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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Sep 10, 2013 at 13:01 review Reopen votes
Sep 10, 2013 at 13:04
Sep 9, 2013 at 12:58 review Reopen votes
Sep 9, 2013 at 13:06
Nov 29, 2012 at 6:33 comment added Gerry Myerson Interesting that this question was closed as "no longer relevant". Does that mean it has been proved that the current foundations of mathematics are consistent? [just kidding - I know the closure options are very limited]
Dec 25, 2011 at 21:24 history closed Felipe Voloch
Andrés E. Caicedo
user6976
user9072
Simon Thomas
no longer relevant
Dec 25, 2011 at 1:24 answer added tab timeline score: 9
Dec 24, 2011 at 22:26 answer added none timeline score: 5
Sep 13, 2011 at 1:58 comment added Thomas Riepe Dear Pierre-Yves, thanks for correcting that! I wonder how that happened.
Sep 12, 2011 at 14:41 comment added Pierre-Yves Gaillard Dear @Thomas: Is this golem.ph.utexas.edu/category/2011/09/voevodsky_on_fom.html the link you wanted to give?
Sep 2, 2011 at 3:35 answer added Taylor Dupuy timeline score: 24
Sep 1, 2011 at 14:44 answer added Mathieu Vidal timeline score: 7
Aug 28, 2011 at 9:22 history edited Pierre-Yves Gaillard CC BY-SA 3.0
removed a sentence
May 19, 2011 at 7:21 history edited Thomas Riepe CC BY-SA 3.0
added 154 characters in body
May 18, 2011 at 22:51 answer added Monroe Eskew timeline score: 6
Feb 20, 2011 at 10:33 comment added David Roberts I have added a vote to close, for being no longer relevant. I feel this question is attracting more answers that are not 'adding value', as they say. Discussion at meta if necessary.
Feb 20, 2011 at 8:52 answer added Anixx timeline score: -20
Feb 19, 2011 at 22:09 answer added Sergey Melikhov timeline score: 6
Feb 8, 2011 at 0:14 answer added Paul timeline score: 6
Nov 11, 2010 at 23:13 comment added Carl Mummert @Steven Gubkin: ZFC minus infinity plus its negation can be interpreted into Peano arithmetic, in the formal logical sense of interpreted. Since the latter has been proven consistent (in multiple independent ways, in fact), the former must also be consistent.
Nov 6, 2010 at 0:59 comment added Steven Gubkin ZFC without the axiom of infinity is known to be consistent?
Oct 6, 2010 at 23:33 answer added Daniel Tausk timeline score: 33
Oct 6, 2010 at 17:37 comment added Carl Mummert Given that ZFC without the axiom of infinity is known to be consistent, it would make as much sense to say that the axiom of infinity would be the first to go, since that will certainly solve the problem :)
Oct 6, 2010 at 15:57 history edited Pierre-Yves Gaillard CC BY-SA 2.5
EDIT clearly indicated.
Oct 6, 2010 at 4:08 vote accept Pierre-Yves Gaillard
Oct 4, 2010 at 23:25 comment added Timothy Chow @Kevin: Replacement will be chopped only if doing so fixes whatever inconsistency happens to shows up. It doesn't make much sense to speculate on what the fix will be when it ain't broke yet.
Oct 4, 2010 at 22:10 comment added Peter LeFanu Lumsdaine Actually, my error: Todd Trimble points out to me off-list that you can indeed construct the free monoid on a set $X$ without replacement, with a bit more care (as eg a subset of the set of partial functions from $\mathbb{N}$ to $X$). What I had in mind is that without replacement, one can’t always make constructions of the form $\bigsqcup_n F^n(X)$, (or $\bigprod$, $\bigcup$, $\varinjlim$ etc.) where $F$ is an arbitrary function on sets. I'm fairly sure I've seen at some point a simple, mathematically well-known example which isn't salvageable in the way “free monoid” was; but now I forget…
Oct 4, 2010 at 17:03 comment added Peter LeFanu Lumsdaine @Kevin Buzzard: as the (non-joking) background in Todd's link points out, while Bourbaki don't include Replacement in the same form that ZF has it, they do include an essentially equivalent axiom: that the union of any set-indexed family of sets forms a set. ZF with replacement dropped (aka Zermelo Set Theory) is not powerful enough for much mathematics at all: for instance, iirc, it can't construct the free monoid on an arbitrary set, nor hence eg polynomial algebras in noncommuting variables.
