On 08/05/2012 09:02 PM, Kevin Wampler wrote:
> On 5/8/2012 12:44 PM, Orchid Win7 v1 wrote:
>> Does the set of all sets that do not contain their own complement
>> contain itself?
>
> Yes

That's great. Now tell me, is the set of all sets that list themselves 
listed in itself?

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On Tue, 08 May 2012 20:44:36 +0100, Orchid Win7 v1 wrote:

>>> If you set out to fail, and succeed, what have you done?
>>
>> You've taken a ride down the fail trail? ;)
> 
> Does the set of all sets that do not contain their own complement
> contain itself?

You do realise that I'm razzing you about your spelling again, right? ;)

Jim

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On 5/8/2012 1:23 PM, Orchid Win7 v1 wrote:
> On 08/05/2012 09:02 PM, Kevin Wampler wrote:
>> On 5/8/2012 12:44 PM, Orchid Win7 v1 wrote:
>>> Does the set of all sets that do not contain their own complement
>>> contain itself?
>>
>> Yes
>
> That's great. Now tell me, is the set of all sets that list themselves
> listed in itself?
>

I assume by "list" you mean "contain"?  If so, then I think the answer 
is no.

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>>>> If you set out to fail, and succeed, what have you done?
>>>
>>> You've taken a ride down the fail trail? ;)
>>
>> Does the set of all sets that do not contain their own complement
>> contain itself?
>
> You do realise that I'm razzing you about your spelling again, right? ;)

Yeah, I had noticed. :-P

I was hoping to blind you with library science...

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>> That's great. Now tell me, is the set of all sets that list themselves
>> listed in itself?
>
> I assume by "list" you mean "contain"?

Yes.

> If so, then I think the answer is no.

And I think the answer is "Russell's paradox".

Interestingly, in Portal 2 GLaDDOS poses nearly the exact same question 
to Wheatly, who also responds with "false".

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On 5/8/2012 1:52 PM, Orchid Win7 v1 wrote:
> And I think the answer is "Russell's paradox".

That answer is also correct.  I'm being overly pedantic in my answers 
but here's my reasoning:

----

1) Does the set of all sets that do not contain their own complement
contain itself?

There is no such set, and thus the statement is vacuously true.

2) Now tell me, is the set of all sets that list themselves listed in 
itself?

There are no sets which contain themselves, thus the set of all these 
sets it in fact the empty set.  The empty set does not contain itself, 
so the answer is "no".

----

Of course both of these answers are assuming modern "post Russel's 
paradox" set theory, but I think they work.


> Interestingly, in Portal 2 GLaDDOS poses nearly the exact same question
> to Wheatly, who also responds with "false".

Maybe an in-joke by some math-savvy writers? Assuming my reasoning is 
correct that is.

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On 08/05/2012 7:05 PM, Orchid Win7 v1 wrote:
> On 08/05/2012 05:04 PM, Jim Henderson wrote:
>
>> The error is when you've reached the end of the trail and keep going.
>
> If you set out to fail, and succeed, what have you done?

Become a gummy parrot. <Boom boom>

-- 
Regards
     Stephen

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On Tue, 08 May 2012 21:48:43 +0100, Orchid Win7 v1 wrote:

>>>>> If you set out to fail, and succeed, what have you done?
>>>>
>>>> You've taken a ride down the fail trail? ;)
>>>
>>> Does the set of all sets that do not contain their own complement
>>> contain itself?
>>
>> You do realise that I'm razzing you about your spelling again, right?
>> ;)
> 
> Yeah, I had noticed. :-P

Oh, good - sometimes stuff like that goes right past you - but you /are/ 
getting better at detecting it. ;)

> I was hoping to blind you with library science...

I'm not that easy to blind. ;)

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> 1) Does the set of all sets that do not contain their own complement
> contain itself?
>
> There is no such set, and thus the statement is vacuously true.

I'm not sure how you came to the conclusion that no such set exists.

Since /by definition/ a set can never contain its own complement (that's 
what I complement /is/), the set above contains all possible sets - 
including itself.

> 2) Now tell me, is the set of all sets that list themselves listed in
> itself?
>
> There are no sets which contain themselves, thus the set of all these
> sets it in fact the empty set. The empty set does not contain itself, so
> the answer is "no".

Again, I'm not sure how you come to the conclusion that a set cannot 
contain itself. Even if that were true, it would imply that the set in 
question /does not exist/, rather than that it is empty.

The answer of course is that the definition of the set is inconsistent. 
If the set does not contain itself, that implies that it does contain 
itself, and vice versa. So the answer is neither "yes" nor "no". The 
answer is "undefined".

>> Interestingly, in Portal 2 GLaDDOS poses nearly the exact same question
>> to Wheatly, who also responds with "false".
>
> Maybe an in-joke by some math-savvy writers? Assuming my reasoning is
> correct that is.

The actual statement was the liar paradox, to which Wheatly, being 
scientifically engineered to be the greatest moron ever, replies

"Er... false. Yes, definitely false. Although I think I may have heard 
this one before, to be fair."

The exasperated GLaDDOS retorts "There IS no answer, you idiot! It's a 
paradox!!"

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On 5/9/2012 1:21 AM, Invisible wrote:
>> 1) Does the set of all sets that do not contain their own complement
>> contain itself?
>>
>> There is no such set, and thus the statement is vacuously true.
>
> I'm not sure how you came to the conclusion that no such set exists.
>
> Since /by definition/ a set can never contain its own complement (that's
> what I complement /is/), the set above contains all possible sets -
> including itself.

Right, and such a "set of all sets" does not exist within standard set 
theory.  It appears, however, that you are unfamiliar with standard 
(ZFC) set theory, the your misconception is understandable.

I am, by the way, glossing over your use of "compliment" which strictly 
speaking isn't well defined the way you've used it.

>> 2) Now tell me, is the set of all sets that list themselves listed in
>> itself?
>>
>> There are no sets which contain themselves, thus the set of all these
>> sets it in fact the empty set. The empty set does not contain itself, so
>> the answer is "no".
 >
> Again, I'm not sure how you come to the conclusion that a set cannot
> contain itself.

Oh come on, it's one of the axioms of ZFC set theory! 
http://en.wikipedia.org/wiki/Axiom_of_regularity

> Even if that were true, it would imply that the set in
> question /does not exist/, rather than that it is empty.

Which is exactly what I said, you'll notice.  I'll quote to save you the 
trouble of reading: "thus the set of all these sets it in fact the empty 
set".

> The answer of course is that the definition of the set is inconsistent.
> If the set does not contain itself, that implies that it does contain
> itself, and vice versa. So the answer is neither "yes" nor "no". The
> answer is "undefined".

That was the answer at around 1901 before modern set theory existed, and 
it was sufficient to destroy the notion of "set" which led to it.  You 
appear to be using a notion of "set theory" like this which hasn't 
really been popular for over a century.  In ZFC set theory there is no 
paradox because you can't create the sort of sets needed for the paradox 
to happen.

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