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

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?

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

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

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

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?

• 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. – lytoooo0 Jul 24 '19 at 2:05
• @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 $T\subseteq\mathcal{P}(T)$). – JunderscoreH Jul 24 '19 at 3:36
• @lytoooo0: Take any set $x$ other than the empty set. What is $\mathcal P(\{x\})$ and is $x$ an element of it? – Asaf Karagila Jul 24 '19 at 6:33
• @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. – Clive Newstead Jul 24 '19 at 12:26