# If $x$ is a transitive set, $\bigcup x^+ = x$, where $x^+ = x\cup\{x\}$

I can't understand a sentence in a textbook: if $x$ is a transitive set, then $\bigcup x^+=x$? Could someone help me to understand?

added: $x^+=x\cup\{x\}$

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It helps if you tell us what the textbook is. –  Paul Jan 11 '12 at 8:13
It is a chinese book. Maybe it is wrong. –  Paul Jan 11 '12 at 8:16

Let us deconstruct this:

1. $\bigcup A = \{z\mid\exists y\in A: z\in y\}$.
2. $x^+=x\cup\{x\}$

From this we have: $\bigcup x^+ = \{z\mid\exists y\in x\cup\{x\}: z\in y\}$. We then follow the definition of a transitive set:

The set $x$ is transitive if for every $y\in x$, $y\subseteq x$

Now we want to show that $x\subseteq\bigcup x^+$ and that $\bigcup x^+\subseteq x$, the axiom of extensionality will ensure equality.

For the first one, it is nearly trivial. Since $x\in x^+$ we have that $y\in x$ then $y\in\bigcup x^+$ immediately (recall the definition of this union).

For the second one, we recall that $y\in x^+$ then $y\in x$ or $y=x$. Therefore $y\subseteq x$. So for every $z\in\bigcup x^+$ we have some $y\subseteq x$ such that $z\in y$, therefore $z\in x$ as wanted.

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The set $x$ is transitive if $z\in y \in x$ implies $z\in x$. Moreover, $x^+=x\cup\{x\}$. So let $x$ be a transitive set.

We have $x\subseteq\bigcup x^+$ since $x\in\{x\}$. Now let $z\in \bigcup x^+$. Then $z\in\bigcup x\cup x$. If $z\in\bigcup x$, then there exists $y\in x$ with $z\in y$. But since $x$ is transitive, $y\subseteq x$ and hence $z\in x$.

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There was a typo I corrected, please take a look at the edited proof. –  Michael Greinecker Jan 11 '12 at 9:06