Re: Orlow cardinality question

In article <MPG.1d2a0af4f113032d989eec@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:

> Virgil said:

> > We claim that the number of members in an ordered set is finite if and
> > only if every non-empty subset of it, including the set itself, has a
> > first member and last member relative to that ordering.
> >
> >
> > We observe that all of the sets {1}, {1,2}, {1,2,3}, ...,{1,2,3,...,n},
> > which have obvious first an last members, are finite by this definition,
> > and we call each of their last members a finite natural.
> >
> > We claim that the union of all such sets is a set N which has no last
> > member, and is, by the above definition, not a finite set.
> >
> > And unless someone can prove that such a union contains a last or
> > largest member, it satisfies the Cantor definition of not-finite, too.
> >

> Well, obviously none of this satisfies me,

Fortunately for the future progress and health of mathematics,
satisfying someone like TO is not an object.

> as I have said, since the Cantor definition takes back seat to actual
> axioms of arithmetic and value.

TO still has to back up his sayings with anything that a mathematician
can take seriously before they are anything more than pipe dreams.




> Is it your position that if a set has a first and last member, that
> it is finite, or only that a finite set must have a first and last
> member?

It depends on the type of ordering. If the ordering is discrete or
sequential, meaning that each member except a last one has a unique
successor in the set and each member except a first one has a unique
predecessor, then yes!

But if one allows dense orders, or other non-discrete orders, then no!

It is certainly true for any subset of the naturals, but not for such
sets as closed intervals of reals, like [0,1].
.

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