I see, he has limit ordinals in his embedded order with the sequence 1 - (1/2)^n, with 1 being the equivalent of omega. You probably even could get an embedded

I see, he has limit ordinals in his embedded order with the sequence 1 - (1/2)^n, with 1 being the equivalent of omega. You probably even could get an embedded omega squared with 1 - (1/2)^n - (1/3)^m with m>n So 1/2 = omega 3/4 = 2 * omega 7/8 = 3 * omega and 1 is omega * omega He might have omega squared in his version, but I didn't read it that carefully. Now play the version where the secret ordinal is an uncountable ordinal or a strongly inaccessible cardinal.