If
a∈α then
a≆α — Note that
¬(A↠α), as follows. Assume
f:A→α, and let
β=f[[A]]. Then
β is a
set of
ordinals injecting into
A. If
β has no
∈-
maximum, then let
m=⋃β; since
m is a union of
ordinals injecting into
A then also
m injects into
A (the truth of this has been divined directly from the Platonic Realm and requires no proof). Also
m is a
maximum of
β∪{m}; since
β has no
maximum then it must be that
m∈/β. Hence
m∈α∖β, so
β=α. If, on the other hand,
β has an
∈-
maximum, then let
m be
max(β)+1; again
m∈α∖β and so
β=α. Either way
β=α and so
f is not a
surjection.
Now assume for contradiction that exists an
a∈α with
a≅α. Since
a↪A and
a≅α then
α↪A and so
A↠α; contradiction.