So here’s a really weird “functional equation” I came up with. Since a function $g:X\rightarrow Y$ can be defined as a subset $g\subseteq X \times Y$ such that for all $ x \in X$, there is a unique $y\in Y$ such that $(x,y)\in g$. Since a function is then just a set, we could feasibly have a function who’s domain and range is itself. So I wanted to find a function $f: f\rightarrow f$.
The first example that comes to mind is $\emptyset : \emptyset \rightarrow \emptyset$, which works, but is kind of boring. For a while, I thought that might be the only solution. But then I came up with a really pathological set that I’m not sure what to make of, but I think it works.
Let $A_0=\{0\}$, we’ll say $A_{n+1}=A_n \times A_n$. So:
$$A_1=\{(0,0)\}$$
$$A_2=\big\{\big((0,0),(0,0)\big)\big\}$$
$$A_3=\Big\{\Big(\big((0,0),(0,0)\big),\big((0,0),(0,0)\big)\Big)\Big\}$$
$$\vdots$$
None of these satisfy the functional equation as $A_n\not \subseteq A_n \times A_n = A_{n+1}$. However, I did come up with a weird idea: could there be an $A_\infty$ that does satisfy the functional equation? It would look something like this:
$$A_\infty = \{(((\cdots(((0,0),(0,0)),((0,0),(0,0))),\cdots,(0,0)))\cdots)))\}$$
Where it has a sort of “infinite binary tree” structure to it.
Another way to interpret it, is that if $a\in A_\infty$ is the sole element of $A_\infty$, then $A_\infty=\{a\}=\{(a,a)\}$. Though one thing I don’t like about using this as a definition is that it ignores the $0$s “at the end” of the ordered pairs (whatever that means.) My other issue with it is that it seems to hinge on the assumption that $A_\infty$ exists, because it doesn’t construct $A_\infty$ directly.
I can’t really wrap my head around this set, and it doesn’t feel especially rigorous, but if it does seem to satisfy my functional equation. Since $A_\infty=\{a\}=\{(a,a)\}$, it seems to be the case that $A_\infty = A_\infty \times A_\infty$. Therefore, $A_\infty : A_\infty \rightarrow A_\infty$ seems to be a valid solution to the functional equation.
I don’t know what to make of this. While the solution to the functional equation $f=\emptyset$ makes complete sense to me, I can’t even tell if I’ve created a well-defined object with $A_\infty$. Is there a way to put this set on firm ground, or is it just complete nonsense?
I’m also curious about other solutions to this functional equation. I don’t think there are any other “normal” solutions besides $\emptyset$. We’d need the really weird property that $f\subseteq f\times f$, and I can’t think of any other sets that work. What other sets, if any, work?
Best Answer
Suppose $f \subseteq f \times f$. By the definition of $\subseteq$,
$$\forall x \in f \quad x \in f \times f$$
Thus
$$\forall x \in f \quad \exists a, b \in f \quad x = \langle a, b \rangle$$
The rank of a Kuratowski ordered pair is strictly greater than the rank of its elements. Thus
$$\forall x \in f \quad \exists a, b \in f \quad \operatorname{rank} x > \operatorname{rank} a$$
That is,
$$\forall x \in f \quad \exists a \in f \quad \operatorname{rank} a < \operatorname{rank} x$$
Thus, if there exists an element of $f$, there exists an infinite sequence of elements of $f$ with strictly decreasing rank. But ranks are well-ordered. Thus $f$ is empty.