# What are all the finite transitive sets?

In set theory, we come across the definition of an ordinal as a well-ordered transitive set. And then we show that $\omega$ and all its elements are ordinals.

My question whether there are finite transitive sets which are not ordinals. Also, whether we know all the finite transitive sets.

• An answer mentions the hereditarily finite sets. It may perhaps be useful to add that these are precisely the elements of $V_\omega=\bigcup_{n\in\mathbb N}V_n$ where $V_0=\emptyset$ and $V_{n+1}=\mathcal P(V_n)$ for all $n$. – Andrés E. Caicedo Oct 11 '15 at 12:27
• There has been some recent attention given to (combinatorial and graph theoretic properties of) these sets. Note, for instance, that $V_\omega=\bigcup_n A_n$, where $A_0=\{\emptyset\}$, and $A_{n+1}=\{\emptyset\}\cup\{x\cup\{y\}\mid x,y\in A_n\}$. If you let $a_n=|A_n|$, the sequence $a_0,a_1,a_2,\dots$ is $1,2,4,12,112,11680,\dots$ You can find some work on it and related references in Enumeration of the adjunctive hierarchy of hereditarily finite sets, by Audrito, Tomescu, and Wagner, J Logic Computation (2015) 25 (3): 943-963. – Andrés E. Caicedo Oct 11 '15 at 12:40

Yes, there are finite transitive sets which are not ordinals. For example, take $X = \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\}$.
You can check quickly that $X$ is transitive, but it is not an ordinal, since it is not linearly ordered under $\in$. We have $\emptyset\not\in \{\{\emptyset\}\}$ and $\{\{\emptyset\}\}\not\in\emptyset$. Of course, the only ordinal of size $3$ is $3 = \{0,1,2\} = \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}$.
As for your second question, every set has a transitive closure, and the transitive closure of $X$ is finite if and only if $X$ is hereditarily finite. So the finite transitive sets are exactly the transitive closures of hereditarily finite sets. I don't think you're going to find a more concrete characterization than that, unfortunately.