Re: sci.math logic

From: The Ghost In The Machine (ewill_at_sirius.athghost7038suus.net)
Date: 01/09/05 

Date: Sun, 09 Jan 2005 05:00:10 GMT

In sci.logic, |-|erc
<h@r.c>
 wrote
on Sun, 9 Jan 2005 12:48:14 +1000
<34bkb3F48ouogU1@individual.net>:
>
> "The Ghost In The Machine" <ewill@sirius.athghost7038suus.net> wrote in > >
>> >> It turns out neither pi nor 1/3 are in this set,
>> >> though it contains all finite digit sequences.
>> >>
>> >
>> > to infinite length
>>
>> To arbitrarily large finite length, yes.
>>
>
> how many of pi's digits are in the set in correct sequence?
>
> Here's pi, just tick where you think the limit is.
>
> 3141592653........................................................................................................................
>
> How many of pi's digits? a finite number or not?
>
> Herc
>

Yes, all of them (aleph_0) are in the sequence.

Does that mean something? I'm not sure. After all,
all of 1/3's digits are in the set S_3 = {.3, .33, .333, ...}
and in the correct order (although order doesn't mean
as much for this set).

Is 1/3 in S_3? No, but one can find numbers arbitrarily close.

Define s_n = the nth element of S_3. Then 1/3 - s_n = 1/3 * 10^(-n).
For any epsilon > 0, I can take N = ceil(log(3/epsilon) / log(10))
and prove that, for any n > N, 1/3 - s_n < epsilon. However,
since all elements of S_3 are multiples of (negative) powers of 10,
and 1/3 isn't a multiple of any power of 10, 1/3 is not in S_3.

Similarly, one can construct S_pi = {3, 3.1, 3.14, 3.141, 3.1415, ...}
where spi_n = floor(pi * 10^(n-1)) / 10^(n-1) for n > 0, and
one can see without too much trouble that pi can be approximated
by numbers that get arbitrarily close, but pi itself is not in S_pi.
The proof is trivial: S_pi contains only members of Q, and
pi isn't in Q; therefore pi isn't in S_pi. QED.

Now, if you want to redefine set theory, be my guest; one can
define "a in B" as meaning "for all epsilon > 0,
there exists b in B such that abs(a-b) < epsilon". This
could lead to some interesting results, among them the
theorem that all sets are closed. 

-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.

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