# A model of extensionality

Suppose I have the set

$$X = \{\varnothing ,\{ \varnothing\} ,\{\{\varnothing\}\} , \{\{\{\varnothing\}\}\} , ..., x_n , \{x_n\}, ... \}$$

My book asks which finite and which infinite subsets are models of extensionality? If $$Y \subseteq X$$ then $$Y \vDash$$ extensionality iff $$Y \cap ((x\setminus y) \cup (y\setminus x)) \ne \varnothing$$ for all distinct $$x, y \in Y$$. It seems to me that all non-empty subsets are models of extensionality, whether they are finite or infinite (except that finite sets must have at least two elements). If $$x=x_m$$ and $$y=x_n$$ (the $$m$$ and $$n$$ indicate the number of braces around the empty set), then $$x_m \setminus x_n = x_{m-n-1} \ne \varnothing$$ for $$m > n$$; and $$x_m \setminus x_n = x_m \ne \varnothing$$ for $$m < n$$.

Note that extensionality holds when $$(x\setminus y) \cup (y\setminus x)$$ is not empty for $$Y$$ for $$x\ne y$$, that means that if $$y\ne x$$, then there exists some $$z\in Y(!)$$ such that $$z\in x\setminus y$$ or $$z\in y\setminus x$$.

Take $$Y=\{\emptyset,\{\{\emptyset\}\}\}$$ for example, in this case both $$\emptyset$$ and $$\{\{\emptyset\}\}$$ contains no elements in $$Y$$, so both are empty sets for $$Y$$, and so if $$x=\emptyset,y=\{\{\emptyset\}\}$$, we have $$\left((x\setminus y) \cup (y\setminus x)\right)^Y$$ is empty so extensionality does not hold.

• And I see that this argument would fail if I had chosen the set of von Neumann natural numbers $X = \{ 0, 1, 2, ... \}$ with $Y=\{ 0, 2 \}$ since $0 \in 2$. Is this why the original $X$ (above) was abandoned as a model for the natural numbers in favor of von Neumann? – Robert Singleton Apr 13 '19 at 13:28
• @RobertSingleton one of the reasons, in general transitive sets are extremely nice(like $\in$ well ordering the set) – ℋolo Apr 13 '19 at 13:56
• I'm beginning to appreciate transitive sets. In general, it seems that if a subset of $X$ is not transitive, then it will not model extensionality. I'm now trying to prove that the only subsets of $X$ that model extensionality are transitive. – Robert Singleton Apr 13 '19 at 14:06

• The only finite subsets of $$X$$ that model extensionality are the sequential subsets $$Y=\{x_0, x_1, ..., x_n \}$$ where $$x_0=\varnothing$$ and $$x_{m+1}=\{x_m\}$$. If an element $$x_m$$ for $$m < n$$ is missing from $$Y$$, then $$x_m \notin Y$$ but $$x_{m+1}\in Y$$, in which case $$Y \cap (x_{m+1} \setminus x_0) = Y \cap x_{m+1} = \varnothing$$ (and $$x_0 \setminus x_{m+1}$$ is itself empty).
• The only infinite subset of $$X$$ that models extensionality is $$X$$ itself. Again, this is because the elements must be sequential as in the finite argument.
• This answer assumes $x_0=\varnothing\in Y$. Any set of the form $\{x_k,x_{k+1},\dots, x_n\}$ also satisfies extensionality. – Alex Kruckman Apr 13 '19 at 18:17
• @AlexKruckman By the same token, any infinite set of the form $\{x_k, x_{k+1}, .... \}$ satisfies extensionality? – Robert Singleton Apr 14 '19 at 14:02