By Thoralf Skolem

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**Example text**

Since 2Ko > »0, we have (2^)*° £ N0*°, but (2**°)^° = 2 K ° K ° = 2K°. On the other hand 2Ko i No**0. Hence 2 No _ IA NO - NO Of course we then have for arbitrary finite n ^> and not only that. Let namely N0 < w = 2 °. Then 2 — NO = in whence m*° = 2*°, In a similar way we obtain for an arbitrary I for all m > 1 and ^ 2 From our axioms, in particular the axiom of choice, we have derived that every cardinal is an aleph. Therefore 2^ a is an aleph. We can also prove by the axiom of choice that 2 a > $a+i or perhaps = Na+i .

Proof. Indeed, if (a,0,b) were eP, b =(= 0, only have the property 1), which is immediately (a,fty) be eP - {(a,0,b)} and (r,a,6) e S. Then (y,a,6) eS yields (a,/3',6) eP, whence (a,/3',6)e cannot coincide with (a,0,b). then P - {(a,0,b)} would not seen, but also 2). Let (a,fty) e P together with P - {(a,0,b)} because (o,j8f,6) Theorem 54. If (a,b',c) ep, then (Ez)((a,b,z)e P & (z,a,c) e S). Proof. Let us assume that we had (&,b\c) e P, while for all z either (a,b,z)e~P or (z,a,c)eS. Let us consider the set Pf = P - {(a,bf,c)}.

Every subset of an inductive finite set u is inductive finite. Proof. Let v be £u. I consider the set x of subsets w of u such that w n v is inductive finite. It is obvious that Oex, because the set 0 is inductive finite. Let y be ex and zeu. Then y n v is inductive finite and (y U {z}) 0 v is either y n v, namely when zev, or (y n v) + {z}, namely if zev. But by the preceding theorem also (y 0u) + {z} is inductive finite. Thus as often as yex, zeu, we have y u{z}ex. Since u is inductive finite, it follows that uex.