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This is an exercise my set theory teacher suggested we try and I feel like the answer is yes (because it seems to me that no set other than $\emptyset$ satisfies $x\times x=x$, then the collection would be exactly $\{\emptyset\}$, which is a set), but I am stuck trying to prove that if $x\neq\emptyset$ then $x\neq x\times x$. I have tried using the axiom of regularity and cardinality supposing $x=x\times x$ but I reach no contradiction... could you give me any clues? (the theory is ZFC)

JMP
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valkyriebel
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  • @epsilon-emperor explain to me what the set of all sets is then? Or ${x:x=x}$? – C Squared Jun 07 '21 at 16:34
  • how are you defining $x\times x$ as of now? – C Squared Jun 07 '21 at 16:36
  • we have defined $x\times x={<a,b>|a,b\in x}$, where $<a,b>={{a},{a,b}}$ is the ordered pair of $a$ and $b$ – valkyriebel Jun 07 '21 at 16:39
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    @CSquared There is no set of all sets! – David C. Ullrich Jun 07 '21 at 17:08
  • @DavidC.Ullrich haha yes I know. user epsilon-emperor commented that everything was a set, and I was challenging his comment. I believe what you call the set of all sets is a class, however... whatever that means – C Squared Jun 07 '21 at 17:11
  • dear @CSquared, in ZFC, the formal definition of a "class" is an equivalence class of the set of all first-order formulas in the language ${\in}$ under the relation of logical equivalence (modulo the axioms of ZFC.) see eg here for some discussion – Atticus Stonestrom Jun 07 '21 at 17:23

1 Answers1

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You are absolutely correct that it suffices to show that, for any $X$, if $X\neq\varnothing$ then $X\times X\neq X$. To see this, suppose for contradiction that $X$ is non-empty and equal to its Cartesian square. The idea here is that we want to construct an infinite $\in$-descending chain, which will contradict the axiom of regularity. How might we do this? Well, since $X$ is non-empty, there exists $x_0\in X$, and since $X=X\times X$ there exists $x_1,y_1,\in X$ such that $x_0=\langle x_1,y_1\rangle$. Now we have $x_1\in\{x_1\}\in x_0$, and since $x_1\in X=X\times X$ there exist $x_2,y_2\in X$ such that $x_1=\langle x_2,y_2\rangle$, whence $$x_2\in\{x_2\}\in x_1\in\{x_1\}\in x_0.$$ Hopefully it is intuitively clear that continuing in this way will give us the desired $\in$-chain. So, how do we capture the idea of this argument rigorously? Well, note that all of the elements in our chain are either elements of $X$ or of form $\{x\}$ for some $x\in X$. So, with this in mind, define $Y\subseteq\mathcal{P}(X)$ to be the set of all subsets of $X$ of size $1$. ($Y$ exists by restricted comprehension.) Now, let $Z=X\cup Y$. Since $X$ is non-empty, $Z$ is non-empty too, so by the axiom of regularity there exists $z\in Z$ such that $z\cap Z=\varnothing$. By definition of $Z$, we have two cases.

  1. If $z\in Y$, then by definition of $Y$ we have $z=\{x\}$ for some $x\in X$, and then $x\in z\cap X\subseteq z\cap Z$.
  2. If $z\in X$, then since $X=X\times X$ we have $z=\langle x,y\rangle$ for some $x,y\in X$. But now $\{x\}\in z\cap Y\subseteq z\cap Z$.

In either case, we see that $z\cap Z$ is non-empty, which gives the desired contradiction.

Atticus Stonestrom
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