Date: Mar 24, 2013 5:35 PM
Author: Virgil
Subject: Re: Matheology � 224

In article 
WM <> wrote:

> On 24 Mrz., 11:19, William Hughes <> wrote:
> > On Mar 24, 11:04 am, WM <> wrote:
> >
> > Your proof covers all lines.  We have for all lines l
> > of the list.
> >
> >   if l and all its predecessors are removed
> >   and no other line is removed,
> >   then the union of all lines is not changed"
> >
> > However, there is no information about what will
> > happen if you try to apply this to two
> > lines e.g. l along with all its predecessors
> > and m along with all its predecessors.

> Nothing will "happen". Induction holds for every line, so it holds for
> all lines of the set of finite lines.

> >
> > Now it is easy to see what will happen in this
> > case.   Since we can replace l and m with
> > one of either l or m, we know what will happen
> > if we remove two lines.
> >
> > Since we can replace l,m and p with
> > one of either l or m or p, we know what will happen
> > if we remove three lines.

> This way is not necessary, since the proof holds for every line
> including all its predecessors.

The theorem does not cover what will transpire when two or more lines,
along with all their predecessors, are removed.

So it is of some interest to note that for any set of lines having a
maximal line in it, removing each line in the set and all its
predecessors has the same effect as removing the maximal member and all
of its predecessors.
> >
> > It is easy to see we know what
> > will happen if we remove a natural
> > number of finite lines.
> >
> > However, we do not know what will happen
> > if we remove an infinite number of
> > finite lines.

> That's why we use induction.

Except that no inductive argument will go from removing a finite set of
lines to removing an infinite set of lines, at least not outside

> Induction holds for all elements of the
> infinite set of natural numbers.

Only if used properly, a skill that WM's claims show clearly that he
does not have.