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Here the problem 1.4.26 from Sohrab, Basic Real Analysis

Duplicate

This are not duplicate of this question:

Problem

Show that, if $A$ is an infinite set, then $|A\times\mathbb{N}|=|A|$. Hint: Let $\mathcal{F}$ denote the set of all bijective maps $f:S\times\mathbb{N}\to S$, where $S\subset A$. Since $|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$, we have $\mathcal{F}\neq \varnothing$ (Why?) Show that Zorn's Lemma can be applied in $\mathcal{F}$ to produce a maximal bijection $h:B\times\mathbb{N}\to B$, with $B\subset A$, and that we must have $B=A$, by examining the cases where $S\setminus B$ is finite or infinite.

Proof

Let $\mathcal{F}$ denote the set of all bijective maps $f:S\times\mathbb{N}\to S$, where $S\subset A$. Since $|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$ (see exercise 1.4.10(i)), we have $\mathcal{F}\neq \varnothing$ (From Exercise 1.4.22(d) we know that for a infinite set $A$, we have $|A|\geq\aleph_0$, and $|\mathbb{N}|=\aleph_0$, thus there is an injection from $g:\mathbb{N}\to A$, and, by defining $S:=g(\mathbb{N})\subset A$, we get a bijective map (composition of two bijective maps) $S\times\mathbb{N}\to\mathbb{N}\times\mathbb{N}\to\mathbb{N}\in\mathcal{F}$).

If we consider the inclusion as partial order, then every chain $S_i\subset A$, $i\in I$, has $A$ as upper bound and applying Zorn's Lemma, we get the existence of a maximal $\bar{S}$ and hence a bijection $h:\bar{S}\times\mathbb{N}\to\bar{S}$, with $\bar{S}\subset A$.

Now we prove that $\bar{S}=A$. $R:=A\setminus \bar{S}$. $\bar{S}$ is infinite, since the example of element of $\mathcal{F}$, that we gave above is bijective to $\mathbb{N}$. We have to do this. The second is trying to follow the hint and is incomplete:

  • Consider that $A\neq\bar{S}$, then there is $r\in R:=A\setminus\bar{S}$. We chose an element $s\in\bar{S}$ and define the bijection $h':\bar{S}\cup\{r\}\times\mathbb{N}\to \bar{S}\cup\{r\}$ as follow

    $h'(a,n)=\begin{cases}h(s,2n-1)&\text{for }a=s\\ r&\text{for }a=r,n=1\\ h(s,2n-2)&\text{for }a=r,n\neq 1\\ h(a,n)&\text{otherwise}\end{cases} $

    Since $\bar{S}\subsetneq \bar{S}\cup\{r\}$ we get a contradiction to the maximality of $\bar{S}$.

  • Now either $R$ is finite or infinite. 1) $R$ finite, let say $R=\{r_1,\cdots,r_n\}$. Then chose $n$ elements $s_1,\cdots,s_n\in\bar{S}$ and we define the bijection $h':A\times\mathbb{N}\to A$ as follow $h'(a,n)=\begin{cases}h(a,2n-1)&\text{for }a\in\{s_1,\cdots,s_n\}\\ a&\text{for }n=1,a\in\{r_1,\cdots,r_n\}\\ h(s_i,2n-2)&\text{for }a=r_i\\ h(a,n)&\text{otherwise}\end{cases}$

    In case $n>0$ we get a contradiction with $\bar{S}$ being the maximal element.

    Thus we have only to exclude the case 2) $R$ infinite. HOW TO DO THAT?

Question

Above I tried two proofs. The first part is the same. The two second parts are given by the two bullets.

The second proof tries to follow the hint.

My questions are: - Is my proof correct? - In which direction should go the proof from the hint?

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Your attempt to apply Zorn’s lemma doesn’t actually work, because you’re applying it to the wrong partial order. You’re taking as your partial order the family of subsets $S$ of $A$ such that there is some bijection from $S\times\Bbb N$ to $S$. If $\mathscr{C}$ is a chain in this partial order, $A$ is not known to be an upper bound for $\mathscr{C}$ in the partial order, because you don’t know that there is a bijection between $A\times\Bbb N$ and $A$: that, in fact, is what you’re trying to prove.

In order to follow the hint, you should consider the partial order $\langle\mathscr{F},\subseteq\rangle$. Let $\mathscr{C}$ be a chain in this partial order, let $f=\bigcup\mathscr{C}$. For each $f\in\mathscr{C}$ there is an $S_f\subseteq A$ such that $f$ is a bijection from $S_f\times\Bbb N$ to $S_f$; let $S=\bigcup_{f\in\mathscr{C}}S_f$.

  • Show that $f$ is a bijection from $S\times\Bbb N$ to $S$. Conclude that $h\in\mathscr{F}$ is an upper bound for $\mathscr{C}$.

Now apply Zorn’s lemma to get a maximal $h\in\mathscr{F}$, and let $B\subseteq A$ be such that $h$ is a bijection from $B\times\Bbb N$ to $B$.

Once you have this, the argument at your first bullet point works fine. I see no need to make two cases out of this depending on whether $A\setminus B$ is finite; that’s an unnecessary complication.

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  • $\begingroup$ Do I understand correct? Your special font for F (latex command?) is my $\mathcal{F}$. Your order is the order of the functions in $\mathcal{F}$ as a cartesian product, that is $A\times\mathbb{N}\times A$? $\endgroup$ – PeptideChain Aug 18 '16 at 19:59
  • $\begingroup$ @LiPo: Yes to both: all of these functions are subsets of $A\times\Bbb N\times A$. (I prefer \mathscr{} to \mathcal{}.) $\endgroup$ – Brian M. Scott Aug 18 '16 at 20:01
  • $\begingroup$ wow, very interesting, thank you very much $\endgroup$ – PeptideChain Aug 18 '16 at 20:05
  • $\begingroup$ @LiPo: You’re very welcome. $\endgroup$ – Brian M. Scott Aug 18 '16 at 20:09
  • $\begingroup$ The first bullet does not work, as there is no contradiction. The restriction to $\overline{S}$ is not the correct one, so the bijection you create is not bigger than the previous one. $\endgroup$ – Jérémy Blanc Oct 29 '17 at 9:53

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