# Recursive set $A = \lbrace 1,A \rbrace$

You have given $A := \{ 1,A \}$ (which presents a weird way of defining a set. Is this recursive?).

The question is $|A| = \ldots$ ? Is $|A|= 2$ because $A = \{ 1 , \{ 1 , \{ 1,\{ 1 , ... \} \} \} \}$? What can you say about $A$?

• In ZFC this set is not defined because of the axiom of regularity Sep 4, 2018 at 17:55
• Sep 4, 2018 at 17:55
• It doesn't compile so it is an undefined declaration. Recursion operates over the same function not over a declaration. Even if you declare a recursive function If It has no terminator that ends the definition It is undefined too. Sep 4, 2018 at 18:03

The easy answer is that assuming the axioms of $\sf ZF$, no set is a member of itself, and therefore the equation $A=\{1,A\}$ admits no solutions, so the cardinality of such $A$ is as relevant as the representation of $\sqrt2$ as a ratio of two integers.
The longer answer is that let's assume that $1$ is not $A$ itself, then if such $A$ exists, then $|A|=2$. The reason we need to say that is that $1$ is not inherently part of the set theoretic language, it is coded as some sort of set. And therefore it can a priori be any given set. If $1=A=\{1,A\}$, then $|A|=1$, since in that case $A=\{1,A\}=\{A\}=\{1\}=1$.
Finally, it is consistent that $A=\{1,A\}$ admits a solution. This requires the failure of the axiom of regularity (or foundation, as it is sometimes called). But since naively speaking the axiom of regularity is not discussed nor its negation, it is impossible to naively declare whether or not a solution exists.