A set S such that $S \subset S^{S}$

Question

I'm looking for a (all) set(s) S such that $S \subset S^{S}$ where $S^{S}$ is the set of all applications from $S$ to $S$. An obvious solution is $\emptyset$ : no element can be found neither in $\emptyset$, nor in $\emptyset^{\emptyset}$, so the are equal.

But I realized that we can define, for $x \in \mathbb{R}$, the function $ \tilde{x}_{1} : \mathbb{R} \to \mathbb{R}$, such that $\forall t \in \mathbb{R}, \tilde{x} (t) = xt$. Then, we can consider $\tilde{\mathbb{R}}_{1} = \{\tilde{x}_{1}, x \in \mathbb{R} \}$.
Then, the function $ \tilde{x}_{2} : \mathbb{R}_{1} \to \mathbb{R}_{1}$ such that $\forall \tilde{t} \in \tilde{\mathbb{R}}_{1}, \tilde{x}_{2} (\tilde{t}_{1}) = \tilde{(xt)}_{1}$ and the set $\tilde{\mathbb{R}}_{2} = \{\tilde{x}_{2}, x \in \mathbb{R} \}$.
By induction, we can now create, for $n \in \mathbb{N}^*$, the set $\tilde{\mathbb{R}}_{n} = \{ \tilde{x}_{n}, x \in \mathbb{R} \}$ We can easily see that, for $n \in \mathbb{N}^*$, we have $\tilde{\mathbb{R}}_{n+1} \subset \tilde{\mathbb{R}}_{n}^{\tilde{\mathbb{R}}_{n}}$ (and $\tilde{\mathbb{R}}_{0} = \mathbb{R}$).

My question is, (well, my questions are) :
- can we talk of $\tilde{\mathbb{R}}_{\infty}$ (I don't really understand what would be this kind of set's family's limit);
- if we can, do we have $\tilde{\mathbb{R}}_{\infty} \subset \tilde{\mathbb{R}}_{\infty}^{\tilde{\mathbb{R}}_{\infty}}$ (because $\infty = \infty +1$);
- I made this with multiplication in $\mathbb{R}$, can we do it in any group having a binary operator (like $(\mathbb{Z},+)$;
- if the answer is yes, can we have such sets not coming from groups.

P.S.: I found some interesting properties to these sets. If we define $\phi_{n \to m} : \tilde{\mathbb{R}}_{n} \to \mathbb{R}_{m}$ being the application transforming any $\tilde{x}_{n}$ into $\tilde{x}_{m}$, it is a bijective function. So $\tilde{(xy)}_{n} = \tilde{x}_{n} \circ \tilde{y}_{n} = \phi_{n-1 \to n} ( \tilde{x}_{n}(\tilde{y}_{n-1}))$. I guess it's nearly currying, because I turn a two-argument function (product) into a function mapping numbers on an (almost-)real function.

Answer 1

Suppose $S$ is a non-empty set with $S\subseteq S^S$. Say, $s\in S$. Then $s\in S^S$, so $s$ must be a map from $S\to S$, in particular $s$ must be a subset of $S\times S$ that contains an ordered pair $\langle s,y\rangle$. With the usual Kuratowski definition for odered pairs, $\langle s,y\rangle = \{\{s\},\{s,y\}\}$, we find that $$ s\in\{s\}\in\{\{s\},\{s,y\}\}\in s,$$ contradicting the Axiom of Regularity (aka. Foundation).

We conclude that $S$ with $S\subseteq S^S$ must be empty, and indeed $\emptyset\subseteq \emptyset^\emptyset$ because $\emptyset $ is a subset of every set. Note however that $\emptyset^\emptyset=\{\emptyset\}\ne\emptyset$ (just like $0^0=1\ne 0$).

Answer 2

The only example is $S=\emptyset$, at least in the standard system of set theory where we accept the axiom of foundation. This axiom is equivalent to the statement that we can assign to each set a rank (an ordinal) with the property that whenever $a\in b$ then the rank of $a$ is strictly smaller than that of $b$. There is no infinite decreasing sequence of ordinals, so if there are sets with some property, we can always limit ourselves to sets of minimal rank with the property.

Now, suppose $S\ne\emptyset$, and let $a\in S$ have smallest possible rank. If $S\subset S^S$, then $a$ is itself a function from $S$ to $S$, which in the standard formalization of functions in set theory, is a subset of $S\times S$ (we identify functions with their graphs), so $a$ is itself a set whose elements are ordered pairs $(b,c)$ with $b,c\in S$. The usual formalization of ordered pairs identifies $(b,c)$ with the set $\{\{b\},\{b,c\}\}$, so $b\in\{b\}\in(b,c)\in a$ and the rank of any such $b$ is strictly smaller than that of $a$, contradicting that $a$ is minimal. There are of course other ways of formalizing ordered pairs, but any reasonable formalization has that the rank of $b,c$ is less than or equal to that of $(b,c)$, so the specific choice here does not matter.

Finally, the use of the axiom of foundation is unavoidable. For instance, it is consistent with the other axioms of set theory that there are "Quine atoms", that is, sets $\Omega$ such that $\Omega=\{\Omega\}$. It is a silly exercise to check that in that case $\Omega=\{(\Omega,\Omega)\}=\Omega^\Omega$.