# RE: The seven step-Mathematical preliminaries

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> Date: Fri, 5 Jun 2009 08:33:47 +0200
> From: tor...@dsv.su.se
> Subject: Re: The seven step-Mathematical preliminaries
>
>
> Brian Tenneson skrev:
>>
>>
>> On Thu, Jun 4, 2009 at 8:27 AM, Torgny Tholerus <tor...@dsv.su.se
>> <mailto:tor...@dsv.su.se>> wrote:
>>
>>
>>     Brian Tenneson skrev:
>>    >
>>    >
>>    > Torgny Tholerus wrote:
>>    >> It is impossible to create a set where the successor of every
>>     element is
>>    >> inside the set, there must always be an element where the
>>     successor of
>>    >> that element is outside the set.
>>    >>
>>    > I disagree.  Can you prove this?
>>    > Once again, I think the debate ultimately is about whether or not to
>>    > adopt the axiom of infinity.
>>    > I think everyone can agree without that axiom, you cannot "build" or
>>    > "construct" an infinite set.
>>    > There's nothing right or wrong with adopting any axioms.  What
>>     results
>>    > is either interesting or not, relevant or not.
>>
>>     How do you handle the Russell paradox with the set of all sets
>>     that does
>>     not contain itself?  Does that set contain itself or not?
>>
>>
>> If we're talking about ZFC set theory, then the axiom of foundation
>> prohibits sets from being elements of themselves.
>> I think we agree that in ZFC, there is no set of all sets.
>
> But there is a set of all sets.  You can construct it by taking all
> sets, and from them doing a new set, the set of all sets.  But note,
> this set will not contain itself, because that set did not exist before.
>
>>
>>
>>
>>
>>     My answer is that that set does not contain itself, because no set can
>>     contain itself.  So the set of all sets that does not contain
>>     itself, is
>>     the same as the set of all sets.  And that set does not contain
>>     itself.
>>     This set is a set, but it does not contain itself.  It is exactly the
>>     same with the natural numbers, *BIGGEST+1 is a natural number, but it
>>     does not belong to the set of all natural numbers.  *The set of
>>     all sets
>>     is a set, but it does not belong to the set of all sets.
>>
>> How can BIGGEST+1 be a natural number but not belong to the set of all
>> natural numbers?
>
> One way to represent natural number as sets is:
>
> 0 = {}
> 1 = {0} = {{}}
> 2 = {0, 1} = 1 union {1} = {{}, {{}}}
> 3 = {0, 1, 2} = 2 union {2} = ...
> . . .
> n+1 = {0, 1, 2, ..., n} = n union {n}
> . . .
>
> Here you can then define that a is less then b if and only if a belongs
> to b.
>
> With this notation you get the set N of all natural numbers as {0, 1, 2,
> ...}.  But the remarkable thing is that N is exactly the same as
> BIGGEST+1.  BIGGEST+1 is a set with the same structure as all the other
> natural numbers, so it is then a natural number.  But BIGGEST+1 is not a
> member of N, the set of all natural numbers.
Here you're just contradicting yourself. If you say BIGGEST+1 "is then a
natural number", that just proves that the set N was not in fact the set "of
all natural numbers". The alternative would be to say BIGGEST+1 is *not* a
natural number, but then you need to provide a definition of "natural number"
that would explain why this is the case.
> The biggest advantage is that everything is finite, and you can then
> really know that the mathematical theory you get is consistent, it does

Even if you define "natural number" in such a way that there are only a finite
number of them (which you haven't actually done, you've just asserted it
without providing any specific definition), you still could have an infinite
number of *propositions* about them if you allow each proposition to contain an
unlimited number of AND and OR operators. For example, even if I say that the
only natural numbers are 1,2,3, I can still make arbitrarily long propositions
like ((3>1) AND (2>1)) OR (3>1)) AND ((2>3) OR (3>1)) AND ((2>3) OR ((1>3) OR
((2>1) OR ((1>3) OR (3>1))))). Of course a non-finitist would be able to prove
that these infinite number of propositions are consistent, but I don't know if
an ultrafinitist would (likewise a non-finitist can accept a proof that
something like the Peano axioms are consistent based on an understanding of
their application to a model dealing with rows of dots, even if the Peano
axioms cannot be used to formally prove their own consistency).
Jesse
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