Zermelo's/WellOrdering Theorem Proof question I've been looking for proofs of the wellordering theorem and found an article that provided a pretty concise proof, but there is a step that I simply cannot understand. You can find it in the image
http://postimg.org/image/temuivngx/full/
T' being comparable to every other element is apparently the contradiction, but I don't understand why. Any proper subset of T (that is an element of $\phi$) MUST be comparable to every element, otherwise it's included in the intersection definition of T and so T is a subset of it. Any help, please? It's driving me kind of crazy.
 A: We have $\phi=\overline\phi$, so for every $Z\in\phi$, either $Z\supseteq T$, or $T'\supseteq Z$. Since $T'\subseteq T$, it follows that every $Z\in\phi$ is comparable with both $T$ and $T'$. Now recall that $T$ is the minimal member of $\phi$ containing every element of $\phi$ that is incomparable with some element of $\phi$. Since $T'\subsetneqq T$, $T'$ does not contain every member of $\phi$ that is incomparable with some element of $\phi$. But the only member of $\phi$ that is a subset of $T$ but not of $T'$ is $T$ itself, so $T$ must be incomparable with some $Z\in\phi$: otherwise $T'$ would contain every member of $\phi$ that is incomparable with some member of $\phi$. On the other hand, we can clearly see from the definition that $T$ is comparable with every $Z\in\phi$, so we have a contradiction.
Added: Let $\mathscr{I}=\{I\in\phi:\exists J\in\phi(I\nsubseteq J\text{ and }J\nsubseteq I)\}$ ; $\mathscr{I}$ is the set of $I\in\phi$ that are incomparable with some element of $\phi$. Let $\mathscr{B}=\{Z\in\phi:\forall I\in\mathscr{I}(Z\supseteq I)\}$; $\mathscr{B}$ is the set of $Z\in\phi$ that contain all members of $\mathscr{I}$ as subsets. Then the set $T$ is defined by $T=\bigcap\mathscr{B}$. The statement in the proof that Then $T$ itself is also such a $Z$ is saying that $T\in\mathscr{B}$: for every $I\in\mathscr{I}$, $I\subseteq T$. And because $T=\bigcap\mathscr{B}$, $T$ is the minimum member of $\mathscr{B}$: if $Z\subsetneqq T$, then $Z\notin\mathscr{B}$. In particular, $T'\notin\mathscr{B}$, and there is therefore some $I_0\in\mathscr{I}$ such that $I_0\nsubseteq T'$. But $T\in\mathscr{B}$, so $I_0\subseteq T$. Remember that $\mathscr{I}\subseteq\phi$, so $I_0\in\phi=\overline\phi$, and the only subset of $T$ that is in $\overline\phi$ and is not a subset of $T'$ is $T$ itself. Thus, $I_0=T$ and therefore $T\in\mathscr{I}$, meaning that $T$ is incomparable with some element of $\phi$, which we know is false.
