How to Prove that Two Defintions of `Transitive' are Equivalent

Question

The definition of Transitive: A set $T$ is transitive if every element of $T$ is a subset of $T$. (Equivalently, $\cup T \subset T$, or $T \subset P(T)$). In Set Theory by THOMAS JECH. Definition 2.9 pp.19

How to proof that the definitions are equivalent?

$\cup T \subset T \iff T \subset P(T)$

Answer 1

To expand on the answer already given, here's a wordy, but hopefully complete, proof.

First we prove the "if" direction. So we assume that $T\subset \mathcal{P}(T)$. Let $X\in\bigcup T$ be given; we need to show that $X\in T$. Well by the definition of union, we know there exists a set $A\in T$ such that $X\in A$. But by assumption, $A\in T$ implies that $A\in \mathcal{P}(T)$, so $A\subset T$. Hence $X$, which is an element of $A$, is also an an element of $T$.

Now for the "only if" direction. Let us assume that $\bigcup T\subset T$. Let $X\in T$ be given and let $x$ be some element of $X$. Since $x\in \bigcup T$, by assumption we have $x\in T$. This means that $X\subset T$, i.e. $X\in \mathcal{P}(T)$. So $T\subset \mathcal{P}(T)$.

Answer 2

To say that $T \subseteq \mathcal{P}(T)$ it to say that every element $A \in T$ is a subset of $T$.

To say that $\cup T \subseteq T$ is to say that every element $a \in A$ of an element $A \in T$ is itself an element of $T$.

Can you see why these are equivalent?