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Topic: Matheology § 201
Replies: 32   Last Post: Jan 28, 2013 2:26 PM

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 Jesse F. Hughes Posts: 9,776 Registered: 12/6/04
Re: Matheology S 201
Posted: Jan 27, 2013 11:56 AM

William Hughes <wpihughes@gmail.com> writes:

> On Jan 27, 2:33 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>> On 27 Jan., 14:12, William Hughes <wpihug...@gmail.com> wrote:
>>
>>
>>
>>
>>
>>
>>
>>
>>

>> > On Jan 27, 10:40 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>> >  <snip>

>>
>> > <... |N contains more than all (finite initial segments)
>>
>> > Piffle. |N does not contain more that all FISONS
>> > nor has anyone claimed this.

>>
>> > the correct statement is
>>
>> >   For every initial seqment f, there is an element of |N, n(f)
>> >   such that n(f) is not in f.  (note that n(f) may change if you
>> > change
>> >   f)

>>
>> > However,
>>
>> >   for ever n(f) there is an initial segment g
>> >   such that g contains n(f)

>>
>> Of course. But these all are relative definitions. As  I said, it is
>> impossible to define infinity absolutely. Same is valid for Cantor's
>> diagonal: For every line n, there is an initial segment of the
>> diagonal that differs from the first n lines. But that does not imply
>> that there is a diagonal that differs from all lines.

>
> It does imply that there is a diagonal that differs
> from each line. We start by noting that if a list L has a finite
> definition, so does the antidiagonal of L, call it l. By induction we
> show that l differs from each line of L.

Frankly, it seems to me that induction is utterly unnecessary.

Once we define the anti-diagonal, it is a triviality to show that it
differs from each line of L.

A minor point, perhaps.

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