In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> 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:
> > > Here you again intermingle potential and actual. We are restricted to > > > the domain of terminating decimals.
You may be but no one else is.
> > > If you cannot understand that, > > > perhaps a formal argument may help: Assume that we are restricted to > > > the well-defined set of terminating decimals.
Whyever assume that?
> > > If you see any evidence > > > that we should leave that domain, say "stop!".
> > You're throwing about terms (potential, restricted to the domain..., > > etc.) > > In ZFC you have no domains?
The domains of ZFC are sane, unlike the ones in Wolkenmuekenheim.
> > 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.
Since all of your sequences are finite, the first one in any list will have some finite length n, which will mean that you have no squares of size greater than that n by n, at which point your argument has crashed.
> That could be proved by AC if it was necessary, but it is not, since > the set of squares is countable.
If every sequence is finite, then WM's "set of squares" is finite also.
> 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.
But not the union of all FISONs.
And thus not necessarily the length of sequences. > > > > 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.
Analysis can be based on any number of set theories, it is just that ZF and ZFC are the most thoroughly studied of them.
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.
What is not clear is what a domain is in Wolkenmuekenheim. Nor what anything else is. --