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I've been reading on set theory and I've found some interesting things.

First, the set of all natural numbers is countably infinite in cardinality. This infinity is denoted by $\aleph_0$. All good.

You can show a one to one correspondence between these natural numbers and integers in general, thus there are as many naturals as integers.

How is that possible?! If you can create a set similar to $\mathbb{N}$ from the negative values of $\mathbb{Z}$ then there is no way they can have the same cardinality. Unless of course, the notion of infinity in $\aleph_0$ dilutes this fact. Please elaborate.

Mauro ALLEGRANZA
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Nεo Pλατo
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  • The notion of cardinality of a set is a generalization to infinite sets of the usual notion of "number of elements". For finite sets, the properites of "cardinality" are the usual one: we can count elements with arithmetic. For infinite sets, the arithmetic of cardinality has specific results. – Mauro ALLEGRANZA Mar 19 '20 at 17:35
  • "If you can create a set similar to $\mathbb{N}$ from the negative values of $\mathbb{Z}$ then there is no way they can have the same cardinality." Citation needed. This is true for finite sets, but not all sets in general - as this example shows. – Noah Schweber Mar 19 '20 at 17:38
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    Just wait until you learn that $\Bbb Q$ is also of the same cardinality... – Asaf Karagila Mar 19 '20 at 17:38
  • You have found a "paradox" that one infinite set, $\Bbb Z$, can be written as the disjoint union of two infinite sets, $\Bbb N$ and $-\Bbb N$, that each have the same size as the original set. (Let's ignore $0$ for this discussion.) That indeed never happens with nonempty finite sets.

    However, the "paradox" is only because our intuition about sets is formed primarily from interacting with finite sets. Mathematically, it is entirely possible—indeed, commonplace—for an infinite set to be written as the disjoint union of two subsets that each have the same size as the original set.

     – Greg Martin Mar 19 '20 at 17:39
  • I think the key new feature of infinite sets in this context is that not all (nice) maps are the same. If I have two finite sets $A$ and $B$, then if some injection $A\rightarrow B$ is surjective then every injection $A\rightarrow B$ is surjective. So it just suffices to show one injective non-bijection to conclude that $\vert A\vert<\vert B\vert$. For infinite sets this breaks down: we can have a non-surjective injection from $A$ to $B$ but also a bijection from $A$ to $B$. – Noah Schweber Mar 19 '20 at 17:40
  • So when I claim $\vert A\vert<\vert B\vert$ for possibly infinite $A,B$, it's not enough to exhibit a single non-surjective injection $A\rightarrow B$, I need to show that every injection $A\rightarrow B$ is non-surjective. – Noah Schweber Mar 19 '20 at 17:43
  • Let me ask: Does it similarly disturb you that the set of natural numbers $\mathbb N$ and the set of even natural numbers $2\mathbb N = {2n \mid n \in \mathbb N}$ are in one-to-one correspondence? – Lee Mosher Mar 19 '20 at 17:59
  • I ask because I could reword your most important sentence like this: "If you can create a set in $\mathbb N$ from the odd values of $\mathbb N$ then there is no way they can have the same cardinality." – Lee Mosher Mar 19 '20 at 18:01
  • @Lee Mosher Yes and definitely. Is this why Cantor's work was rejected by earlier mathematicians? – Nεo Pλατo Mar 19 '20 at 18:10
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    Regarding early mathematicians, I have no idea. Nowadays, mathematicians have no problem with a set, such as $\mathbb N$, which is in bijective correspondence with a proper subset of itself. In fact, there is a theorem in set theory which tells you exactly which sets have that property: they are the infinite sets. – Lee Mosher Mar 19 '20 at 18:12
  • Which theorem is that? I am now deeply interested in this. – Nεo Pλατo Mar 19 '20 at 18:13
  • @Plato https://math.stackexchange.com/questions/911370/how-to-prove-that-a-set-is-infinite-iff-it-is-dedekind-infinite – spaceisdarkgreen Mar 19 '20 at 19:23
  • @Plato Note that that theorem relies on (a small fragment of) the axiom of choice: it's consistent with ZF (= set theory without choice) that, for example, there are infinite sets which can't be split into two infinite pieces ("amorphous" sets), and it's easy to show that such a set can't be in bijection with any of its proper subsets. But this is a very technical topic. – Noah Schweber Mar 19 '20 at 19:56

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Being infinite is equivalent, for a given set, to being in bijection with a proper subset. Thus the notion of infinity does dilute our intuition.

Cantor himself said "I see it, but I can't believe it", in a letter to Dedekind, referring to the existence of a bijection between $[0,1]$ and $[0,1]^2$.

For perhaps the simplest example of this phenomenon, consider the shift map, $x\to x+1$, which is a bijection of $\Bbb Z_{\ge0}$ with $\Bbb N$.

  • The first part of your answer is somehow murky, as the usual definition of infinite would require the axiom of choice to be equivalent to what you're saying. But fine. The second part is entirely wrong. Cantor wrote that to Dedekind with regards to $[0,1]$ and $[0,1]^2$ having a bijection. Not about $x\mapsto x+1$, which is something that was already noted in the 13th century. – Asaf Karagila Mar 19 '20 at 17:45
  • Thanks @AsafKaragila. I was indeed incorrect. –  Mar 19 '20 at 17:56
  • Now what's a bijection? – Nεo Pλατo Mar 19 '20 at 18:11
  • It's a map that is both $1-1$ and onto. –  Mar 19 '20 at 18:23
  • @Plato: You cannot start talking about cardinals before you talk about bijections. It might be a good idea to go back and revise, or if there is nothing about bijections, find a better source. – Asaf Karagila Mar 19 '20 at 19:40