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How do you prove that any collection of sets $\{X_n : n \in \mathbb{N}\}$ such that for every $n \in \mathbb{N}$ the set $X_n$ is equinumerous to the set of natural numbers, then the union of all these sets, $\bigcup_{i\in \mathbb{N}}$ $X_i$ is also equinumerous to the set of natural numbers? (By equinumerous I mean there exists a one-to-one and onto function $f:X_n \to \mathbb{N}$.)

Is this statement false as it stands?

user26857
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Mark
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    By "equivalent"... what exactly do you mean? – Arturo Magidin Aug 02 '11 at 19:37
  • @Arturo: There exist a one-to-one and onto function $f$ $:$ $A$ $\to$ $N$ – Mark Aug 02 '11 at 19:39
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    i.e., you want show that "the union of denumerable sets is denumerable". The statement, as given, is false, then; if you take "enough" sets that are bijectable with $\mathbb{N}$, the union may fail to be bijectable to $\mathbb{N}$; for example, $\mathbb{N}\times\mathbb{R}$ is bijectable with $\mathbb{R}$, and $\mathbb{N}\times\mathbb{R}$ is the union $\cup (\mathbb{N}\times{r})$ over all $r\in\mathbb{R}$. Each $\mathbb{N}\times{r}$ is bijectable with $\mathbb{N}$, but the union is not. – Arturo Magidin Aug 02 '11 at 19:41
  • This question is not well defined. How big is the collection of the sets? Each of the sets? Do you work in ZFC, or some other theory? – Asaf Karagila Aug 02 '11 at 19:42
  • I'm assuming $N$ is $\mathbb{N}$ (you can type it with \mathbb{N}); what is $Z_n$, though? – Arturo Magidin Aug 02 '11 at 19:43
  • @Arturo: Please give me a minute to edit the question – Mark Aug 02 '11 at 19:44
  • @Arturo: I am done editing – Mark Aug 02 '11 at 19:48
  • @Mark: And is $N$ the set of natural numbers, $\mathbb{N}$? – Arturo Magidin Aug 02 '11 at 19:50
  • @Arturo: yes it is, I have corrected it – Mark Aug 02 '11 at 19:50

2 Answers2

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The answer depends on your set theory.

If your set theory includes the Axiom of (Countable) Choice, then you can proceed as follows:

  1. For each $n\in\mathbb{N}$, select a bijection $f_n\colon X_n\to\mathbb{N}$. (This step requires the Axiom of Countable Choice);
  2. Select a bijection $g\colon\mathbb{N}\times\mathbb{N}\to\mathbb{N}$; there are several explicit examples of this. For example, the Cantor pairing function $g(p,q) = \frac{(p+q)(p+q+1)}{2}+q$.
  3. Define $f\colon \bigcup\limits_{n\in\mathbb{N}}(X_n\times\{n\})\to \mathbb{N}$ by mapping $(x,n)$ to $g(f_n(x),n)$.

This defines a bijection between the disjoint union of the $X_n$ onto $\mathbb{N}$. To get a bijection in the case where the $X_n$ are not disjoint, note that $\bigcup\limits_{n\in\mathbb{N}} X_n$ embeds into the disjoint union (map $x$ in the union to $(x,m)$ where $m$ is the smallest $n\in\mathbb{N}$ such that $x\in X_n$), which is bijectable to $\mathbb{N}$; then use the Cantor-Bernstein Theorem applied to this embedding and to embedding that maps $\mathbb{N}$ to $X_1$ into the union to get a bijection.

However, if your set theory does not include the Axiom of Choice, then the answer may be that the union need not be bijectable with $\mathbb{N}$. In particular, it is consistent with ZF that the real numbers are a countable union of countable sets, and of course the real numbers are not bijectable with $\mathbb{N}$.

sbj
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Arturo Magidin
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If you want to show that the countable union of countable subsets is countable, you can use Cantor-Schroeder-Bernstein (I don't think it uses AC --even in summer :) ), and set up injections between $\mathbb N$ and $\mathbb N \times \mathbb N $, and the other-way-around , by generalizing this:

take any two primes , say 2,3, and map : $(a,b)\rightarrow 2^a3^b$ ( you can see that, to generalize to a product of k-copies of $\mathbb N$, just take k different primes; if you want an actually countably-infinite product, this is maybe more delicate), and an injection in the opposite direction is given by , e.g., n->(n,0,0,...).

