5

I want to prove that if $A$ in an infinite set, then the cartesian product of $A$ with 2 (the set whose only elements are 0 and 1) is equipotent to $A$.

I'm allowed to use Zorn's Lemma, but I can't use anything about cardinal numbers or cardinal arithmetic (since we haven't sotten to that topic in the course).

I read a proof of the fact that if $a$ is an infinite cardinal number, then $a+a=a$, which is something similar to what I want to prove.

Any suggestions will be appreciated :)

hmakholm left over Monica
  • 286,031
Juan Carlos Capultitla
  • 59
  • A proof that $a+a=a$ is not just similar to what you want but precisely what you want, since there is an obvious bijection between the disjoint union of two copies of $A$, and the set $A\times 2$. – Andrés E. Caicedo Nov 28 '14 at 00:17
  • 4
    To the people making the edits: In set theory, it is $2$, not $\mathbb Z_2$ or ${2}$ or $[2]$. It is standard notation. Leave it as is. – Andrés E. Caicedo Nov 28 '14 at 00:26

2 Answers2

3

Consider the collection $S$ of pairs $(X,f)$ where $X\subseteq A$ is infinite and $f:X\times 2\hookrightarrow X$. Then $S$ is nonempty, since $A$ being infinite contains a copy of $\Bbb N$, and it is known $\Bbb N\times 2\simeq \Bbb N$. It should be evident that under the ordering of extension, this set always has $(X,f)\in S$ if $X=\bigcup X_i$ and $f=\bigcup f_i$ with each $(X_i,f_i)\in S, i\in I$ and $I$ a total order. It follows by Zorn's lemma that there is a maximal element $(A',g)$ in $S$. If we show that $A'$ has the same cardinality as $A$, we're done. Now it is clear $\# A'\leqslant \# A$ since $A'$ is a subset of $A$. So assume $\#A'<\#A$. Then $A\setminus A'$ must be infinite, since else we would have by $A=A'\sqcup (A\setminus A')$ an equality (recall that if $\mathfrak a$ is infinite and $\mathfrak b$ finite, $\mathfrak a+\mathfrak b=\mathfrak a$) , so there is some $A''\subset A$ disjoint from $A'$ with cardinality $\aleph_0$. But then by patching things together, we would be able to extend the injection $A'\times 2\to A'$ to a larger $(A'\sqcup A'')\times 2\to (A'\sqcup A'')$.

Pedro
  • 122,002
  • Some minor comments: I would require that the sets $X$ as considered in your collection $S$ must be infinite, and add that the fact that $\mathbb N$ injects into any infinite set $A$ is also a consequence of Zorn's lemma, and that this implies the equality $\mathfrak a+n=\mathfrak a$ that you use later. – Andrés E. Caicedo Nov 28 '14 at 00:52
  • For similar arguments along the same lines, see here. – Andrés E. Caicedo Nov 28 '14 at 00:55
  • @AndresCaicedo I think I said the sets must be infinite, but then I accidentally edited it out. Thanks. – Pedro Nov 28 '14 at 00:56
  • I think that you can also directly consider the pairs $(X,f)$ where $X\times 2$ injects to $A$. Then the end of the prood is easier because you have to add only one point instead of a countable infinite. – Laurent Claessens Feb 06 '20 at 06:02
  • 1
    Answer to myself : no. Because after using Zorn lemma, we have no garantee that $g:2\times A'\to A$ is not surjective. So it is not clear that one could find 2 elements left in $A$. (or, at least, I am not able) – Laurent Claessens Feb 06 '20 at 06:17
3

Here's a kind of brute-force way to do it.

Lemma. Assume that $A$ is infinite. Then there is a set $B$ such that $A$ is equipotent to $B\times\mathbb N$.

Assuming this lemma, we only need to prove your property when $A$ is in fact $B\times \mathbb N$. But then, obviously, this is a bijection: $$ f:B\times\mathbb N\times \{0,1\} \to B\times\mathbb N : (b,n,m)\mapsto(b,2n+m) $$

Proof of lemma. Let a "cool" function mean an injective function $g:B\times \mathbb N\to A$ for some $B\subseteq A$, where $g(b,0)=b$ for all $b\in B$. Order the cool functions by set inclusion; Zorn's lemma then applies, and we get a cool function $g$ that cannot be extended to a larger cool function.

Now if $A\setminus \operatorname{Rng} g$ is infinite, then it has a countably infinite subset (I assume this is known, as a consequence of AC or Zorn), and then it is easy to see that $g$ cannot have been maximal. (Here it is useful that $g(b,0)=b$ such that $B$ is disjoint from $A\setminus \operatorname{Rng} g$). So $A\setminus \operatorname{Rng} g)$ is finite; call its size $M$.

Let therefore $h:\{0,1,\ldots,M-1\}\to A\setminus \operatorname{Rng} g$ be a bijection. Furthermore $B$ must be nonempty (because $A$ is infinite), choose a $b_0\in B$. Now $$ (b,n) \mapsto \begin{cases} g(b,n) & \text{if }b\ne b_0 \\ h(n) & \text{if }b=b_0 \land n<M \\ g(b,n-M) & \text{if }b=b_0 \land n\ge M \end{cases}$$ is a bijection $B\times\mathbb N\to A$.

hmakholm left over Monica
  • 286,031
  • 1
    Oops, took so long to type that Pedro Tamaroff beat me by a long time. His solution is simpler, but I'll leave this up anyway since it isn't wrong (I think). – hmakholm left over Monica Nov 28 '14 at 01:01
  • You may want to comment that the fact that $\mathbb N$ injects into any infinite set is a consequence of Zorn's lemma (it is worth making this explicit, since it is the key fact that allows both Pedro's solution and yours to get started). For instance, given $A$ infinite, we can look at the set of injections from a subset of $\mathbb N$ into $A$, ordered by extension, and Zorn's lemma gives us that one such injection is infinite. – Andrés E. Caicedo Nov 28 '14 at 01:05
  • @Andres: Comment added. – hmakholm left over Monica Nov 28 '14 at 01:11