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Topic: Van Neumann Hierarchy
Replies: 13   Last Post: Apr 28, 2014 10:45 AM

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 David C. Ullrich Posts: 3,555 Registered: 12/13/04
Re: Van Neumann Hierarchy
Posted: Apr 28, 2014 10:45 AM

On Sun, 27 Apr 2014 18:05:15 +0100, Sandy <sandy@hotmail.invalid>
wrote:

>dullrich@sprynet.com wrote:
>> On Sat, 26 Apr 2014 20:01:26 +0100, Sandy <sandy@hotmail.invalid>
>> wrote:
>>

>>> dullrich@sprynet.com wrote:
>>>> On Fri, 25 Apr 2014 20:13:43 -0700, William Elliot <marsh@panix.com>
>>>> wrote:
>>>>

>>>>> The Van Neumann hierarchy is defined with transfinite induction as
>>>>> V_eta = \/{ P(V_xi) | xi < eta }.

>>>>
>>>> Huh> That reminds me of the Von Neumann hierarchy...

>>>
>>> May I ask how the case eta = 0 is to be interpreted? My vague thoughts:
>>> There is no xi < 0 so V_xi can be assigned no meaning. So P(V_xi) is
>>> also without meaning, and \/{etc} is without meaning.
>>>
>>> The version of the definition of von Neumann's hierachy that I have seen is:
>>>
>>> V_0 = 0

>>
>> Where 0 is the empty set...

>
>I take it that when von Neumann on sets is being discussed, the first
>ordinal is the empty set and thus 0 may denote them both.

Yes, I was just clarifying for anyone else who might be confused

>
>>> V_{alpha+1} = P(V_alpha)
>>> V_{lamba} = U_{alpha < lambda} V_alpha (lambda a limit ordinal).
>>>
>>> I have not seen a one-clause definition before reading William Elliot's
>>> post.

>>
>> What he wrote is ok, at least in that regard.
>>
>> Officially in set theory a single set has a union:
>> If S is a set then U S is {x : x in y for some y in S}.
>>
>> It follows that U 0 = 0. Since as you point out there is
>> no xi < 0 we have
>>
>> V_0 = U {V_xi : xi < 0} = U 0 = 0.

>
>I think I get it. My worry was with the V_xi in {P(V_xi) | xi < 0}.
>What, I asked myself, is V_xi if there is no xi < 0?. And if I don't
>know what V_xi is, I cannot know what P(V_xi) is. I need to look at the
>set in question as {...| xi < 0} (never mind what ... is) which is empty
>since xi < 0 is false whatever xi is.
>
>So to a general question about sets: suppose that t is a well-formed
>term and F is a well-formed formula, then {t : F} is empty if F is
>identically false *even if t denotes nothing*. That is correct, is it?

I'm not sure whether I should say yes or no to this, because it's
not clear to me that it's possible for a term in the language to
"dennote nothing" - not clear to me exactly what that even means.

To handle the present case a slightly modified version is enough:
If t is a well-formed term and F is a well-formed formula which
us identically false then {t:F} is the empty set, regardlless of t.

>
>Example: {the rational square root of 2 : 0=/=0} is the empty set even
>though 'the rational square root of 2' doesn't denote anything.

This makes less sense to me than the example we started with,
because it's not at all clear that there _is_ a term that "means"
"the rational square root of 2".

Because if there were such a term t then we could also
form {t : x = 0}, which would be a set containing exactly
one element, that being the rational square root of 2.

On the other hand, P(S) _is_ a well-formed term. To
figure out what {P(V_xi) : xi < 0} we don't need to
worry about the meaning of P(V) when V does not exist.
We simply look at the definition: This set has one element
for each xi < 0. For a total of no elements.
The notion of the power set of a non-existent
set simply doesn't come into it.

Date Subject Author
4/25/14 William Elliot
4/26/14 David Hartley
4/27/14 William Elliot
4/27/14 David Hartley
4/27/14 William Elliot
4/28/14 William Elliot
4/26/14 David C. Ullrich
4/26/14 Sandy
4/26/14 William Elliot
4/27/14 Sandy
4/27/14 William Elliot
4/27/14 David C. Ullrich
4/27/14 Sandy
4/28/14 David C. Ullrich