Hа a level of metatheory ZFC is inconsistent. The proof of it
The unexpected fact, leans on that standard in the theory
Models the assumption, that set of all formulas of any theory
The first order, is countable ZFC-set.
The proof.
Let's designate a symbol x$Y a predicate: x there is an element of
set Y.
Let's designate a symbol x#Y a predicate: (1) x there is an element
countable sets Y which definable by means of some ZFC-formula F (x)
(2) Existence of set Y certain by means of formula F (x), together
with axioms ZFC.(3) (x$Y) it is demonstrable in ZFC.
Let's designate a symbol x~#Y a predicate: (1) x there is no element
countable sets Y which definable by means of some ZFC-formula F (x)
(2) Existence of set Y certain by means of formula F (x), together
with axioms ZFC. (3)(x$Y) it is demonstrable in ZFC.
Let's assume, that the set of all ZFC-formulas can be considered as
usual ZFC-set. Then by virtue of axioms of substitution in ZFC
existence is demonstrable set W which elements will be all ZFC-
definable sets. Then by virtue of an axiom of allocation in ZFC
existence so-called wild Rassel's set RW which is certain as follows
will be deduced: x$RW<->x~#x. For set RW by obvious image it is
received, that RW#RW<->RW~#RW.
Thus for RW in ZFC at a level of a metatheory it will be
demonstrable, that x$Y it is demonstrable in ZFC and it is
simultaneously indemonstrable in ZFC.
http://planetmath.org/?op=getobj&from=papers&id=329
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