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.


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

You received this message because you are subscribed to the Google Groups 
"Everything List" group.
To post to this group, send email to everything-list@googlegroups.com
To unsubscribe from this group, send email to 
For more options, visit this group at 

Reply via email to