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Let us say that a set $X$ is transitive if for all $x \in X$, if $x$ is a set then $x \subseteq X$.

(This is all from a naive set theory perspective; I'm assuming, for instance, that $\{1, 2, 3\}$ is a transitive set because none of its elements are sets.)

Now, let us define the transitive closure $T(X)$ of a set $X$ to be the intersection of all transitive sets of which $X$ is a subset. Roughly speaking, the transitive closure of $X$ is $X \cup$ the elements of the elements of $X \cup $ the elements of the elements of the elements of $X$, and so on.

If $T(X)$ is a finite set, then obviously $X$ is a finite set. Moreover, there's a sense in which $X$ is "definitely" a finite set, in a way that e.g. $\{\mathbb{N}\}$ is not, since even though $|\{\mathbb{N}\}| = 1$, we have $|T(\{\mathbb{N}\})| = |\{\mathbb{N}\} \cup \mathbb{N}| = \aleph_0$.

With that in mind, is there a standard name (analogous to "the cardinality of $X$") for the cardinality of $T(X)$?

Is there a standard term (analogous to "X is a finite set") for sets $X$ such that $T(X)$ is a finite set?

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    The transitive closure of $x$ is finite iff $x\in V_\omega$. – 2017-02-04
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    I should mention that the modern set theory defines the natural numbers as sets starting from the empty set and iterations: for example, $0=\varnothing$, $1=\{0\}$, $2=\{0,1\}$ and so on. Thus the transitive closure of $\{1,2,3\}$ is $\{0,1,2,3\}$. – 2017-02-04
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    Yes, I know; perhaps I'll edit my question to remove the comment about "naive set theory". – 2017-02-04
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    By the way, the term "transitive closure" is completely standard, so you don't need to put it in quotes. – 2017-02-04

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I didn't heard any name of the cardinality of $T(x)$. It is just usually called the cardinality of the transitive closure of $x$. However, if $T(x)$ is countable, $x$ is often called a hereditary countable set.

Now I am going to answer your second question:

Claim. $T(x)$ is finite if and only if $x\in V_\omega$, where $V_\omega$ is the $\omega$th von Neumann hierarchy.

Proof. Note that $V_n$ is transitive for all $n$. If $x\in V_\omega$, then there is some $n<\omega$ such that $x\in V_n$. Since $V_n$ is transitive and $x\in V_n$, $x\subseteq V_n$ and $T(x)\subseteq V_n$, by your definition of the transitive closure. Therefore $T(x)$ is finite.

The proof of the converse uses an induction: assume that if $|T(y)|

This is the reason why the set $V_\omega$ is called the set of hereditary finite sets.

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    Thank you, this is exactly what I was looking for. – 2017-02-04