Re: CANTOR's theorem

In article <1117636759.340243.148750@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> *** T. Winter wrote:
> > In article <1117355736.611070.264940@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> > > *** T. Winter wrote:
> > > > Did you not see my article where I defined a large number of elements
> > > > of P(N) for any given map f from N to P(N), such that those elements
> > > > are not in the map?
....
> > But I will do it step by step and ask you to tell me which of my steps
> > is exactly wrong, and why:
> > 1: Let be given any injective map g from N to N and a map f from N to P(N).
> > 2: Define as H(g) the subset of N such that {h in H(g) if g(h) not in f(h)}.
> > 3: This is a proper subset of N. Now define M(g) = g(H(g)).
> > 4: Because H(g) is a proper subset of N (for all g), M(g) is also a proper
> > subset of N, and so an element of P(N).
> > 5: You can verify that none of the M(g) are in the image of f, and so f is
> > not surjective. Because:
> > 6: if f is surjective, M(g) must be the image of some element k of N.
> > 7: Suppose that M(g) is the image of some element k of N.
> > 8: The question now is: is g(k) in M(g)? This can not be answered, and
> > so we have a contradition. ( g(k) in M(g) iff g(k) not in M(g).)
> > 9: The only assumption made was in (7), so that assumption is false, M(g)
> > is not the image of some element k of N.
> > 10: And so the map is not surjective.
>
> I don't see what your mapping g is good for.

You asked me for a repeat of the proof I had of a big family of sets that are
not the image of some particular map. Pray pay attention; that is what the
mapping g is good for.

> The mapping f is the same as that of Hessenberg. The principle does not
> change, by introducing g.

That is not a reply. What particular step above is wrong?

> > So none of the M(g) are in the image of f. If I am wrong, tell me
> > exactly which step above is wrong. And do not come up with a different
> > map f, because that would make step 6 wrong.
>
> I said already that the triple {m, M, g} cannot exist.

That is no reply. What particular step is wrong?

> But perhaps here
> is a better argument to support my claim. I have not yet had the time
> to make it watertight, but it sounds promising. The best method to find
> a leak is to show it to you.

That the triple cannot exist is irrelevant (and yes, it cannot exist, that
is the whole purpose of the definition). But what particular step in my
proof is wrong?

> Define a relation (it is not a mapping, but that does not matter) from
> N to P(N).

I asked you to tell me what particular step in my proof is wrong. You
still evade the question. Pray tell me. What particular step in my
proof is wrong?

> In this relation each n e N is related to ("mapped to") two elements of
> P(N). For instance 1 --> {1} and 1--> {2,3}. In effect, N is used
> twice. But set theorists would argue that doesn't matter, because even
> twice the naturals are not enough to map onto P(N). But in doing so, we
> have no set of non-generators, because the same natural n can be a
> generator and a non-generator (like 1 in above example).

You are seriously hooked on the concept of non-generators. But I will play
M = {n in N | n notin f1(n) and n notin f2(n)}
is this set in your map?

> Can you prove the non-surjectivity of that set? Don't try it with 1a
> and 1b! It is the same number 1 which is related "mapped" to two
> elements of P(N).

Is the above sufficient? But pray tell me which step of my proof is wrong.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
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