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From: Peter Percival on 20 May 2007 09:19
Aatu Koskensilta wrote:
>
> As everyone and his sister knows, in Zermelo set theory the stages of the
> cumulative hierarchy we can provably reach are all below the omega + omega'th
> level. In terms of the conception of the world of sets given by the cumulative
> hierarchy this doesn't really make much sense; there are well-orderings
> within V_omega + omega of order type way beyond omega + omega, and it thus seems
> a bit odd why we should not iterate the "set of" operation more than omega +
> omega times.
>
> This line of thought suggests the following mathematical question. What is the
> least ordinal alpha with the property that alpha > omega and if there is a
> well-ordering of type beta in V_alpha, beta < alpha? (Hint: consider the
> fixed point of a suitable ordinal function.)

Sorry you had no takers, not in the ng at least. So what's the answer?

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Remove "antispam" and ".invalid" for e-mail address.
From: Rupert on 22 May 2007 02:56
On Apr 19, 1:01 am, Aatu Koskensilta <aatu.koskensi...(a)xortec.fi>
wrote:
> As everyone and his sister knows, in Zermelo set theory the stages of the
> cumulative hierarchy we can provably reach are all below the omega + omega'th
> level. In terms of the conception of the world of sets given by the cumulative
> hierarchy this doesn't really make much sense; there are well-orderings
> within V_omega + omega of order type way beyond omega + omega, and it thus seems
> a bit odd why we should not iterate the "set of" operation more than omega +
> omega times.
>
> This line of thought suggests the following mathematical question. What is the
> least ordinal alpha with the property that alpha > omega and if there is a
> well-ordering of type beta in V_alpha, beta < alpha? (Hint: consider the
> fixed point of a suitable ordinal function.)
>
> --
> Aatu Koskensilta (aatu.koskensi...(a)xortec.fi)
>
> "Wovon man nicht sprechen kann, daruber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus

It's the first fixed point of the beth function, isn't it?

From: Aatu Koskensilta on 22 May 2007 08:26
On 2007-05-22, in sci.logic, Rupert wrote:
> On Apr 19, 1:01 am, Aatu Koskensilta <aatu.koskensi...(a)xortec.fi>
> wrote:
>>
>> This line of thought suggests the following mathematical question. What is the
>> least ordinal alpha with the property that alpha > omega and if there is a
>> well-ordering of type beta in V_alpha, beta < alpha? (Hint: consider the
>> fixed point of a suitable ordinal function.)
>
> It's the first fixed point of the beth function, isn't it?

Yep.

In an attempt to have discussion about logic in sci.logic on a not utterly
inane level, let's try another exercise. As everyone knows NBG (with global
choice) is a conservative extension of ZFC. This is not difficult to prove,
but in fact I don't know of any place one could actually read a proof. So
the exercise is to prove this.

Here's what is to be done. Start with a countable model of ZFC and add to it
as classes all definable collections. This gets us a model of NBG without
global choice. The only subtlety in showing this is in showing that classes
defined with conditions with free class variables are also included. Alas,
this doesn't (necessarily) get us a model in which global choice holds, since
there is no guarantee that there is a choice function for the universe among
the classes we added. So, to remedy this apply a bit of forcing, using the
(proper class) of choice functions, ordered by inclusion -- or reverse
inclusion, depending on whether you like the "less-is-more" convention or not
-- as the forcing notion. To finish the proof, show that the set portion of
the model was not disturbed by the forcing, effectively by noting that the
intersection of any set and the new global choice function is already included
in the original model (it's just one of the set-sized choice functions).

This exercise is a rather nice introduction to forcing.

(I'll give some credit also to anyone who points me to a proof in the
literature.)

