Transitive closure of an ordinal is the ordinal itself?

Question

Let $\lambda\in \mathbb{ON}$, then is it true that $\text{trcl}(\lambda)=\lambda$?

I think this is true, since an ordinal is already a transitive set, so it contains all of its subsets, and their subsets etc.

Thank you very much in advance!

Answer 1

The transitive closure is defined in several ways depending also on if you are working with the axiom of regularity or not.

One definition is \begin{equation} \text{trcl}(X)=\bigcap\{T: X\subseteq T\wedge T \text{ is transitive } \} \end{equation} That is the intersection of all transitive set containing $X$. One must verify that this is indeed a set or that there exists for every set $X$ a transitive set containing $X$ and that the intersection of transitive sets is transitive. Using this definition we have that if $M$ is transitive then the intersection of all transitive sets containing $M$ is exactly $M$.

For the other definitions of transitive closure namely:

\begin{equation} \text{trcl}(X)=\bigcup\{\bigcup^n X:n\in \omega \} \end{equation} where we recursively define $\bigcup^0 X= X$ and $\bigcup^{n+1}X=\bigcup (\bigcup^n X)$ \begin{equation} \text{trcl}(X)=\bigcup\{\text{trcl }(Y)\cup\{Y\}:Y\in X \} \end{equation} then one actually shows that these definitions are equivalent to the first. If you are taking any introductory set theory course they usually prove these equivalences.