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Let $X$ be infinite. Then $X$ and $X\times\Bbb N$ are equinumerous.

Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!

My attempt:

We denote $A$ and $B$ are equinumerous by $A\sim B$.

Let $Y=\{(A,f) \mid A \in\mathcal{P}(X) \text{ and } f:A \to A\times\Bbb N \text{ is bijective }\}$. We define a partial order $<$ on $Y$ by $$(A_1,f_1) < (A_2,f_2) \iff A_1 \subseteq A_2 \text{ and } f_1\subseteq f_2$$

Since $X$ is infinite, there exists $A\subseteq X$ such that $A \sim \Bbb N$ (Here we assume Axiom of Countable Choice). Thus $A\times \Bbb N\sim \Bbb N\times \Bbb N\sim \Bbb N\sim A$.

Hence there exists a bijection $f:A\to A\times \Bbb N$ and consequently $f\in Y$. Hence $Y\neq\emptyset$.

Let $Z$ be a chain in $Y$, $F=\{f \mid \exists A \in\mathcal{P}(X),(A,f)\in Z\}$, and $f^\ast=\bigcup\limits_{f\in F}f$.

  1. $f^\ast$ is a mapping

For $(a,x_1),(a,x_2)\in f^\ast$. Then there exists $(A_1,f_1),(A_2,f_2) \in Z$ such that $(a,x_1)\in f_1$ and $(a,x_2)\in f_2$. Since $Z$ is a chain, we can safely assume $(A_1,f_1) < (A_2,f_2)$. It follows that $f_1 \subseteq f_2$. Thus $(a,x_1),(a,x_2)\in f_2$ and consequently $x_1=x_2$ by the fact that $f_2$ is a mapping.

Let $A^\ast=\operatorname{dom}f^\ast$. Then $A^\ast=\bigcup\limits_{f\in F} \operatorname{dom}f$.

  1. $\operatorname{ran}f^\ast=A^\ast\times\Bbb N$

For all $f\in F$, $f$ is a bijection from $\operatorname{dom}f$ to $\operatorname{dom}f\times\Bbb N$, and thus $\operatorname{ran}f = \operatorname{dom}f\times\Bbb N$.

$\operatorname{ran}f^\ast=\bigcup\limits_{f\in F} \operatorname{ran}f =\bigcup\limits_{f\in F} (\operatorname{dom}f\times\Bbb N)=(\bigcup\limits_{f\in F} \operatorname{dom}f)\times\Bbb N=\operatorname{dom}f^\ast\times\Bbb N=A^\ast\times\Bbb N$.

  1. $f^\ast$ is injective

Assume $(a_1,x),(a_2,x)\in f^\ast$. Then there exists $(A_1,f_1),(A_2,f_2) \in Z$ such that $(a_1,x) \in f_1$ and $(a_2,x) \in f_2$. Since $Z$ is a chain, we can safely assume $(A_1,f_1) < (A_2,f_2)$. It follows that $f_1 \subseteq f_2$. Thus $f_1(a_1)=f_2(a_1)$ and consequently $f_2(a_2)=x=f_1(a_1)=f_2(a_1)$. Hence $f_2(a_2)=f_2(a_1)$ and consequently $a_2=a_1$ by the fact that $f_2$ is injective. It follows that $f^\ast$ is injective.

To sum up, $f^\ast:A^\ast\to A^\ast\times\Bbb N$ is bijective and hence $f^\ast \in Y$. Furthermore, $f^\ast$ is an upper bound of $Z$ by definition. Thus $Y$ satisfies the requirement of Zorn's Lemma and hence has a maximal element $(\bar{A},\bar{f})$.

  1. $X\sim X\times\Bbb N$

First, I will prove that $X\sim\bar{A}$. If not, then $X\setminus \bar{A}$ is infinite and consequently there exists $C\subseteq X\setminus \bar{A}$ such that $C\sim \Bbb N$. Thus there exists a bijection $\underline f:C \to C\times\Bbb N$. Hence $g=\bar{f} \cup \underline f$ is a bijection from $\bar{A}\cup C$ to $(\bar{A}\times\Bbb N)\cup (C\times\Bbb N)=(\bar{A}\cup C)\times \Bbb N$ since $C\cap \bar{A}=\emptyset$.

Thus, $g:\bar{A}\cup C\to (\bar{A}\cup C)\times \Bbb N$ is bijective and $\bar{f}\subsetneq g$. This contradicts the maximality of $(\bar{A},\bar{f})$.

Hence $X\sim\bar{A}\sim \bar{A}\times\Bbb N\sim X\times\Bbb N$ and consequently $X\sim X\times\Bbb N$.

Akira
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  • You need a mapping with domain $A$. Just saying $f^*$ is the union of some chain doesn't imply that - instead you want to apply Zorn's lemma to get a maximal element of $Y$. – David C. Ullrich Sep 13 '18 at 17:44
  • @DavidC.Ullrich Can you be more specific? I have gone through step by step to prove that $f^\ast \in Y$ and $f^\ast $ is an upper bound of $Z$, i.e. the requirement of Zorn's lemma is satisfied by: 1. $f^\ast$ is actually a mapping. 2. $\operatorname{ran}f^\ast = \operatorname{dom} f^\ast \times\Bbb N$. 3. $f^\ast$ is injective. – Akira Sep 13 '18 at 22:14
  • Your proof looks correct to me. Just one minor detail: $Y$ has elements of the form $(A,f)$, but sometimes you seem to neglect that. – SEBASTIAN VARGAS LOAIZA Sep 14 '18 at 00:15
  • Sorry, I didn't read carefully. I thought that $f^$ was the supposed bijection from $X$ to $X\times\Bbb N$. Actually $f^$ is just part of the application of Zorn's lemma that I thought was missing. Sorry. – David C. Ullrich Sep 14 '18 at 00:17
  • @SEBASTIANVARGASLOAIZA Please point out where I neglected that! – Akira Sep 14 '18 at 00:23
  • Hi @DavidC.Ullrich Does my attempt look fine or contain any error? – Akira Sep 14 '18 at 00:24
  • I haven't read it in detail. I thought there was an obvious huge error but I was wrong about that. – David C. Ullrich Sep 14 '18 at 12:08

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