I've never seen an ultrafinitist definition of the natural numbers.
The usual definition via Peano's axioms obviously rules out there being
a largest number. I would suppose that an ultrafinitist definition of
the natural numbers would be something like seen in a computer (which is
necessarily finite). The successor operation would be defined such that
Successor (Biggest) = 0 or -Biggest.
Brent
Quentin Anciaux wrote:
> If you are ultrafinitist then by definition the set N does not
> exist... (nor any infinite set countably or not).
>
> If you pose the assumption of a biggest number for N, you come to a
> contradiction because you use the successor operation which cannot
> admit a biggest number.(because N is well ordered any successor is
> strictly bigger and the successor operation is always valid *by
> definition of the operation*)
>
> So either the set N does not exists in which case it makes no sense to
> talk about the biggest number in N, or the set N does indeed exists
> and it makes no sense to talk about the biggest number in N (while it
> makes sense to talk about a number which is strictly bigger than any
> natural number).
>
> To come back to the proof by contradiction you gave, the assumption
> (2) which is that BIGGEST+1 is in N, is completely defined by the mere
> existence of BIGGEST. If BIGGEST exists and well defined it entails
> that BIGGEST+1 is not in N (but this invalidate the successor
> operation and hence the mere existence of N). If BIGGEST in contrary
> does not exist (as such, means it is not the biggest) then BIGGEST+1
> is in N by definition of N.
>
> Regards,
> Quentin
>
> 2009/6/4 Torgny Tholerus <tor...@dsv.su.se>:
>
>> Brian Tenneson skrev:
>>
>>>> How do you know that there is no biggest number? Have you examined all
>>>> the natural numbers? How do you prove that there is no biggest number?
>>>>
>>>>
>>>>
>>>>
>>> In my opinion those are excellent questions. I will attempt to answer
>>> them. The intended audience of my answer is everyone, so please forgive
>>> me if I say something you already know.
>>>
>>> Firstly, no one has or can examine all the natural numbers. By that I
>>> mean no human. Maybe there is an omniscient machine (or a "maximally
>>> knowledgeable" in some paraconsistent way) who can examine all numbers
>>> but that is definitely putting the cart before the horse.
>>>
>>> Secondly, the question boils down to a difference in philosophy:
>>> mathematicians would say that it is not necessary to examine all natural
>>> numbers. The mathematician would argue that it suffices to examine all
>>> essential properties of natural numbers, rather than all natural numbers.
>>>
>>> There are a variety of equivalent ways to define a natural number but
>>> the essential features of natural numbers are that
>>> (a) there is an ordering on the set of natural numbers, a well
>>> ordering. To say a set is well ordered entails that every =nonempty=
>>> subset of it has a least element.
>>> (b) the set of natural numbers has a least element (note that it is
>>> customary to either say 0 is this least element or say 1 is this least
>>> element--in some sense it does not matter what the starting point is)
>>> (c) every natural number has a natural number successor. By successor
>>> of a natural number, I mean anything for which the well ordering always
>>> places the successor as larger than the predecessor.
>>>
>>> Then the set of natural numbers, N, is the set containing the least
>>> element (0 or 1) and every successor of the least element, and only
>>> successors of the least element.
>>>
>>> There is nothing wrong with a proof by contradiction but I think a
>>> "forward" proof might just be more convincing.
>>>
>>> Consider the following statement:
>>> Whenever S is a subset of N, S has a largest element if, and only if,
>>> the complement of S has a least element.
>>>
>>> By complement of S, I mean the set of all elements of N that are not
>>> elements of S.
>>>
>>> Before I give a longer argument, would you agree that statement is
>>> true? One can actually be arbitrarily explicit: M is the largest
>>> element of S if, and only if, the successor of M is the least element of
>>> the compliment of S.
>>>
>>>
>> I do not agree that statement is true. Because if you call the Biggest
>> natural number B, then you can describe N as = {1, 2, 3, ..., B}. If
>> you take the complement of N you will get the empty set. This set have
>> no least element, but still N has a biggest element.
>>
>> In your statement you are presupposing that N has no biggest element,
>> and from that axiom you can trivially deduce that there is no biggest
>> element.
>>
>> --
>> Torgny Tholerus
>>
>>
>
>
>
>
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