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




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 oneclause 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 wellformed >term and F is a wellformed 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 wellformed term and F is a wellformed 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 wellformed 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 nonexistent set simply doesn't come into it.



