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By fix any element, I mean $f(x)\ne x$ for all $x\in A$. I saw this question in Pinter's Set theory book and I'm wondering if my answer is correct. This question was in an exercise after the section where Zorn's Lemma was discussed.

My attempt: Let $\mathcal{P}$ be the set of bijective functions on subsets of $A$ which do not fix any element. Clearly $\mathcal{P}$ is nonempty because we can pick two different elements $a_1$ and $a_2$ from $A$ and we can consider the function which sends $a_1$ to $a_2$ and $a_2$ to $a_1$. Now, consider $\mathcal{P}$ as a partially ordered set with inclusion as the partial order.

We invoke Zorn's Lemma to show the existence of such a function. Let $\mathcal{C}$ be a chain in $\mathcal{P}$. Now, we claim that $h = \bigcup \mathcal{C}$ is an upper bound for $\mathcal{C}$. But for that, we need to first how that $h$ is a bijection on some subset of $A$. Let $D=\bigcup \left\{ B\subseteq A \, : \, (\exists f\in \mathcal{C}) (f\text{ is a function on } B)\right\}$. It is not too hard to see that $h$ is a function from $D$ to $D$.

We now show that $h$ is a bijection. Let $\langle x_1,y \rangle,\langle x_2,y \rangle \in h$. Then $\langle x_1,y \rangle\in f$ and $\langle x_2,y \rangle \in g$ for some $f\in \mathcal{C}$ and some $g\in \mathcal{C}$. Since $\mathcal{C}$ is a chain, we may assume $f\subseteq g$ and so $\langle x_1,y \rangle\in g$. Since $g$ is one to one, $x_1=x_2$. Now we show that $h$ is onto. Let $y\in D$. There is some $B\subseteq A$ satisfying that there is a function $f\in \mathcal{C}$ on $B$ and $y\in B$. Since $f$ is onto, $\langle x , y \rangle \in f$ for some $x\in B$. So, $\langle x , y \rangle \in h$. This shows that $f$ is onto. Now, if $h$ fixed some element in $D$, then there must be some function in $\mathcal{C}$ which would fix an element but that is not possible. So,this shows that $h$ is an upperbound for the chain $\mathcal{C}$. Thus, there must be a maximal element $f\in \mathcal{P}$.

Now, we claim that $f$ is a function on $A$. Suppose not, there must be some $a\in A$ which is not element of $\operatorname{dom} f$. Now, pick an element $a' \in \operatorname{dom } f$. Then consider the function $g$ on $\operatorname{dom } f \cup \{ a\}$ such that $g(a)=f(a')$, $g(a')=a$ and $g(x)=f(x)$ for all $x\ne a, a'$. Clearly, $g$ is a bijection and $f\subsetneq g$ which is a contradiction that $f$ is maximal. So, $f$ is a function on $A$ which does not fix any points.

Is this proof correct?

ashK
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1 Answers1

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No; the argument breaks down in the last step. Note in particular that $f$ is actually not contained in $g$, since $g(a')=a$ is different from $f(a')$ (which must be an element of $\operatorname{dom} f$ since $f$ is bijection from $\operatorname{dom} f$ to itself). Moreover, $g$ is not a bijection, since there is some element $b\neq a'$ of $\operatorname{dom} f$ such that $f(b)=a'$, and then $g(b)=a'=g(a)$ so $g$ is not injective.

In fact, it is not necessarily true that $f$ is defined on all of $A$. If $f$ were defined on all of $A$ except for just one point $a$, then $f$ would still be maximal: any proper bijective extension of $f$ would have to fix $a$, since there is nowhere else you can map $a$ to.

Eric Wofsey
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  • It was too huge a mistake to be overlooked. Is there any way I could possibly fix this by changing $\mathcal{P}$? – ashK Oct 02 '20 at 15:32
  • There are a lot of different ways to fix it, but off the top of my head, I don't see any direct way to use Zorn's lemma (i.e., a use of Zorn's lemma to a poset consisting of bijections on subsets such that a maximal element is then defined on all of $A$). Instead, you'll probably want to do a more indirect argument: first use Zorn's lemma to get some maximal element that is close to the bijection you want, and then modify it in some small way to make it actually a bijection defined on all of $A$. – Eric Wofsey Oct 02 '20 at 15:42
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    @AshishK: The simplest modification that comes to mind is to let $C={a_n:n\in\Bbb N$ be a countably infinite subset of $A$ and take $\mathcal{P}$ to be the set of bijections on subsets of $A\setminus C$. A maximal element element of $\mathcal{P}$ will then be defined on all of $A$ except $C$ and possibly one point of $A\setminus C$, and in either case the problem is essentially reduced to finding a permutation of $\Bbb N$ with no fixed points. – Brian M. Scott Oct 02 '20 at 23:00