Re: Why? [was Re: Cantor`s powerset theorem is false?]


apoorv wrote:
Let us agree, as you say ,that the axiom of infinity is equivalent to
1A i.e the set N exists.

This is just plain not true.
What the axiom of infinity does require is that there exist
an inductive set; it then follows from Separation that there exists
a least inductive set. We usually take this set to be N.
But it doesn't have to be. It could be bigger. You cannot
define N at all in first-order logic, not with a recursive axiom-set
anyway.

So we have the null set, its successors and
the set N. We want to deduce the existence of a set containing N in a
situation where we do not have powerset and regularity.

No, you DON'T want to do that. You have ALREADY done that.
The axiom of infinity ALREADY says and does that.
The axiom of infinity asserts the existence of a set
containing N (as a subset, not as a member; if you want a set
containing N as a member then just use pairing on N and get {N} ).
To ensure the existence of at least one set containing all the naturals
is the whole purpose of the axiom of infinity. It doesn't say that
this
set IS N, but by taking the intersection of all the sets that "are
infinite"
in this sense, you wind up with N (or your model's approximation of N,
anyway).

Now N does not
belong to N if and only if S(N)=NU{N} is different from N . To assert
that omega does not belong to omega is equivalent to asserting the
existence of the successor of omega---that is to add that as an
additional axiom to our system.

Oh, bull***. That the successor of EACH AND EVERY
set exists IS A THEOREM. It is NOT an axiom. EVERY set
has a successor. You don't need an new axiom to get the
successor of omega. Once omega exists at all, its successor
AUTOMATICALLY exists. That's true for EVERY set.

.

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