Cantor Confusion
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From: Arturo Magidin on 28 Sep 2006 19:10 In article <1159484526.290920.66150(a)h48g2000cwc.googlegroups.com>, <the_wign(a)yahoo.com> wrote: >Most of the responses point out that a simpler proof or a proof >that avoids the problem of the self-reference is to consider ANY >list of real numbers. However, that doesn't seem to simplify >anything. Now instead of assuming the list contains all reals, >we must make no assumptions. But that means we must >consider incomplete lists and complete lists, because the >construction that occurs in step 2 must be shown to be valid >in all cases. There is no need to consider any kind of list separately. Step 2 can be shown to be valid for ANY function f:N->R, regardless of whether it is one-to-one or not, regardless of whether we assume it to be surjective or not. All we need to assume is that it is a ->function<-. > In the case of lists that contain all reals, which >we must consider unless we are assuming no such list exists >(What good is the proof in that case?) That you ask the question suggests you do not understand the argument at all. If one proves that for EVERY function f:N->R, there exists x in R (which depends on f) such that x is not in f(N), then this proves that EVERY list of real numbers is incomplete. This proves that NO list can be complete, and that's what you want the proof for in the first place. > then we immediately >recognize that the real number described in step #2 as >ill-defined and meaningless. If you say so. Sure. Wanna buy a bridge? -- ====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes" by Bill Watterson) ====================================================================== Arturo Magidin magidin-at-member-ams-org
From: Poker Joker on 28 Sep 2006 19:27 "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message news:efhkpt$3136$1(a)agate.berkeley.edu... >> In the case of lists that contain all reals, which >>we must consider unless we are assuming no such list exists >>(What good is the proof in that case?) > > That you ask the question suggests you do not understand the argument > at all. It was rhetorical. Sometimes people ask questions that they don't really expect to be answered. They do so when they are truing to make a point. Wikipedia has a whole section on it if you would like to learn more. http://en.wikipedia.org/wiki/Rhetorical_question
From: Poker Joker on 28 Sep 2006 19:35 "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message news:efgfhd$261u$1(a)agate.berkeley.edu... > In article <1159410937.013643.192240(a)h48g2000cwc.googlegroups.com>, > <the_wign(a)yahoo.com> wrote: >>Cantor's proof is one of the most popular topics on this NG. It >>seems that people are confused or uncomfortable with it, so >>I've tried to summarize it to the simplest terms: >> >>1. Assume there is a list containing all the reals. >>2. Show that a real can be defined/constructed from that list. >>3. Show why the real from step 2 is not on the list. >>4. Conclude that the premise is wrong because of the contradiction. > > This is hardly the simplest terms. Much simpler is to do a ->direct<- > proof instead of a proof by contradiction. > > 1. Take ANY list of real numbers. > 2. Show that a real can be defined/constructed from that list. > 3. Show that the real from step 2 is not on the list. > 4. Conclude that no list can contain all reals. > How can it be simpler if the list can be ANY list instead of a particular one. ANY list opens up more possiblities than a single list. Also, if its true for ANY list, then it must be true for a specific list. So if considering a single specific list shows a flaw, then looking at ANY (ALL of them) list doesn't help.
From: Poker Joker on 28 Sep 2006 19:39 "Peter Webb" <webbfamily-diespamdie(a)optusnet.com.au> wrote in message news:451b5130$0$28952$afc38c87(a)news.optusnet.com.au... > You produce ANY list of all Reals, I can show you a missing real. > Therefore I can do it for ALL lists, and hence there is no complete list > of Reals. If he shows you a list of all reals you can't. I thought that would be obvious to even the casual observer.
From: cbrown on 28 Sep 2006 20:19
Virgil wrote: > In article <1159417542.425540.214160(a)i3g2000cwc.googlegroups.com>, > cbrown(a)cbrownsystems.com wrote: > > > the_wign(a)yahoo.com wrote: > > > Cantor's proof is one of the most popular topics on this NG. It > > > seems that people are confused or uncomfortable with it, so > > > I've tried to summarize it to the simplest terms: > > > > > > 1. Assume there is a list containing all the reals. > > > 2. Show that a real can be defined/constructed from that list. > > > 3. Show why the real from step 2 is not on the list. > > > 4. Conclude that the premise is wrong because of the contradiction. > > That is a proof by contradiction, which many constructionists object to. > I assume you are replying to the OP here, not me. Cheers - Chas |