Re: limitation to induction on finite bounds

From: The Ghost In The Machine (ewill_at_aurigae.athghost7038suus.net)
Date: 06/25/04 

Date: Fri, 25 Jun 2004 08:00:16 GMT

In sci.logic, |-|erc
<gotcha@beauty.com>
 wrote
on Fri, 25 Jun 2004 00:25:21 GMT
<RhKCc.63303$sj4.186@news-server.bigpond.net.au>:
> "The Ghost In The Machine" <ewill@aurigae.athghost7038suus.net> wrote
>>
>> Does this set cover [0,1) ?
>>
>> T = {0, 0.1, 0.2, ... 0.9, 0.01, 0.02, ,,, 0.99, 0.001, 0.002, ... 0.999, ... }
>>
>> All finite decimal prefixes exist in that set, by construction.
>>
>> Is 2/5 in that set? (Yes, it's the fifth element, 0.4.)
>>
>> Is 1/3 in that set?
>>
>> Is pi/4 = 0.78539816339744830962... in that set?
>>
>> Is the number of times you've claimed 1/3 is in {0.3, 0.33, 0.333, ...}
>> divided by the total number of times you've posted to Usenet
>> in that set? :-)
>
> Effectively yes. If you cannot provide a number inbetween any of the above numbers
> and the members then the *set difference* is 0.
>
> I'll define set_minus.
>
> SetA = {2, 3, 4}
> NumA = 3.2
>
> SetA SetMinus Numa = 0.2

That's not much of a definition. Are you defining

A SetMinus n = min(a in A) abs(n - a)?

Or, if one prefers, the computational form:

func setMinus(Set<Number> a, Number b) returns Number
{
    Number diff = positiveInfinity;

    foreach el (a)
    {
       if(diff > abs(el - b)) diff = abs(el - b);
    }
}

Or the recursive form:

func setMinus(Set<Number> a, Number b) returns Number
{
    if(a == emptySet) return positiveInfinity;
    else return min(abs(firstEl(a) - b), setMinus(a - {firstEl(a)}, b));
}

assuming Set<> operator-(Set<>, Set<>) is appropriately defined.

If so, that's fine. (The positiveInfinity is a bit of a kludge.)

>
> Now what is T SetMinus Pi ? 0

Actually, 2.14159265... but that's mostly because T only covers [0,1).
Had you asked about T SetMinus pi/4, or T SetMinus (pi - 3),
the value would be 0.

What is T SetMinus any r in [0,1)? 0.

I fail to see the point of defining this operation but it's not
inconsistent with current mathematics. However, it also doesn't
prove, denumerate, or show evidence for the denumerability of [0,1).

[rest snipped] 

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

Relevant Pages

  • Re: limitation to induction on finite bounds
    ... Or, if one prefers, the computational form: ... func setMinusreturns Number ... I fail to see the point of defining this operation but it's not ... prove, denumerate, or show evidence for the denumerability of [0,1). ...
    (sci.math)
  • Re: limitation to induction on finite bounds
    ... On Fri, 25 Jun 2004 08:00:16 GMT, The Ghost In The Machine ... >Or, if one prefers, the computational form: ... >I fail to see the point of defining this operation but it's not ... >prove, denumerate, or show evidence for the denumerability of [0,1). ...
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    ... On Fri, 25 Jun 2004 08:00:16 GMT, The Ghost In The Machine ... >Or, if one prefers, the computational form: ... >I fail to see the point of defining this operation but it's not ... >prove, denumerate, or show evidence for the denumerability of [0,1). ...
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