Re: Orlow cardinality question

In article <MPG.1d29f221e6122cb9989eeb@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:

> Virgil said:

> > In the above TO makes the assumption that a set of finite strings
> > cannot be infinite. The set S = {"x", "x", "xxx", ...} is made up
> > only of finite strings, each string being 1 character longer than
> > its predecessor. Unless TO can prove that such a set has a longest
> > string in it, one can construct the Function on S which appends an
> > character to each member and thus injects S into a proper subset of
> > itself. Thus the set S of finite strings is Cantor_infinite, even
> > if it were TO_finite.

> Yep Cantor screwed up.

Since everyone else is distinguishing between finiet sets and infinite
sets by Cantor's criterion of whether the set allows injection to a
proper subset, if TO is using a different standard, he should not keep
it secret as he has been doing, but publish it here and now>

What is your criterion for deciding whether a set if finite or
infinite, TO?

Your failure to give us any other criterion leaves us to assume
Cantor's is the only valid one.
> >
> > Thus TO's assumption that an infinite set of strings must contain
> > an infinite string is shown to be false, at least by the Cantor
> > standard of finite versus infinite for sets.
> An infinite set of strings constructed from a finite set of symbols
> must include infinitely long strings.

Often claimed, never proved, and frequently disproved.

N=S^L => (finite N <=> finite S and finite
> L) and (infinite N <=> infinite S or infinite L).

Wrong! If S and L are both naturals, then S^L is also and the set of
all S^L's in not bounded any more than N is bounded.

If TO claims a bound on N or on S^L, let him display it, proving
(1) that it is a natural number and (2) that there is no larger
natural number.

N is finite if and only if {S^L : S in N and L in N} is finite, and
neither of them is finite, since each allows injection into a proper
susbset.

> > TO assumes, possibly because it holds for any finite set of
> > strings, one can find a longest string, allowing TO to identify an
> > L, that the same must hold for infinite sets of strings, which is
> > esssentially assuming his result, that a set of all finite strings
> > must be finite.


> Hey, you can have an infinite set of finite strings if you want, you
> know. Just use an infinite set of symbols!

I can do it with only one character, as long as there is no limit to how
many times I can use it.

That's actually related to
> something I put forth just recently. Can you guess what it was?
> >
> > 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.

> Well isn't that special? I am so happy for you!
> >
> >
> > 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.

> You "observe". Is this biology? Why don't you try PROVING that every
> finite set MUST have a first and last element?


I have proved that every ordered set allowing an injection into a proper
subset has at least one subset which lacks either a first of a last
member, and if the set is well-ordered as are tne naturals it, some
subset lacks a last member. By contraposition it follows that every
non-empty subset of every finite ordered set must have a first and last
member.
> >
> > 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 yet, that "reasoning", based on first and last members (which
> doesn't allow for circular sets

Circular sets are not ordered sets, so TO's objection is irrelevamnt to
the actual issue.


> >
> > And unless someone can prove that such a union contains a last or
> > largest member, it satisfies the Cantor definition of not-finite,
> > too.
> >
> No, we simply deflate your "largest finite" into the pile of
> stretched rubber it is.
It is TO's "largest finite natural", since no one of any sense believes
there is any limit, in the naturals or in the reals which contain them,
on the size of naturals.
.

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