LudovicoVan
Posts:
3,379
From:
London
Registered:
2/8/08


Re: Cantor's argument and the Potential Infinite.
Posted:
Nov 17, 2012 2:24 AM


"Uirgil" <uirgil@uirgil.ur> wrote in message news:uirgilC5CD34.15035616112012@BIGNEWS.USENETMONSTER.COM... > In article <k851n8$fgs$1@dontemail.me>, > "LudovicoVan" <julio@diegidio.name> wrote: >> "Uirgil" <uirgil@uirgil.ur> wrote in message >> news:uirgil8D50A0.02310116112012@BIGNEWS.USENETMONSTER.COM... >> > In article <k850hm$a03$2@dontemail.me>, >> > "LudovicoVan" <julio@diegidio.name> wrote: >> >> "Uirgil" <uirgil@uirgil.ur> wrote in message >> >> news:uirgil981B6A.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

