Re: Definition of finite.

On May 3, 2:38 pm, G. Frege <nomail@invalid> wrote:

I guess, his point is the following: Just assume _for the sake of
the argument_ that there were infinite sets in w (here I'll use an
intuitive notion of infinite!) - since we don't have a definition
of finite/infinite so far, this might very well be the case, at the
present stage of the development of our formal theory (i.e. we
don't have a proof that this is not the case).

The question doesn't make sense UNLESS we have a definition of
'finite'. So suppose, we have a definition of 'finite' BEFORE having
proven the existence of w. Then, after we've proven the existence of
the set that we name 'w', we prove:
new <-> n is finite.

If we _define_ now
/finite/ (in our theory) the way we did, we could _prove_ (i.e. we
would get), that all sets in w are _finite_ (which _actually_ would
not be the case).

We'd be using 'finite' in a sense that is not equivalent to the
definitions we could give (and not Dedekind finite, either) prior to
proving the existence of the set we call 'w'. And we'd KNOW that.

So proving that all n in w are finite by "assuming" that they are
finite is "circular" (from this point of view).

And that is a confused point of view, which is cleared up by
considering what definitions are.

Actually, *I* personally would consider this as asome sort of
argument for NOT using the mentioned definition as our "basic"
definition for "finite" in set theory (say ZFC) - despite its
simplicity (and hence its attractiveness).

So one might claim (if you are to follow the argumentation from
above) that our "basic" definition of /finite sets/ should be via
_Dedekind's_ definition of /infinite sets/.

No, that doesn't follow even under the terms of the argument you
mentioned, because we can define an equivalent of the 'member of w'
definition even before proving the existence of the set we call 'w'.
Specifically, even before introducing the axiom of infinity and thus
even before proving the existence of a least successor-inductive set,
we can define 'finite' so that, after we do prove the existence of a
least successor-inductive set and name it 'w', our previous definition
will be equivalent to 'is a member of w'.

Of course, then we can show (at least in ZF_C_) that these two
definitions of "finite" are equivalent. So we might claim that the
natural numbers are _really_ finite. (And this might be considered
as a justification for using of the definition mentioned above.)

But we can also show in Z that 'member of w' is equivalent to certain
definitions of 'finite' that are prior to definiing 'w'.

MoeBlee

.

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