Date: Mar 24, 2013 7:22 AM
Subject: Re: Matheology § 224

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. Therefore it holds for l, m, and p
(because they belong to a set of predecessors. This is a property of
the natural numbers and dos not require any further attention.

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

Regards, WM