Cantor Confusion
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From: MoeBlee on 29 Sep 2006 20:02 Poker Joker wrote: > "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. By 'any' we mean an arbitrary one. The way we talk about an arbitrary object is to choose a variable not free in any previous line in the argument nor free in the conclusion we will eventually draw and then use the rule of universal generalization to draw our eventual conclusion. So if we want to prove something about an arbitrary enumeration of denumerable sequences of digits, we say, "Suppose f is an enumeration whose range is a subset of the set of denumerable sequences of digits" (and, of course, we will presume NOTHING about f other than that it is an enumeration whose range is a subset of the set of denumerable sequences of digits. > Also, if its true for ANY list, then it must be > true for a specific list. If the property holds for any list, then, if there exists a list, then the property holds of any such list that exists. > So if considering a single specific list > shows a flaw, then looking at ANY (ALL of them) list doesn't > help. If there is a specific list that does not have the property, then we will not be able to prove that the property holds of an arbitrary list. But I don't know what specific list you think is "flawed". Nor do I see what your point has to do with Arturo's point that we don't have to adopt a reductio ad absurdum assumption, since we can just show for an arbitrary list that it does not list every real number. MoeBlee
From: Virgil on 29 Sep 2006 20:03 In article <UeiTg.25580$QT.16960(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message > news:efj3bk$120f$1(a)agate.berkeley.edu... > > In article <N_YSg.1208$3E2.403(a)tornado.rdc-kc.rr.com>, > > Poker Joker <Poker(a)wi.rr.com> wrote: > >> > >>"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. > > > > Because a direct proof is simpler than a proof by contradiction. > > > >> ANY list opens up more possiblities than > >>a single list. > > > > Any list does not require you to assume that there is a "single list" > > which some some particular property. > > > > ====================================================================== > > "It's not denial. I'm just very selective about > > what I accept as reality." > > --- Calvin ("Calvin and Hobbes" by Bill Watterson) > > ====================================================================== > > We all noticed you neglected this logic: > > 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. But the construction method, which applies equally well to all lists without modification, if flawless for any list is equally flawless for all lists. And it is flawless for any list.
From: Poker Joker on 29 Sep 2006 20:11 "Virgil" <virgil(a)comcast.net> wrote in message news:virgil-A8F506.18030329092006(a)comcast.dca.giganews.com... > In article <UeiTg.25580$QT.16960(a)tornado.rdc-kc.rr.com>, > "Poker Joker" <Poker(a)wi.rr.com> wrote: > >> "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message >> news:efj3bk$120f$1(a)agate.berkeley.edu... >> > In article <N_YSg.1208$3E2.403(a)tornado.rdc-kc.rr.com>, >> > Poker Joker <Poker(a)wi.rr.com> wrote: >> >> >> >>"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. >> > >> > Because a direct proof is simpler than a proof by contradiction. >> > >> >> ANY list opens up more possiblities than >> >>a single list. >> > >> > Any list does not require you to assume that there is a "single list" >> > which some some particular property. >> > >> > ====================================================================== >> > "It's not denial. I'm just very selective about >> > what I accept as reality." >> > --- Calvin ("Calvin and Hobbes" by Bill Watterson) >> > ====================================================================== >> >> We all noticed you neglected this logic: >> >> 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. > > But the construction method, which applies equally well to all lists > without modification, if flawless for any list is equally flawless for > all lists. > > And it is flawless for any list. Define a real number r between 0 and 1. Denote r_n to be the nth digit of r. The digits of r are defined as follows: r_n = 5 if r_n = 4, otherwise r_n = 4. That cunstruction applies equally well to all lists without modification. If it is flawless for any list it is flawless for all lists. And it is flawless for any list. That's what you're saying. I see where you are coming from and I won't bother answering such "flawless" logic anymore, so don't bother.
From: MoeBlee on 29 Sep 2006 20:12 Poker Joker wrote: > "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. Of course. But so what? No one has shown an enumeration such that every real number is in the range of the enumeration. It's simple: What we mean by a 'list of all reals' is a 1-1 function whose domain is countable and whose range is the set of real numbers. We show that for an arbitrary function f whose domain is countable, there is some real number that is not in the range of f. Therefore, for any arbitrary function f whose domain is countable, the range of f is not the set of real numbers. So, since f is arbitrary, there is NO function whose domain is countable and whose range is the set of real numbers. A fortiori, there is no 1-1 function whose domain is countable and whose range is the set of real numbers. That's just plain logic, plain mathematical reasoning. MoeBlee
From: Poker Joker on 29 Sep 2006 20:16
"MoeBlee" <jazzmobe(a)hotmail.com> wrote in message news:1159574564.041788.85490(a)k70g2000cwa.googlegroups.com... > Poker Joker wrote: >> "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. > > By 'any' we mean an arbitrary one. The way we talk about an arbitrary > object is to choose a variable not free in any previous line in the > argument nor free in the conclusion we will eventually draw and then > use the rule of universal generalization to draw our eventual > conclusion. So if we want to prove something about an arbitrary > enumeration of denumerable sequences of digits, we say, "Suppose f is > an enumeration whose range is a subset of the set of denumerable > sequences of digits" (and, of course, we will presume NOTHING about f > other than that it is an enumeration whose range is a subset of the set > of denumerable sequences of digits. > >> Also, if its true for ANY list, then it must be >> true for a specific list. > > If the property holds for any list, then, if there exists a list, then > the property holds of any such list that exists. > >> So if considering a single specific list >> shows a flaw, then looking at ANY (ALL of them) list doesn't >> help. > > If there is a specific list that does not have the property, then we > will not be able to prove that the property holds of an arbitrary list. > > But I don't know what specific list you think is "flawed". Nor do I see > what your point has to do with Arturo's point that we don't have to > adopt a reductio ad absurdum assumption, since we can just show for an > arbitrary list that it does not list every real number. > > MoeBlee By analogy, what you're saying is: For ANY x there is a procedure to find a y such that x/y = 1. Because we are using the verbage "ANY", we don't have to worry about special cases like when x = 0. That's how mathematicians work? Or are you just saying that you need not look at special cases when we don't want to? Or is it that if a special case is overlooked enough, then it no longer counts? |