First some definitions: $X$ is inductive if $\emptyset\in X$ and $x\in X\implies x\cup\{x\}\in X$
Let $N = \cap\{X : X \text{ is inductive }\}$. Let us use the following notation:
$0 = \emptyset$, $1 = \{0\}$, $2 = \{0 , 1\}$, ...
if $n\in N$ let $n + 1 = n \cup \{n\}$. Let us define $<$ (on $N$) by $n < m$ iff $n\in m$.
A set $T$ is transitive if $x\in T$ implies $x\subset T$.
Then I have :
If $X$ is inductive, then the set $\{ x\in X : x$ is transitive} is inductive. Hence every $n\in N$ is transitive.
Solution. Suppose $X$ is inductive, it follows that $\emptyset\in X$. The empty set is also vacuously transitive, hence $\emptyset\in \{ x\in X : x$ is transitive}. Now suppose that $x\in\{ x\in X : x$ is transitive} it follows that $x\in X$ and hence $x\cup\{x\} \in X$. Similarly it follows that $x$ is transitive. Now, let $a\in x\cup\{x\}$, then there are two cases either $a\in x$ or $a\in \{x\}$. In the first case it follows that since $a\in x$ and $x$ is transitive $a\subset x$, and hence $a\subset x\cup\{x\}$. In the second case since $a\in\{x\}$ and $\{x\}$ is a singleton set it follows that $a = x$, thus $a \subset x\cup\{x\}$. Thus $x\cup\{x\}$ is transitive, and hence $x\cup\{x\} \in\{ x\in X : x$ is transitive}. Therefore we can conclude that $\{ x\in X : x$ is transitive} is inductive. $\Box$
Now the problem, I really cannot see how the corollary "Hence every $n\in N$ is transitive." follows from this.
But I do note that I know $N$ is transitive, and for each $n\in N$, $n = \{m\in N : m < n\}$ by the previous question.
So I was wondering if someone could help shed light on how the corollary follows from the initial question, or point out the flaw in my proof thus suggesting a reason it doesn't currently follow.
Thank you very much.