Re: An easy argumentation to show a severe limitation of set theory

On 27 Mar 2007 16:24:06 -0700, Albrecht wrote:
Dave Seaman schrieb:

On 27 Mar 2007 12:55:02 -0700, Albrecht wrote:
An easy argumentation to show a severe limitation of set theory


Let's consider the infinite sequence S of sets with the general form
of the n-th element
S(n) = {{1}, {1, 2}, {1, 2, 3}, ..., {1, 2, 3, ..., n}}:

1 -> {{1}}
2 -> {{1}, {1, 2}}
3 -> {{1}, {1, 2}, {1, 2, 3}}
...

The sequence S consists of elements S(n) which are sets. The sets S(n)
consists of elements E(n, m) with m< n or m=n (n, m in |N), which are
sets too. The set E(n, m) is the m-th element oft the set S(n) which
is the n-th element of the sequence S.

All elements of this infinite sequence of sets are sets with finite
cardinality.

The infinite set of this kind of sets is the set M = {{1}, {1, 2}, {1,
2, 3}, ...}.

Question: Does the infinite set M contain the element |N (the set of
the natural numbers)?

You are the one describing the set. It is up to you to make clear which
elements are included and which are not.

I tried to make clear that the set M is the infinite set which
contains all sets which elements are natural numbers which build
successive sequences starting with number 1.

Are you being deliberately vague? You have not stated whether N is a
member of the set or not.

Then I pose the question if this set M must contain the set |N or not.

If you ask the question that way, then the answer is very easy. No, it
is not true that the set *must* contain N. And, it is equally untrue
that the set *cannot* contain N.

You have not described a set until you have told us exactly what its
members are, and you have not said whether N is a member or not. It's
that simple.

I don't ask, if I am free to include |N. I ask if |N must be in the
set M to be an infinite set.

How many different ways do you want that question answered? No, it is
not true that N must be in the set. And no, it is not true that N cannot
be in the set.

For each natural number n let A_n = { i in Z : 1 <= i <= n }. Let A =
{ A_n : n in N }. Then A is an infinite set, but N is not a member of A.

You say so.

Because I said what A is, and N is not a member of A. You have not said
what your set M is. That is the difference.

Let B = A U {N}. Then B is an infinite set whose members are all the
members of A, plus the single set N.

Argumentation A1:
If the set M contains the set |N as an element, the element |N has to
succeed a finite element of the set (the set M can be considered as a
sequence since the elements can be ordered with increasing
cardinality).

No. N is an element of B, but N is not the successor of any element of
B.

In your set B it must not be a successor. No problem.

But it is impossible to have a sequence of sets of successively
increasing cardinality starting by 1 in which an infinite set succeed
a finite set.

Irrelevant, because N need not be the successor of any set in the
sequence.

But with my proposition you can agree?

You have not stated a proposition. You asked a poorly formulated
question. I pointed out exactly why the question is ambiguous.


The sequence {1}, {1, 2}, {1, 2, 3}, ... doesn't contain |N as an
element.

Argumentation A1 leads to the conclusion, that M can't contain the
element |N.

The set A (above) does not contain N. This is not because it *can't*
contain N, but merely because (unlike B), the set A happens to have been
*defined* not to contain N.

No, it hadn't to be defined like this.

Wrong. I am the one who defined the set, and my definition (see above)
is A = { A_n : n in N }. Since there is no n such that A_n = N, it
follows that N is not a member of A.

Argumentation A2:
Any element of the sequence S of sets of the form {{1}, {1, 2}, {1, 2,
3}, ..., {1, 2, 3, ..., n}} is finite if n is finite. The value n is the
cardinality of the element E(n, n) with the greatest cardinality in
the set S(n). So the infinite set M = {{1}, {1, 2}, {1, 2, 3}, ...} is
only able to be infinite, if it contains a set as element with an
infinite cardinality like |N.

Not every set has a greatest element. Your statement does not even make
sense.

My argument isn't as you assert.

First define your set M, and then we'll discuss your argument. You have
not said whether N is a member of M or not, and therefore I do not accept
M as a set.


Argumentation A2 leads to the conclusion, that M have to contain |N as
an element to be infinite in cardinality.

No. Both A and B are infinite in cardinality. B contains N, and A does
not.

See above.

On that point we can agree. Define your M, and we'll talk.



Overall Conclusion:
Since argumentation A1 contradicts argumentation A2, in ZFC it is not
decidable if the set M contains the element |N or not. The infinite
set M has to be undefined in ZFC.


Nonsense. You simply forgot to specify which set you were talking about,
A or B. Both are infinite sets that meet your incomplete specifications.



I don't see any lack in my specification. Please see my answers to
other postings.

You have failed to specify whether N is a member of M. I will not comment
further until you have removed that ambiguity.



--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case to be heard May 17
U.S. Court of Appeals, Third Circuit
<http://mumia2000.org/>
.

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