Understanding Specker's disproof of the axiom of choice in New Foundations

First notice that one can carry out in NF Zermelo's proof that the axiom of choice implies that all sets can be well-ordered. It follows that the relation on cardinals defined in your quote from Specker satisfies the conditions for a linear ordering. (It's trivially reflexive and transitive. Antisymmetry is the Cantor-Schröder-Bernstein theorem. Linearity follows from well-ordering because, for any two well-ordered sets, one is in (order-preserving) bijection with an initial segment (possibly improper) of the other.) It remains to show that, in any nonempty collection $X$ of cardinals, there is a smallest one. I'll use the fact that there is a set $U$ so big that every cardinal in $X$ occurs as the cardinality of a subset of $U$. In ZFC-style set theories, I'd get $U$ by choosing a representative set for each cardinality in $X$ and taking the union of these representatives; in NF it's easier, since the universe $V$ is a set that can serve as $U$. Well-order $U$. By what I already said, each cardinal $x\in X$ is the cardinality of either $U$ or a proper initial segment of $U$. If all are the cardinality of $U$, then $X$ contains just one cardinal, which is obviously least. So suppose some $x\in X$ is represented by an initial segment of $U$. (The same $x$ may correspond to several initial segments; that's OK.) Among all those initial segments, for all $x\in X$, there is a shortest because $U$ is well-ordered. The correponding $x$ is the smallest cardinal in $X$.

There is no assurance that $T(c) \leq c$ from $T$ being a "nonincreasing function". It is quite possible for $T(k) \gt k$ to hold for some cardinals.

It is possible to show that if $|\phi(c))|$ so is $|\phi(T(c))|$ and also that if $|\phi(c)|$ is finite and $T^{-1}(c)$ exists (as it would if $T(c) \gt c$) then so is $|\phi(T^{-1}(c))|$. So $T(c) \lt c$ is impossible because $c$ is the smallest cardinal of the kind indicated, and $T(c) \gt c$ is impossible NOT because "$T$ is a nonincreasing function" but because we would then have $T(c) \gt c \gt T^{-1}(c)$ and $T^{-1}(c)$ would be a cardinal of the given sort smaller than $c$.

THUS we have $c=T(c)$ and so $|\phi(c)| = |\phi(T(c))|$. But also we can show that $|\phi(T(c))| = T(|\phi(c)|)+$ (1 or 2) [NOT $|\phi(c)|+$ (1 or 2)]. Standard natural numbers are fixed by the $T$ operation, but general natural numbers do not have to be. We can show that $|\phi(c)| \equiv T(|\phi(c)|) \pmod 3$, and this is enough to get the contradiction.

Unfortunately, the argument just isn't intuitive.

``Work in NF, and assume the axiom of choice. Then we can prove that in general, the cardinality of a set A has greater or equal cardinality than the set of singletons drawn from A.''

I doubt this very much, except it be by some artifice such as the ex falso. The general view among NFistes is that the refutation of choice just is nasty, and it is not clear what is going on.

One way in is to consider the proof of the axiom of infinity in NF. That does make some sort of sense.

Merry Xmas, by the way