Oct 4, 2010 at 15:07 answer added Timothy Chow timeline score: 96
Oct 4, 2010 at 13:27 comment added Todd Trimble @Andras: Joking aside, there is a serious point behind Taylor's post: that the replacement axiom is indeed enormously powerful, even if not strictly speaking necessary for most of core mathematics (which can be developed within a universe of sets whose rank is less than $\omega+\omega$, where replacement fails). In fact, Taylor's book Practical Foundations of Mathematics culminates in a discussion of how to formulate the replacement axiom in the context of dependent type theory.
Oct 4, 2010 at 12:45 comment added András Salamon @Todd: not fair, you are only allowed to resurrect those jokes once a year!
Oct 4, 2010 at 0:23 comment added Todd Trimble Um, yes, I know all this, Andras?! Was joking around myself?!
Oct 4, 2010 at 0:09 comment added András Salamon @Todd: Lawvere points out a few messages in that this was a (slightly irresponsible) April Fool. The final comment of the thread is appropriate: "I suppose that it is possible that somebody, browsing at random, *might* find it in the CATEGORIES archives, in years to come, and not read ahead to find out what the world had had to say about this discovery back in the mad, exciting days of the late C20." Note that this does not detract from the larger issue that ZF may well be inconsistent.
Oct 3, 2010 at 23:10 answer added Felipe Voloch timeline score: 72
Oct 3, 2010 at 21:30 comment added Todd Trimble Kevin Buzzard: yup. See mta.ca/~cat-dist/catlist/1999/zf-010499.
Oct 3, 2010 at 20:54 comment added Kevin Buzzard Bourbaki dropped the axiom(-scheme) of replacement in their development of mathemetics, so they don't, I think, have enough mathematics to build the ordinals. However their work seems to indicate that they had enough to do an awful lot of mathematics (probably all of the mathematics I've ever done and will do won't need replacement). My guess is that if ZFC is inconsistent then replacement will be the first axiom for the chop.
Oct 3, 2010 at 19:53 history edited Kaveh
retag
Oct 3, 2010 at 19:43 answer added Todd Trimble timeline score: 12
Oct 3, 2010 at 19:09 answer added Pietro Majer timeline score: 22
Oct 3, 2010 at 18:31 comment added The Mathemagician I think he needs to be a little more specific about what he means by this question.WHICH foundations? ZFC set theory? Categorical foundations a la Lawvere? What?
Oct 3, 2010 at 16:16 answer added Rob Simmons timeline score: 24
Oct 3, 2010 at 13:50 answer added Jim Conant timeline score: 28
Oct 3, 2010 at 13:42 answer added Dick Palais timeline score: 83
Oct 3, 2010 at 13:36 comment added muad I don't understand his objection to Gentzen's proof at 29:00. Why would someone be skeptical about well foundedness of $\epsilon_0$?
Oct 3, 2010 at 12:37 comment added Pierre-Yves Gaillard @Gerry Myerson - Sorry about the confusion. The question is "What if the current foundations of Mathematics are inconsistent?" A subsidiary question is "Has this kind of opinion been expressed before?"... At least that's how I see it. You're welcome to answer any of the questions, to ask others, to edit the question... - An implicit question is "What do you think of Vladimir Voevodsky's talk?"
Oct 3, 2010 at 12:33 comment added Michael Bächtold +1 just for pointing out this talk.
Oct 3, 2010 at 12:15 comment added Gerry Myerson I'm confused. Is the question, "What if the current foundations of Mathematics are inconsistent?" Or is the question, "Has this kind of opinion been expressed before?"
Oct 3, 2010 at 12:11 comment added José Figueroa-O'Farrill Does anyone know what Voevodsky means by the "current troubles between mathematics and physics" in his talk?!
Oct 3, 2010 at 11:45 comment added Todd Trimble What opinion? That ZFC or even Peano Arithmetic is in fact inconsistent? The trouble is that this sort of thing is not too likely to be put in print by a reputable mathematician, even if he expresses his (perhaps occasional) doubts in private. But yeah, I can think of a few reputable mathematicians who sometimes express sentiments ("opinion" is maybe too strong a word) along these lines. Backing that up is another story altogether.
Oct 3, 2010 at 11:20 comment added Piero D'Ancona Then the unreasonable effectiveness of mathematics would become slightly more unreasonable
Oct 3, 2010 at 10:24 history edited Pierre-Yves Gaillard CC BY-SA 2.5
fixed grammar
Oct 3, 2010 at 10:17 history asked Pierre-Yves Gaillard CC BY-SA 2.5