Cantor's Diagonal?
in [Math]
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From: Arturo Magidin on 3 Feb 2010 00:36 On Feb 2, 10:18 pm, zuhair <zaljo...(a)gmail.com> wrote: > Back again to my point which is: the Diagonal argument do not seem to > constitute a proof like the other argument of uncountability of N. > > Let me clarify, suppose you had the Diagonal argument working for all > countable sets and in all arrangements of these countable sets, i.e. > suppose we could have found a diagonal always with any countable set > like w+1, w+2 ,.etc, then I would say that this is an argument general > enough to make me conclude that the set of all infinite binary > sequences is uncountable, So... the standard here is not "correctness", but "enough to force you to admit it"? > but Cantor's Diagonal argument do not work > like that, and I showed that we CANNOT have a diagonal with w+1, You showed that *you* couldn't come up with a diagonal. The standard, again, seems to be personal uniwllingness to accept something. >the > diagonal only works with sets that are isomorphic to w, Your definiition of "works" seems to be as surjective as your definition of "proof". > this make this > argument restricted and not general enough to conclude something like > uncountability of the set of all infinite binary sequences. Your definition of "restricted", "general", and "enough to conclude" seems to be uttelry subjective (and utterly worthless). > Anyhow I must be mistaken really. Ya think? If alpha is any countable ordinal and your "arrangement" is a map f:alpha --> B (where B is the set of binary sequences), then let t:omega-->alpha be a bijection between omega and alpha. Then ft is a map from omega to B. Define "the diagonal" in terms of ft; that f is not surjective follows from the fact that t is bijective and ft is not surjective. If you original alpha was omega, and t is not the identity, the argument still holds. -- Arturo Magidin
From: Tim Little on 3 Feb 2010 00:38 On 2010-02-03, zuhair <zaljohar(a)gmail.com> wrote: > No, that is a different argument altogether from the Diagonal > argument. I was speaking specifically about the Diagonal argument and > not about the other proof, so don't mix the two arguments. I was trying to work out why you brought totally superfluous ordinals into the discussion. > I showed that we CANNOT have a diagonal with w+1, the diagonal only > works with sets that are isomorphic to w, this make this argument > restricted and not general enough to conclude something like > uncountability of the set of all infinite binary sequences. "X is uncountable" is defined as "there does not exist a surjection from N to X". The diagonal argument shows that every function f:N->2^N fails to be a surjection, which *by definition* proves that 2^N is uncountable. The End. - Tim
From: Arturo Magidin on 3 Feb 2010 00:39 On Feb 2, 11:26 pm, zuhair <zaljo...(a)gmail.com> wrote: > Cantor's claim is of *uncountability* of the set of all infinite > binary sequences, so the matter is not limited to showing the failure > of that with a specific arrangement of a specific countable set N. Of course it is. If A is *any* countable set, then by definition there is a bijection f:A-->N. If t:A-->B is any map from A to the set of binary sequences, then since there is no bijection from N to B then tf^ {-1} (composing right to left) is not surjective. Since f^{-1} is bijective, nonsurjectivity of tf^{-1} implies that t is not surjective, thus proving that there can be no surjection from A to B either. > I am attempting to disprove Cantor's claim of course, by showing that > it is limited to specific situation, by showing that it is not general > enough to make such a claim of uncountability. By showing that, after so many years, you can still exhibit an incredible amount of projection and an overeliance on the Argument From Personal Ignorance. -- Arturo Magidin
From: Tim Little on 3 Feb 2010 00:43 On 2010-02-03, zuhair <zaljohar(a)gmail.com> wrote: > However I do really think that I am mistaken, but I don't know were > is my mistake exactly. You seem to have lost sight of the definition of countability. - Tim
From: Virgil on 3 Feb 2010 00:49
In article <38df2f48-b006-4005-ba49-e21d47a154b3(a)b2g2000yqi.googlegroups.com>, zuhair <zaljohar(a)gmail.com> wrote: > Hi all, > > I have some difficulty digesting the diagonal argument of Cantor's. > > The argument is that the set of all infinite binary sequences cannot > have a bijection to the set of all natural numbers, thereby proving > that the former set is uncountable? > > However the argument looks to me to be so designed as to reach to that > goal? > > One can look at matters from an alternative way such as to elude > Cantor's conclusion! > > Examine the following: > > Lets take the infinite binary sequences of the letters O and H > > so for example we have the sequence > > X = OHOHOH........ > > in which O is coupled to the even naturals and H coupled to the odd > naturals. > > so the sequence above is > > X= {<O,i>,<H,j>| i is an even natural, j is an odd natural} > > so X is just an example of a infinite binary sequence. > > However lets try to see weather we can have a bijection between > the set of all infinite binary sequences and the set w+1 > which is {0,1,2,....,w} > > so we'll have the following table: > > 0 , 1 , 2 , 3 , ... > 0 H , O , O , H ,.... > 1 O , H , H , O ,.... > 2 H , H , H , H ,.... > 3 O ,O , H , H ,.... > . > . > . > . > > w O, O , O, O ,... > > > Now according to the above arrangement one CANNOT define a diagonal ! > since the w_th sequence do not have a w_th entry > to put H or O in it. > > So if we can have a diagonal then this would be of the set of all > infinite binary sequences except the w_th one, i.e. of the following > > 0 , 1 , 2 , 3 , ... > 0 H , O , O , H ,.... > 1 O , H , H , O ,.... > 2 H , H , H , H ,.... > 3 O ,O , H , H ,.... > . > . > . > . > > Suppose that the diagonal of those was D=HHHHH.... > i.e. D={<H,n>| n is a natural number} > > Now the counter-diagonal would be D' = OOOO... > i.e. D' = {<O,n>| n is a natural number} > > However there is nothing to prevent the w_th infinite binary sequence > from being D' ? > > So neither we can have a diagonal of all the infinite binary > sequences, nor the diagonal of a subset of these sequences would yield > a successful diagonal argument such as to conclude that the set of all > infinite binary sequences is uncountable? > > Thereby Cantor's argument fail in this situation! > > What I am trying to say is that this Diagonal argument seems to be > purposefully designed to reach the goal of concluding that > the set of all infinite binary sequences is uncountable, by merely > selecting a particular bijection with the set {0,1,2,3,....} > in a particular arrangement, such as to make a diagonal possible, such > as to conclude the uncountability of these infinite binary sequences, > While if we make simple re-arrangement like the one posed above then > this argument vanish! > > There must be something wrong with the way I had put things here, but > I would rather want to read the full proof of this diagonal argument > in Zermelo's set theory. > > Zuhair IIRC, the Cantor arguments proves "given ANY ennumeration of binary sequences there are binary sequences not included in it." And it does this by showing that if one assumes that one HAS an ennumeration, one can construct binary sequences not included in it. I do not see that what you say above is relevant to that argument. |