How to show that for every set, $T$ there is a transitive set, $T'$ such that $T\subseteq T'$?
Thank you again!
How to show that for every set, $T$ there is a transitive set, $T'$ such that $T\subseteq T'$?
Thank you again!
First note that $T'=\varnothing$ is transitive and it is a subset of every other set.
The more likely situation is that you are asked to find a transitive $T'$ such that $T\subseteq T'$. Let $T_0=T$ and $T_{n+1}=\{x\mid\exists y\in T_n: x\in y\}$. Now take $T'=\bigcup_{n=0}^\infty T_n$, and show it is transitive.
This $T'$ is called the transitive closure of $T$. Generally speaking if $(X,R)$ is an ordered set we can talk about the transitive closure of $R$, and in the context of sets we simply take the transitive closure of $(T,\in)$.
The smallest such set $T'$ is the transitive closure of $T$, which is constructed recursively. Let $T_0=T$, and for $n\in\omega$ let $T_{n+1}=\bigcup T_n$. Then let $T'=\bigcup_{n\in\omega}T_n$; clearly $T\subseteq T'$, and it’s not hard to show that $T'$ is transitive: you’ve thrown in everything that’s needed to make it so.