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Topic: set builder notation
Replies: 12   Last Post: Aug 24, 2013 1:38 AM

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David C. Ullrich

Posts: 3,163
Registered: 12/13/04
Re: set builder notation
Posted: Aug 18, 2013 10:03 AM
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On Sat, 17 Aug 2013 18:47:40 +0100, Peter Percival
<peterxpercival@hotmail.com> wrote:

>dullrich@sprynet.com wrote:
>> On Sat, 17 Aug 2013 04:12:34 -0700 (PDT), lite.on.beta@gmail.com
>> wrote:
>>

>>>
>>> S = {x /in A | P(x) }
>>>
>>> For the set builder notation above, what we really means is:
>>>
>>> all things x, such that "x is element of A *and* P(x) is true" correct?

>>
>> Yes and no.
>>
>> Yes:
>>
>> (1) {x in A | P(x)}
>>
>> is the same as
>>
>> (2) {x | x in A and P(x)}.
>>
>> No:
>>
>> No, because (2) is actually not a "legal"
>> construction of a set! (2) is of the form
>>
>> (3) {x | Q(x)},
>>
>> and things of the form (3) are officially not
>> allowed.
>>
>> Not allowed because they lead to contradictions:
>> Let
>>
>> S = {x | x is not an element of x}.

>
>But is 'x is not an element of x' of the form 'x in A and P(x)'?


No.

> I
>suppose what I'm doing is challenging you to reproduce Russell's paradox
>with sets of the form {x | x in A and P(x)}. I'm rather sure (but I
>know nothing) that {x in A | P(x)} is *nothing but* alternative notation
>for {x | x in A and P(x)}.


Informally it's just an alternate notation.

The fact that {x | P(x)} is not allowed seems more important.
If you want you can change things so that {x | P(x)| is
allowed as long as P(x) is of the form "x in A and Q(x)".

>
>>
>> Then S an element of S implies S not an element
>> pf S, and conversely; there is no such set S.
>>
>> Mathhematians other than set theorists use
>> (3) all the time, but officially it has to be (1).
>>
>>

>>>
>>> The vertical bar is essentially conjunction, correct?

>>




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