Cantor Confusion
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From: MoeBlee on 29 Sep 2006 20:36 Poker Joker wrote: > 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's nonsense and has no bearing on uncountability proofs. MoeBlee
From: MoeBlee on 29 Sep 2006 20:41 Poker Joker wrote: > "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. I said no such thing by analogy or otherwise. > 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? No, you've got it reversed. If we say 'any' to mean an arbitrary one, then we don't assume that it falls under a special case or that it doesn't fall under a special case. If there are special cases such that the property does not hold of those objects that are a special case, then the universal genearlization WON'T go through. It is the VERY NATURE of the rule of universal generalization that it won't go through if there is even a single special case that precludes the generalization. You seem completely unfamiliar with such basic principles of mathematical reasoning. Why don't you read a book on the subject? MoeBlee
From: Ross A. Finlayson on 29 Sep 2006 20:44 MoeBlee wrote: > Poker Joker wrote: > > 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's nonsense and has no bearing on uncountability proofs. > > MoeBlee It helps to address all the "uncountability" proofs, or as I call them, rrresults, at once, all three of them. So I do. Why doesn't anybody answer me? They do. Ross
From: MoeBlee on 29 Sep 2006 20:45 Poker Joker wrote: > "Poker Joker" <Poker(a)wi.rr.com> wrote in message > news:WHiTg.25583$QT.7978(a)tornado.rdc-kc.rr.com... > > > > 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? > > Better yet: > > For ANY real number x there is a procedure to find a real number y, such > that x/y = 1. > y isn't a real number when x = 0. > Therefore, 0 isn't a real number. No, that's just nonsense. It is certainly not within the auspices of the logic of such mathematical arguments as those showing the uncountability of the reals. MoeBlee is just an example of the kind of argument that mathematical reasoning does NOT allwo.
From: Alan Morgan on 29 Sep 2006 20:42
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 if it's true for ANY list then it must be true for a specific list. So if considering a single specific list shows a flaw then perhaps that list doesn't really exist. It's as if, after hearing the proof that sqrt(2) is irrational, you reply "But what if I give you a and b, both integers, such that a/b = sqrt(2)? Your argument is demolished!". Yes, if you provide what the proof shows can't exist then the proof is wrong. But you can't just assume that that thing exists. Alan -- Defendit numerus |