On 25 Jan., 12:23, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > WM <mueck...@rz.fh-augsburg.de> writes: > > On 25 Jan., 01:39, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > >> WM <mueck...@rz.fh-augsburg.de> writes: > >> > On 24 Jan., 14:16, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > >> >> WM <mueck...@rz.fh-augsburg.de> writes: > >> >> > You will have have recognized that here the diagonal argument is > >> >> > applied. It is obvious that up to every line = column the list is a > >> >> > square. > > >> >> It is clear that, for all j, d(j) != t_j(j) and hence d != t_j. If > >> >> that's what you mean by the diagonal argument, great! > > >> >> Once again, however, you say something that has no clear meaning to > >> >> me. Can you clarify "It is obvious that up to every line = column the > >> >> list is a square?" I've no clue what it means. > > >> > Then ponder a while about the following sequence > > >> > d > > >> > d1 > >> > 2d > > >> > d11 > >> > 2d2 > >> > 33d > > >> > and so on. In every square there are as many d's as lines. The same > >> > could be shown for the columns. > > >> Yes, in this sequence of three squares, what you say is true. > > > Is there a first square where my observation would fail? > > >> But none of this is relevant, because we've explicitly defined the > >> anti-diagonal d and it is a triviality to see that it is an infinite > >> sequence of non-zero and non-nine digits. And this fact really has > >> nothing at all to do with limits of sequences of squares. It is all > >> perfectly explicit. > > > Here you again intermingle potential and actual. We are restricted to > > the domain of terminating decimals. If you cannot understand that, > > perhaps a formal argument may help: Assume that we are restricted to > > the well-defined set of terminating decimals. If you see any evidence > > that we should leave that domain, say "stop!". But only if you are > > sure. > > You're throwing about terms (potential, restricted to the domain..., > etc.)
In ZFC you have no domains? Isn't ZFC the fundament of analysis? How can analysis have domains then, if they are foreign to ZF?
> that have no obvious meaning in the theory ZF, so I see no > reason to reply to this.
You should try to learn what a domain is.
> We're trying to prove a certain claim in ZF, > so unless you can indicate how to interpret these terms, I have > nothing to say.
In an ordered set like the sequence of squares above, we have for every subset a first element. If you claim to know a square that is not a square, then there must be a first square that is not a square. That could be proved by AC if it was necessary, but it is not, since the set of squares is countable. However, the axiom of choice cannot be circumvented in ZFC. Every set can be well-ordered such that every non-empty subset has a first element.
> > Of course! Why not? Isn't every i in N finite? > > Great. > > Is t_i(i) also defined for every i in N?
Yes, of course, with absolute certainty. > > Assuming you will say yes, then I must ask: > > Is d(i) therefore defined for every i in N?
Yes, of course, with absolute certainty.
And nothing of that takes us out of the domain of all terminating decimals, because ;
> > In ZF every n in N is finite.
and with it every FISON.
> > I know. But it would be nice if you read it again and again. Or try an > > experiment: Write a long sequence of digits d_1, d_2, d_3, ... and do > > not stop. Are you in danger to leave the domain of finite sequences? > > If they have any meaning in the language of ZF, then simply tell me > what the meaning is.
If you deny that analysis is based upon ZFC, then let me know. Then I will be no longer interested in ZFC. But if ZFC is the fundament of analysis, then it should be clear in ZFC what a domain is.