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Topic: Cantor's argument and the Potential Infinite.
Replies: 17   Last Post: Nov 17, 2012 10:59 PM

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LudovicoVan

Posts: 3,201
From: London
Registered: 2/8/08
Re: Cantor's argument and the Potential Infinite.
Posted: Nov 17, 2012 2:24 AM
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"Uirgil" <uirgil@uirgil.ur> wrote in message
news:uirgil-C5CD34.15035616112012@BIGNEWS.USENETMONSTER.COM...
> In article <k851n8$fgs$1@dont-email.me>,
> "LudovicoVan" <julio@diegidio.name> wrote:

>> "Uirgil" <uirgil@uirgil.ur> wrote in message
>> news:uirgil-8D50A0.02310116112012@BIGNEWS.USENETMONSTER.COM...

>> > In article <k850hm$a03$2@dont-email.me>,
>> > "LudovicoVan" <julio@diegidio.name> wrote:

>> >> "Uirgil" <uirgil@uirgil.ur> wrote in message
>> >> news:uirgil-981B6A.02055216112012@BIGNEWS.USENETMONSTER.COM...

>> <snipped>
>>

>> >> > ZFC offers a standard set theory in which actually infinite sets are
>> >> > not
>> >> > only allowed but actually required to exist, and no one yet has been
>> >> > able to show that ZFC is not a perfectly sound set theory.

>> >>
>> >> That is only because you are so incoherent as to insist to call N an
>> >> actual
>> >> infinity.

>> >
>> > In ZFC, the N is an actually infinite set. So until you can show that
>> > ZFC is internally inconsistent, which no one has yet done, we have
>> > actual infinities in ZFC.

>>
>> That's interesting: would you be so kind to show me how/why, technically
>> although informal as it needs be, N is an "actual infinity" in ZFC?

>
> ZFC requires the existence of a set N such that
> {} is a member of N, and
> If x is a member of N, so is x \/ {x}, and
> N is a subset of every set S such that
> {} is a member of S and
> If x is a member of S, so is x \/ {x}
>
> Such a set is provably not finite, as finiteness of a set would require
> that it biject with some MEMBER of such an N, which N provably does not.


Sure, N is the minimal set with 0 and closed under the successor operation.

But that remains a characterization of a *potential infinity*.

Well (not) done.

-LV





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