Re: 1-1/2+1/3-1/4+1/5-1/6+1/7

On Tue, 05 Feb 2008 16:38:52 +0100, Han de Bruijn
<Han.deBruijn@xxxxxxxxxxxxxx> wrote:


"So N is the value of a function S, namely S(0)."

Ehrr??? You mean? :-o

What function do you have in mind here? We may define infinitely many
functions S such that S(0) = N. Says nothing about the values of S at
argument values other than N.

Suppose there is the additional information that S can _only_ consist of
composite successor functions s? Isn't that possible / plausible ?

No, it's not possible/plausible:

"As to Han's query whether we can express omega in terms of 0 and the
successor function the answer is no. In order to see this we need but
note that omega is not the successor of anything [...]." (A. K.)

With other words, we can show/prove (in ZFC) that for no set A

N = s(A).

Proof 1 (using the axiom of regularity):

Assume that for some set A: N = s(A). Hence (by definition of s) N = A u
{A}. Hence A e N. With other words, A is a natural number. But then (by
the properties of N) s(A) e N too. Hence N e N. This contradicts
regularity. Hence for no set A: N = s(A). []

Proof 2 (using the notions finite/infinite instead):

Assume that for some set A: N = s(A). Hence (by definition of s) N = A u
{A}. Hence A e N. With other words, A is a natural number. Now (by the
properties of N) any set in N is finite, hence A is finite, and s(A) is
finite too. On the other hand, N is infinite, contradiction! Hence for
no set A: N = s(A). []

Proof 3 (using neither a reference to regularity nor to the notions
finite/infinite)

Lemma: N !e N

Comment: It's easy to prove (without any reference to regularity
or infinity) from the definition of N, that the following holds:

"No natural number is a subset of any of its elements." (Halmos)

Since for any set S we have S c S, this means that for no
natural number n: n e n. And hence (easy) N !e N.

Assume that for some set A: N = s(A). Hence (by definition of s) N = A u
{A}. Hence A e N. With other words, A is a natural number. But then (by
the properties of N) s(A) e N too. Hence N e N. But N !e N. Contra-
dition! Hence for no set A: N = s(A). []

Finally: Since in ZFC everything is a set, this means that there can't
be a function S in ZFC "only consisting of composite successor functions
s" such that

N = S(0).

qed.


F.

--

E-mail: info<at>simple-line<dot>de
.

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