Cantor's Diagonal?
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From: Peter Webb on 3 Feb 2010 01:15 "zuhair" <zaljohar(a)gmail.com> wrote in message news:0f20debd-10ad-47a1-b80f-56217fd668ee(a)c4g2000yqa.googlegroups.com... On Feb 3, 12:20 am, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > Can you explain this a bit better? What is the sequence? > > What does <> mean? > > Ordered pair. > > X={<O,0>,<H,1>,<O,2>,<H,3>,<O,4>,<H,5>,.....} > > another presentation might be like that > > X= O_0, H_1, O_2, H_3,......... > > which is simply X=OHOHOH....... > > __________________________________ > > If X=OHOHOH ...., then its not an infinite set; its not a set at all. > > Do you mean: > > X is the set of all sequences a_1, a_2, a_3 ... where each a_n is an > element > of {O,H} ? No > > So OHOH... is an element of X, as are OOOOO..., HHHH..., HOO... etc? No > > Ie, the set of non-terminating binary sequences with H=1 and O=0 ? > > If that is the case, why didn't you just say "the set of non-terminating > binary sequences" ? OHOHOH.... means O followed by H which is followed by O which is followed by H and so on infinitely Zuhair _______________________________ Well, I can form an injective map from X to N as follows: X -> 1.
From: Peter Webb on 3 Feb 2010 01:20 > > Cantors argument applied to this list shows R cannot be bijected with w+1, > or indeed any countable ordinal. Yes of course, BUT the Diagonal argument fails with other arrangements, that's my point. _______________________________ Yeah, and if you try and put peanut butter into your fuel tank your car will probably fail as well. Of course I picked an arrangement where I could use Cantor's argument; picking some arrangement where it doesn't work would be a complete waste of time.
From: zuhair on 3 Feb 2010 02:04 On Feb 3, 12:39 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > 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 Well no doubt I am ignorant of math, that's something that I admit repeatedly. I have a lot of personal arguments, yes, no doubt, they are insignificant , also yes not doubt. However I see that my point (if it can be called as a point) is missed really. Most of the answers here were in my personal opinion irrelevant. I'll try to illustrate my point in a Naive manner. Cantor's argument DOES show that every injection f from N to S *provided that N is ordered in the natural way* and S is the set of all infinite binary sequences, then f is not surjective. I have put an emphasis above on the clause "when N is ordered in the natural way" what I mean by that is N is ordered such that 0 is the first of all members of N then followed by 1 then followed by 2 then by 3 ,.... so this natural order have the following properties (1) there is a starting member, i.e. a member that is not the successor of any member. (2) there is an immediate successor member for each member (3) Except the starting member every member have an immediate predecessor member.. (4) No two members have the same successor. So when N is thrown into this natural order, then if we compare N with S , then we can define a diagonal, and the diagonal argument works, since every injection from N (when thrown into this order ) to S cannot be surjective, because the counter-diagonal would be an infinite binary sequence that is in S but not in the Range of the injective function f from N(as ordered above) to S. But my question is: Is that enough to show that S is uncountable? We only compared S with a certain arrangement of N, is that sufficient to prove that there is not bijection between S and N. It definitely prove that there is no bijection between S and N as N is put in the natural order, but does that mean that we cannot have a bijection between S and N if N is put in another order. It seems that it does really, but I want the proof of that. For example lets rearrange N into an ordinal arrangement that is similar to the set w+1 Lets put the 0 to the end of all natural numbers, so lets order N in the following manner N={1,2,3,.....,0 } Now lets try to find a diagonal with N thrown into that arrangement. ... 0 , 1 , 2 , 3 , ... 1 H , O , O , H ,.... 2 O , H , H , O ,.... 3 H , H , H , H ,.... 4 O ,O , H , H ,.... .. .. .. .. 0 O, O , O, O ,... One can easily see that one cannot define a diagonal here, since 0 is the Omega_th member of N, and there is no Omega_th entry in any infinite binary sequence, so one cannot have a diagonal argument with N thrown into this arrangement. My question is: in the later situation if we don't have a diagonal, then how can we be sure that there cannot be a bijection between S and N. My point is that if we cannot define a diagonal for all arrangements of N, so how can we know from the diagonal argument per se, that no bijections are possible from N to S. Definitely there is something I am missing. Zuhair
From: Marshall on 3 Feb 2010 02:20 On Feb 2, 11:04 pm, zuhair <zaljo...(a)gmail.com> wrote: > > Lets put the 0 to the end of all natural numbers, so lets order N in > the following manner > > N={1,2,3,.....,0 } That's not a sequence. Marshall
From: Jesse F. Hughes on 3 Feb 2010 06:35
Marshall <marshall.spight(a)gmail.com> writes: >> X= {<O,i>,<H,j>| i is an even natural, j is an odd natural} > > Can you explain this a bit better? What is the sequence? > What does <> mean? I would expect a "sequence" to be > a mapping from the naturals to some target set. <,> is ordered pair. Just reverse the ordered pairs he wrote down and you'll see what sequence he means: X = { <0,O>, <1,H>, <2,O>, .... } I assume he accidentally wrote the ordered pairs backwards. > > >> so X is just an example of a infinite binary sequence. > > I don't see how. -- Jesse F. Hughes "Philosophy, Socrates, if pursued in moderation and at the proper age, is an elegant accomplishment, but too much philosophy is the ruin of human life." -- Callicles, in "Gorgias" |