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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?

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

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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 $T\subset \mathcal{P}(T)$.

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    $\begingroup$ My edit was for a typo. In the 2nd part "Let us assume that $\bigcup T\in T$"... I changed $\in$ to $\subset$......+1 $\endgroup$ – DanielWainfleet Aug 4 '19 at 3:38
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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?

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  • $\begingroup$ The second line is understandable, but the first line, how can you say that? Is the $P(T)$ here means a power set? $T \subseteq P(T)$ is always true because of the definition of power set, seems something wrong in my words. $\endgroup$ – lytoooo0 Jul 24 '19 at 2:05
  • $\begingroup$ @lytoooo0 $T$ is always an element of $\mathcal{P}(T)$ (written $T\in\mathcal{P}(T)$) but rarely is $T$ a subset of $\mathcal{P}(T)$ (written $T\subseteq\mathcal{P}(T)$). $\endgroup$ – JunderscoreH Jul 24 '19 at 3:36
  • $\begingroup$ @lytoooo0: Take any set $x$ other than the empty set. What is $\mathcal P(\{x\})$ and is $x$ an element of it? $\endgroup$ – Asaf Karagila Jul 24 '19 at 6:33
  • $\begingroup$ @lytoooo0: $T \subseteq \mathcal{P}(T)$ means that if $A \in T$, then $A \in \mathcal{P}(T)$—that is, every element of $T$ is an element of $\mathcal{P}(T)$. But the elements of $\mathcal{P}(T)$ are subsets of $T$, so $T \subseteq \mathcal{P}(T)$ means that every element of $T$ is a subset of $T$. This is not true of all sets, as Asaf's comment shows. $\endgroup$ – Clive Newstead Jul 24 '19 at 12:26

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