Date: Feb 15, 2013 5:46 PM
Subject: Re: Matheology � 222 Back to the roots
WM <email@example.com> wrote:
> On 15 Feb., 00:44, William Hughes <wpihug...@gmail.com> wrote:
> > > > Two potentially infinite sequences x and y are
> > > > equal iff for every natural number n, the
> > > > nth FIS of x is equal to the nth FIS of y
> > So we note that it makes perfect sense to ask
> > if potentially infinite sequences x and y are equal,
> and to answer that they can be equal if they are actually infinite.
If they must either be equal or not equal, TERTIUM NON DATUR, then even
merely potentially finite sequences, if any such things really exist,
can be proved to be equal, for example by induction.
> But this answer does not make sense.
No one expects WM to make sense.
> You cannot prove equality without having an end, a q.e.d.
Then you must be rejecting induction, which specifically rejects any
> You only prove that we can go on an on and on. And that's potential
> infinity. But it is never proved to be infinite because there is no
> step inrto the infinite.
Induction does not distinguish whether the naturals are an actual set or
only one of WM's pretend sets. It works either way.
> > we have cases where they are not equal and cases
> > where they are equal.
> Potentially infinite sequences are never actually infinite.
Induction does not care which sort of infiniteness one has.
> means: d stretches from d_1 to the d_n of your choice. And exactly the
> same is in infinitely many lines. Of course you can choose whatever n
> you like (because that is the meaning of potentially infinite: you can
> choose whatever n you like).
Induction choses all of them, after the first, simultaneously.
> > We also note that no
> > concept of completed is needed, so equality can
> > be demonstrated by induction.
Not without accepting that the inductive process is complete.
> You cannot prove equality because you would need and end of file
Not using induction! Induction specifically avoids needing any
You can only prove equality up to every desired step.
Induction proves it for ALL n in |N.
> But it can also be stated by induction that d cannot be more than all
> its FISs which also are in the list. And you agreed to that.
But one does not have ALL its FIS's until one has all n's in |N.
> > > Do you claim that the list
> > > 1
> > > 12
> > > 123
> > > ...
> > > does not contain every FIS of d?
If d =214365... then your list contains none of those FIS's.
> > > Do you claim that there are two or more FISs of d that require more
> > > than one line for their accomodation?
In order to be two, they cannot be one.
> But you would be better off not to answer?
We might all be better off merely killfiling WM's nonsense, but he makes
such silly mistakes that we cannot give up the fun of finding them and
pointing them out.