--
Aatu Koskensilta (aatu.koskensilta(a)xortec.fi)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Rupert on 22 May 2007 20:02
On May 22, 10:26 pm, Aatu Koskensilta <aatu.koskensi...(a)xortec.fi>
wrote:
> On 2007-05-22, in sci.logic, Rupert wrote:
>
> > On Apr 19, 1:01 am, Aatu Koskensilta <aatu.koskensi...(a)xortec.fi>
> > wrote:
>
> >> This line of thought suggests the following mathematical question. What is the
> >> least ordinal alpha with the property that alpha > omega and if there is a
> >> well-ordering of type beta in V_alpha, beta < alpha? (Hint: consider the
> >> fixed point of a suitable ordinal function.)
>
> > It's the first fixed point of the beth function, isn't it?
>
> Yep.
>
> In an attempt to have discussion about logic in sci.logic on a not utterly
> inane level, let's try another exercise. As everyone knows NBG (with global
> choice) is a conservative extension of ZFC. This is not difficult to prove,
> but in fact I don't know of any place one could actually read a proof. So
> the exercise is to prove this.
>
> Here's what is to be done. Start with a countable model of ZFC and add to it
> as classes all definable collections. This gets us a model of NBG without
> global choice. The only subtlety in showing this is in showing that classes
> defined with conditions with free class variables are also included. Alas,
> this doesn't (necessarily) get us a model in which global choice holds, since
> there is no guarantee that there is a choice function for the universe among
> the classes we added. So, to remedy this apply a bit of forcing, using the
> (proper class) of choice functions, ordered by inclusion -- or reverse
> inclusion, depending on whether you like the "less-is-more" convention or not
> -- as the forcing notion. To finish the proof, show that the set portion of
> the model was not disturbed by the forcing, effectively by noting that the
> intersection of any set and the new global choice function is already included
> in the original model (it's just one of the set-sized choice functions).
>
> This exercise is a rather nice introduction to forcing.
>
> (I'll give some credit also to anyone who points me to a proof in the
> literature.)
>
> --
> Aatu Koskensilta (aatu.koskensi...(a)xortec.fi)
>
> "Wovon man nicht sprechen kann, daruber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus

There's a widely-used textbook which has a proof - but I won't reveal
it just now, so as not to spoil the exercise for those who want to try
it.

From: Rupert on 22 May 2007 23:56
On May 22, 10:26 pm, Aatu Koskensilta <aatu.koskensi...(a)xortec.fi>
wrote:
> On 2007-05-22, in sci.logic, Rupert wrote:
>
> > On Apr 19, 1:01 am, Aatu Koskensilta <aatu.koskensi...(a)xortec.fi>
> > wrote:
>
> >> This line of thought suggests the following mathematical question. What is the
> >> least ordinal alpha with the property that alpha > omega and if there is a
> >> well-ordering of type beta in V_alpha, beta < alpha? (Hint: consider the
> >> fixed point of a suitable ordinal function.)
>
> > It's the first fixed point of the beth function, isn't it?
>
> Yep.
>
> In an attempt to have discussion about logic in sci.logic on a not utterly
> inane level, let's try another exercise. As everyone knows NBG (with global
> choice) is a conservative extension of ZFC. This is not difficult to prove,
> but in fact I don't know of any place one could actually read a proof. So
> the exercise is to prove this.
>
> Here's what is to be done. Start with a countable model of ZFC and add to it
> as classes all definable collections. This gets us a model of NBG without
> global choice. The only subtlety in showing this is in showing that classes
> defined with conditions with free class variables are also included. Alas,
> this doesn't (necessarily) get us a model in which global choice holds, since
> there is no guarantee that there is a choice function for the universe among
> the classes we added. So, to remedy this apply a bit of forcing, using the
> (proper class) of choice functions, ordered by inclusion -- or reverse
> inclusion, depending on whether you like the "less-is-more" convention or not
> -- as the forcing notion. To finish the proof, show that the set portion of
> the model was not disturbed by the forcing, effectively by noting that the
> intersection of any set and the new global choice function is already included
> in the original model (it's just one of the set-sized choice functions).
>
> This exercise is a rather nice introduction to forcing.
>
> (I'll give some credit also to anyone who points me to a proof in the
> literature.)
>
> --
> Aatu Koskensilta (aatu.koskensi...(a)xortec.fi)
>
> "Wovon man nicht sprechen kann, daruber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus

There's a widely-used textbook which has a proof - but I won't reveal
it just now, so as not to spoil the exercise for those who want to try
it.

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