Re: Cantor and the binary tree



David Kastrup wrote:

> > Look here: The natural number n e N is nothing else than an
> > abbreviation of its initial segment {1,2,3,...,n} c N.
> > N consists exclusively of elements n. Similarly N consists exclusively
> > of subsets = initial segments (all of which include 1). There is no
> > element of N which is not an element of such a subset. And there is not
> > a pair of different elements n and n' of N, which satisfy the following
> > condition:
> > n belongs to an initial segment S which does not contain n'
> > and
> > n' belongs to an initial segment S' which does not contain n
> > in short:
> > n e S and n' !e S and n' e S' and n !e S'.
> > As this requirement is impossible to satisfy, the segment of n includes
> > all elements less than n. This holds for any n e N. Therefore N is a
> > segment.
>
> Nope. Your "Therefore", again, is a piece of [censored] and a non
> sequitur. There is nothing whatsoever to the left of this "Therefore"
> that would justify the conclusion.

The conclusion is justified by the bijection of n on its initial
segment.
There is no largest n, so there is no largest segment. But any n is
finite, so any segment is finite. Outside of segments there are no
natural numbers.
*Therefore* the set of all n, called N, is a segment.

Regards, WM

PS: You will not prevent this obvoious fact by cursing and shouting. So
stop foaming and leave things as they are. But you will keep on
swearing like a trooper, I know. The only question is, why can't you
and Hughes behave like civilized persons?

.

Relevant Pages

  • Re: Cantor and the binary tree
    ... >>> Of course every set of naturals is bounded by a natural. ... >a pair of different elements n and n' of N, which satisfy the following ... >n belongs to an initial segment S which does not contain n' ... >n' belongs to an initial segment S' which does not contain n ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... >>> Of course every set of naturals is bounded by a natural. ... > a pair of different elements n and n' of N, which satisfy the following ... > n belongs to an initial segment S which does not contain n' ... > n' belongs to an initial segment S' which does not contain n ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... >>> Every countable set like N is potentially infinite but not actually ... > n belongs to an initial segment S which does not contain n' ... > n' belongs to an initial segment S' which does not contain n ... Not so (unless WM insists that N is a member of itself, ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... >> Of course every set of naturals is bounded by a natural. ... n belongs to an initial segment S which does not contain n' ... n' belongs to an initial segment S' which does not contain n ... This segment is potentially infinite, ...
    (sci.math)
  • Re: Cantor Confusion
    ... N is a linear set (every finite initial segment (= line) includes ... The potentially infinite sequence of natural numers is an initial ... Every natural number belongs to the sequence and to at least one ...
    (sci.math)