as the other. Such statements are not necessarily fallacious when dealing with cardinalities of infinite sets, even though they might be when dealing with cardinalities of finite sets. – MPW Mar 29 '17 at 03:24
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    Also, note that there is a difference in saying "there exists a bijection" and saying "every map is a bijection". – MPW Mar 29 '17 at 03:29
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    @user3258845, what about the map $(n,m) \mapsto 2n+m$? Because that stuff looks bijective as hell. – mdave16 Mar 29 '17 at 03:30
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    Calling that map $f$, note that $f(0,2)= 2=f(1,0)$, so it is nonbijective as hell. – MPW Mar 29 '17 at 03:36
  • What is $\mathbb{Z}_2$? Is it the two element group $\mathbb{Z}/2\mathbb{Z}$? Is it the set ${0, 1}$? Is it the set $\mathbb{Z}^2 = {(a, b)|a,b \in \mathbb{Z}}$? (Is it the $2$-adic integers?!) Please clarify. – Benjamin Dickman Mar 29 '17 at 04:11
  • I meant as "the set of all integers modulo 2" – user3258845 Mar 29 '17 at 04:35
  • @MPW, I had $m$ supposed to be $0$ or $1$, I could have gone into more detail to have a well defined construction, I just didn't want to for a comment. As you clearly want detail, $(n,m) \mapsto 2n + \inf{r \in \mathbb{N} : m - r \in 2 \mathbb{Z}}$. Now that map looks bijective as hell. – mdave16 Mar 31 '17 at 22:52

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