Re: R: R: R: Why Cantor Was Wrong

In Message-ID: <37af0e45.5458548@news.connix.com> dennis@sculptware.com (Dennis Paul Himes) wrote >cattabri@mailbox.dsnet.it (Cattabriga) wrote: >> Let g:a -> P(a). We will construct a subset b of a >> that is not in ran g. Specifically let >> >> i) b = { x in a | x notin g(x)}. >> >> Then b subseteq a, but for each x in a >> >> ii) x in b <-> x notin g(x). >> >> Hence b neq g(x), > > Or, more precisely, Ax (b != g(x)). (Using != to mean "not equal to".) >The proof of this from ii) has been left off, probably because it's so >obvious, but what you write below makes be suspect that you don't understand >it, so I'll supply it. Why do you give lesson if you are not able to distinguish a proof for contradiction from a constructive proof ? > Prove: Ax (b != g(x)). > Assume its negation: Ey (b == g(y)) > Either y is in g(y) or y is not in g(y) > Case 1: y in g(y) > Since b == g(y), y in b. > Since y in g(y), y not in b (by definition of b). > (y in b) & (y not in b) is a contraditcion. > Case 2: y not in g(y) > Since b == g(y), y not in b. > Since y not in g(y), y in b (by definition of b). > (y not in b) & (y in b) is a contraditcion. > Since either case leads to a contradiction, the assumption is false. > Since its negation is false, Ax (b != g(x)), QED > There are two classical expositions of Cantor's argument: one for contradiction and a constructive one. What irony! I began this discussion with an attack to a proof for contradiction, and all the answers ordered me to forget the proof for contradiction and any assumption toward a contradiction like yours above, and I was leaded to talk only about the constructive argument. Now you say that the right exposition for Cantor's 'theorem' is the proof for contradiction. The two different expositions are always running after each other, so that what the one doesn't say surely is stated by the other one. I think that the two arguments are distinct from the logical point of view see my posting Subject: Why Cantor Was Wrong Author: Cattabriga <cattabri@mailbox.dsnet.it> Date: 27 Jul 99 08:35:02 -0400 (EDT), and they should not be mistaken one each other. In such posting I clearly distinguished Version I, the proof for contradiction, from Version II, the constructive argument above. I think that the melting of "not(B=g(b))" in Version I with "b neq g(x)" in Version II is scientifically incorrect. Your > Prove: Ax (b != g(x)). is not properly like the constructive argument "given a set A and a function g:A -> P(A) there is a B such that B subseteq A and it is not a member of the range of g". (Please observe the two distinct versions have distinct notations, and usually I don't like any blending) >> iii) is the relative complement of b in a. Its existence is ensured >> by the axiom of subsets. In fact we are always allowed to assert >> AxAwEyAz(x in y <-> x in w & x notin z) whenever the set w is given, >> and z is a subset of w. This set y is the relative complement of z in >> w. > > No one disputes that c exists. We just fail to see your point in >showing that it exists. This is a very strange wording. In ZF we have AxAwEyAz(x in y <-> x in w & x notin z) and we can state it as *) |-ZF AxAwEyAz(x in y <-> x in w & x notin z). which is by itself enough to state the existence of "c". There is no reason to ask to someone of '''showing that it exists''', unless you have gone out of ZF. *), such as every instance of the axiom of subsets, can always be stated everywhere in ZF. If you deny the possibility to state *) in ZF then you have created your personal subsystem which is not ZF. Even if I don't like your notation we could for example state *) along your 'proof' as follows: > Prove: Ax (b != g(x)). ran g subseteq P(a), then I guess that your "x's" are all subsets of a, and that your "Ax" can be understood "for all subsets x of a", > Assume its negation: Ey (b == g(y)) and that your "Ey" can be understood "exists a subset y of a". Later you will speak of the definition of b then I guess you mean b = { x in a | x notin g(x)}. Since *) |-ZF AxAwEyAz(x in y <-> x in w & x notin z) we can then state Ax(x in c <-> x in a & x notin b) i.e. c = { x in a | x in g(x)}. Do you need some explanation more? Of course all the elements of a that are not in b are all those elements x of a such that x in g(x). > Either y is in g(y) or y is not in g(y) This statement have two interpretations: i) for all subsets y of a, either y is in g(y) or y is not in g(y). ii) exists a subset y of a such that either y is in g(y) or y is not in g(y), and b is the only subset of a such that b == g(y) . From ii) of course follow yours cases 1 and 2. But let us consider i). Since all the elements of a that are not in b are all those member y of a such that y in g(y), i.e. for each y in a y in c <-> y in g(y) then, for each y in a c == g(y). By its definition c is a subset of a. i) for all subset y of a, either y is in g(y) or y is not in g(y). > Case 1: y in g(y) > Since b == g(y), y in b. > Since y in g(y), y not in b (by definition of b). > (y in b) & (y not in b) is a contraditcion. Case 2: y in g(y) Since c == g(y), y in c. Since y in g(y), y in c (by definition of c) and no contradiction follows. > Case 3: y not in g(y) > Since b == g(y), y not in b. > Since y not in g(y), y in b (by definition of b). > (y not in b) & (y in b) is a contraditcion. Case 4: y not in g(y) Since c == g(y), y not in c. Since y not in g(y), y notin c (by definition of c). and no contradiction follows. Not all cases lead to a contradiction. And if the assumption was the statement "no subset of a is equal to g(y)", no conclusion can be drawn. Paola Cattabriga ______________________________________________________________ http://www.serdata.it/cattabriga/ ______________________________________________________________