And, BTW, any choice of injections in CSBernstein allows to construct an actual bijection.

EDIT: I think it is not too hard to show the map (a,b)->$2^a3^b$ is an injection; if we had $2^a3^b=2^{a'}3^{b'}$, it would follow that $2^{a-a'}3^{b-b'}=1$; by simple divisibility arguments, each of the factors on the left-hand side would have to divide 1; it then follows that a-a'=0 and b-b'=0, i.e., a=a', b=b'.

EDIT#2 : Please see some of the caveats in the comments section about concluding that the union of countables is countable.

gary
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  • It is consistent without choice that $\aleph_1$ is a countable union of countable sets. You have to have choice for that, even if Cantor-Bernstein doesn't require it. – Asaf Karagila Aug 02 '11 at 22:44
  • Sorry, but isn't the countable union of countable counatble, i.e., $\aleph_0$? – gary Aug 02 '11 at 23:15
  • @gary: As Asaf says, in the absence of the axiom of choice, it is possible for a countable union of countable sets to be uncountable. For example, there is a model of ZF where $\mathbb{R}$ is a countable union of countable sets. – Zhen Lin Aug 03 '11 at 00:42
  • Yes, my bad, I keep assuming ZFC automatically; I keep imposing my assumptions on questions; I am becoming senile way too early. – gary Aug 03 '11 at 01:18
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    @gary: Basically, the problem is that one needs to choose countably many embeddings/bijections into $\mathbb{N}$ simultaneously. On the other hand, the union of countably many sets that come equipped with embeddings into $\mathbb{N}$ can be shown to be countable in ZF, without having to invoke choice; that is, if you have a family $(X_n,f_n)$, $n\in\mathbb{N}$, with $f_n\colon X_n\to \mathbb{N}$ one-to-one for each $n$, then it is easy to prove that $\cup X_n$ is countable; but without having the $f_n$ given, that's when you need AC (at least countably). – Arturo Magidin Aug 03 '11 at 11:15
  • @Arturo: Thanks for the explanation: So, if I were to do an injection between $\mathbb N^k=\mathbb N \times \mathbb N \times...$ , in which I select the first k primes: {2,3,..,$p_k$}, so that I assign : $(x_1,x_2,...,x_k)\rightarrow 2^{x_1}3^{x_2}....p_k^{x_k}$, and then I select , say, h:n$\rightarrow (n,0,0,..,0)$I don't see clearly under which case this falls; would you say $x_n\rightarrow p_n^{x_n}$ is in the family, so I don't need choice, and this is in your first alternative $(X_n,f_n)$? – gary Aug 05 '11 at 06:39
  • @gary: Given a countable set $X$, you can prove in ZF without choice that $X^{\omega}$ is countable, just like one can prove that an arbitrary cartesian power of a nonempty set is nonempty without having to use the Axiom of Choice, but one cannot prove that an arbitrary cartesian product of nonempty sets is nonempty. Likewise, one can prove in ZF that if $X_1,\ldots,X_n$ are countable, then so is $X_1\cup\cdots\cup X_n$ and $X_1\times\cdots\times X_n$. The problem arises when you have infinitely many sets, and no canonical choice of embeddings into $\mathbb{N}$. – Arturo Magidin Aug 05 '11 at 07:28
  • For the proof that the function mapping NxN to primes is an injection, there might be a faulty assumption. The multiplicative inverse does not exist in N, so i believe there is an issue with performing the algebra leading to the conclusion, however, uniqueness of prime factorizations guarantees that the function is injective...it would probably be safer just to say that. –  Aug 23 '12 at 03:30