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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: 3,033
Registered: 12/13/04
Re: Van Neumann Hierarchy
Posted: Apr 27, 2014 12:11 PM
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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...

> 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.



>
>

>>
>>>
>>> Within ZF, using regularity and replacement, how does one show
>>> that for every set A, there's some ordinal eta, with A in V_eta.
>>>
>>> It would suffice to show for every transitive set A,
>>> there's some ordinal eta, with A in V_eta.
>>>
>>> With that, since every set is a subset of it's transitive

>>
>> You really really really need to stop criticizing others' English.
>> Or learn a little English yourself. It's "its", not "it's".
>>

>>> closure, every set would be inside the Van Neumann hierarchy.
>>




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