>>13784644>explain yourselvesI'm not a set theorists (I mean who is, except for some grumpy spergs on Math StackExchange), but note that you can understand this as a result of incompleteness for a set theory T (containing at least Robinson arithmetic).
Let be the proper class of all sets in your theory T. This class holds contains all sets that jointly fulfill the axioms of your set theory. E.g. for T=ZFC, there exists an empty set in and there exists an infinite set N in , and for all sets x and y in there exits there exists the pair {X,Y} in V. And there exists choice functions in ,
etc. etc.
Now imagine there were a set in such that its elements would jointly fulfill the axioms of T. This would mean that the empty set is in and the naturals N are in , and that all pairs are in , etc.
If that set existed, then the theory T would prove that its own axioms are consistent, as they are modeled by the set object .
So what you can do is assume the existence of the set, forming a new set theory
T' = T + exists.
This new theory T', proves that T is consistent.
We can see how this works with the "large cardinal N".
In ZF, define an injective function f on the naturals N recursively as
This has e.g.
The range of this function is in V and is exactly the hereditarily finite sets.
Set set (countable by f) itself fulfills all axioms of ZF except it doesn't contain an infinite sets.
So T'=ZF proves the theory T="ZF - N exists" consistent.
There's always room to postulate new set.
