Re: limitation to induction on finite bounds

From: |-|erc (gotcha_at_beauty.com)
Date: 07/04/04 

Date: Sun, 04 Jul 2004 03:17:27 GMT

"The Ghost In The Machine" <ewill@aurigae.athghost7038suus.net> wrote in
> In sci.logic, |-|erc
> <gotcha@beauty.com>
> > "The Ghost In The Machine" <ewill@aurigae.athghost7038suus.net> wrote
> >> >> >> >> The plot of the set of {0.3, 0.33, 0.333, ...} will cover 1/3, if
> >> >>one uses your
> >> >> >> >> procedure to draw all real numbers such that 03033 SetMinus r = 0.
> >>
> >> Depends on your definition of "cover". But OK, congratulations.
> >> You've now proved that all sets are closed. (This is because
> >> all sets cover their limit points, according to your definition-set.)
> >>
> >
> > fine, so your set 010203 covers all reals from 0 to 1.
> > cover meaning: Ar 0<r<1, T setminus r=0,
> > not necessarily a bijection.
>
> That it does. Of course it does not *include* all reals,
> merely covers them -- if the definition of "cover"
> includes SetMinus or its equivalent.

therefore, the infinite digit sequence of all irrationals occurs in the set 010203.

Herc

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