Show that $x = \{\{x\}\}$ is not a set using the axioms of $ZFC$

Question

I am currently studying a course on Set Theory with the axioms of $ZFC$ and I understand that the axiom of foundation/regularity prevents self membership, i.e. $x \in x$ is not allowed.

My question is:

How do I show that $x = \{\{x\}\}$ is not a valid definition for a set?

I understand that this is a circular definition and that $\{x\}$ has greater rank than $x$ and so $\{x\} \notin x$. Is it therefore possible to show that $\{x\} \notin x$ directly from the $ZFC$ axioms?

Answer 1

We use the following two axioms to show the desired result:

The Axiom of Foundation tells us that every non-empty set contains elements that are disjoint from itself. This then implies $x \in x$ is not true for any set $x$ (as you state in your question).

The Axiom of Pairing tells us that for any two sets $x,y$ the set $\{ x,y \}$ is a valid set. By extension, the set $\{x \}$ exists for any valid set $x$ (since $\{x,x\}=\{x\}$).

We notice immediately that $x = \{ \{x \} \} = \{ \{ \{ \{ x \} \} \} \} = \cdot \cdot \cdot$ (infinite descending sequence of sets). This should make you question whether such a set can really exist. Therefore, you should be inclined to look for a contradiction of the axioms.

Proof by Contradiction

Begin by assuming that $x$ is a valid set. This means that it should obey all the axioms of $ZFC$ (including the two listed above).

By Pairing, $\{x \}$ exists - and by applying Pairing again, we know that $\color{red}{\{x ,\{x\}\}}$ exists.

Clearly, the above set contains two elements: $\color{blue}{x}, \color{green}{\{x\}}$ and so is not equal to the empty set. Therefore, Foundation tells us that this set must contain at least one disjoint element from this set. However, $$ \color{blue}{x} \cap \color{red}{\{x, \{x\}\}} = \color{blue}{\{ \{ x \} \}} \cap \color{red}{\{x, \{x\}\}} = \{x \}$$ $$\text{and}$$ $$\color{green}{\{ x \}} \cap \color{red}{\{x, \{ x \} \}} = x$$

Therefore, none of the elements of $\color{red}{\{x, \{x \} \}}$ are disjoint from itself (since $\color{green}{\{x\}} \neq \emptyset$ and $\color{blue}{ x} \neq \emptyset$). This violates Foundation so this cannot be a set. But by Pairing this must be a valid set if $x$ is a valid set. Therefore, $x$ cannot be a valid set as we have a contradiction. $\Box$

Answer 2

Suppose $x$ is a set and $\{x\}\in x$. By 2 x applications of the axiom of pairing, we know $y=\{x,\{x\}\}$ is a set. By the axiom of regularity, we know $\exists z((z\in y)\bigwedge((z\cap y)=\phi)$. Now $z$ can be either $x$ or $\{x\}$. If $z=\{x\}$ then $z\cap y=\{x\}$. If $z=x$ then $z\cap y=x\cap\{x,\{x\}\}$. And $\{x\}\in x$ and $\{x\}\in\{x,\{x\}\}$ so $z\cap y\neq\phi$. Hence a contradiction and $\{x\}\notin x.$