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Topic: Matheology S 224
Replies: 16   Last Post: Apr 21, 2013 6:53 PM

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Posts: 2,777
Registered: 12/13/04
Re: Matheology S 224
Posted: Apr 21, 2013 12:13 PM
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On 21/04/2013 10:03 AM, Frederick Williams wrote:
> fom wrote:
>> On 4/20/2013 3:40 PM, Frederick Williams wrote:

>>> Nam Nguyen wrote:
>>>> On 20/04/2013 8:59 AM, fom wrote:

>>>>> On 4/20/2013 5:25 AM, Alan Smaill wrote:
>>>>>> Frederick Williams <> writes:

>>>>>>> Nam Nguyen wrote:
>>>>>>>> On 19/04/2013 5:55 AM, Frederick Williams wrote:

>>>>>>>>> Nam Nguyen wrote:
>>>>>>>>>> On 18/04/2013 7:19 AM, Frederick Williams wrote:

>>>>>>>>>>> Also, as I remarked elsewhere, "x e S' /\ Ay[ y e S' -> y e S]"
>>>>>>>>>>> doesn't
>>>>>>>>>>> express "x is in a non-empty subset of S".

>>>>>>>>>> Why?

>>>>>>>>> It says that x is in S' and S' is a subset of S.

>>>>>>>> How does that contradict that it would express "x is in a non-empty
>>>>>>>> subset of S", in this context where we'd borrow the expressibility
>>>>>>>> of L(ZF) as much as we could, as I had alluded before?

>>>>>>> You really are plumbing the depths. To express that x is non-empty you
>>>>>>> have to say that something is in x, not that x is in something.

>>>>>> but the claim was that x *is in* a non-empty set --
>>>>>> in this case S', which is non-empty, since x is an element of S',
>>>>>> and S' is a subset of S.
>>>>>> (Much though it would be good for Nam to realise that
>>>>>> some background set theory axioms would be kind of useful here)

>>>>> Yes. I thought about posting some links indicating
>>>>> that primitive symbols are undefined outside of a
>>>>> system of axioms (definition-in-use)
>>>>> The other aspect, though, is that Nam appears to be using an
>>>>> implicit existence assumption. So,
>>>>> AxASES'(xeS' /\ Ay(yeS' -> yeS))
>>>>> clarifies the statement and exhibits its second-order nature.
>>>>> This is fine since he claims that his work is not in the
>>>>> object language.

>>>> Right.

>>> If fom's formula is to express "x is in a non-empty subset of S" then it
>>> needs to have both x and S free, so delete the first two quantifiers.

>> Do you have a particular x and S in mind?

> I probably misunderstood. If Nam saying that, for every x and every set
> S, x is in a non-empty subset of S, then your formula expresses that.
> But clearly it is false.

I didn't say "every x and every set S". In the underlying context I was
talking about, x is _an_ individual and S is _a_ set (however general
each might be).

>> Or are we reverting to the distinction between real
>> and apparent variables from the first "Principia
>> Mathematica"?

> I call them free and bound respectively.

>> Or are we interpreting a statement in relation to a
>> general usage over an unspecified domain? My quantifiers
>> are in place to make clear the meaning for general usage.
>> Within any context involving proof, the leading quantifiers
>> obey rules:
>> |AxASES'(xeS' /\ Ay(yeS' -> yeS))
>> |ASES'(teS' /\ Ay(yeS' -> yeS))
>> |ES'(teS' /\ Ay(yeS' -> yeP))
>> ||(xeP' /\ Ay(yeP' -> yeP))
>> The original statement is assumed (hence, is stroked)
>> The existential statement is assumed (hence, a second stroke)
>> Now the presuppostions of use are clear.
>> That was my only purpose.

There is no remainder in the mathematics of infinity.


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