Re: Orlow cardinality question

Virgil said:
> In article <MPG.1d24c44e27a77aa6989eb2@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
>
> > imaginatorium@xxxxxxxxxxxxx said:
> > > Tony Orlow (aeo6) wrote:
> > > > Virgil said:
> > > > > In article <MPG.1d20c4e04d6ab043989e46@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > > > > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> > > > >
> > > > > > > Such an assumption is your idea that there cannot be an infinite
> > > > > > > set
> > > > > > > of finite elements, for example. Or that the limit of a sequence
> > > > > > > must
> > > > > > > somehow become a part of that sequence.
> > > > > >
> > > > > > But I did not ASSUME there cannot be an infinite number of finite
> > > > > > naturals. I
> > > > > > proved it mathematically
> > > > >
> > > > > Those proofs require assming the equivalent of of twhat they purport to
> > > > > prove.
> > > > >
> > > > That's simply not true. Respond to them specifically, without snipping
> > > > and
> > > > otherwise amngling them.
> > >
> > > OK, I will amngle not.
> > I bet you don't even know what amngling is! ;)
> > I guess it's a sort of relf-seferential term.
> >
> > > But anyway, three is too many. If it were we who
> > > were the idiots, and you were trying to explain something, you might
> > > reasonably try three different explanations, but if you claim you have
> > > proofs in the normal mathematical sense, one is enough. If the proof is
> > > valid, the existence or otherwise of other proofs is irrelevant. So
> > > *you* choose one of your proofs, and I (prolly others too; why _do_
> > > cranks attract such attention? why is it all just so tantalisingly
> > > fascinating?) will point out the first error, without mangling or
> > > anything else. Then if you wish to you can revise it and try again, or
> > > just expand the claim that we show is false.
> >
> > Would that that were sufficient. But, since they all get rejected for reasons
> > that don't pan out, I have produced three. If three proofs from different
> > areas
> > all agree, that does bolster the argument. So, you want to hack at one at a
> > time? Okay. Try the information theory one:
> >
> >
> >
> > Given a set of symbols with size S, we can produce a set of all strings using
> > those symbols that have length L, and the size of this set will be S^L.
> > Digital
> > number systems fall into this category, with S being the number base, which
> > is
> > always finite (2 for binary, 10 for decimal).
>
> > If we want to have an infinite
> > set of digital strings, therefore, S^L needs to be infinite, but S is finite,
> > so L, the length of the strings, needs to be infinite to have an infinite set
> > of such strings.
>
> 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.
>
> 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. N=S^L => (finite N <=> finite S and finite L)
and (infinite N <=> infinite S or infinite L).
>
>
>
> >
> >
> > So, where in this argument do you see an error? Can S^L ever be infinite and
> > produce an infinite number of strings, without infinite S and/or L? Keep in
> > mind the relation to trees. S is the number of branchings at each node. L is
> > the number of levels in the tree, or branchings in each path.
>
> 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! 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?
>
> 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, now does it?) conflated to finiteness, is at odds with other
areas of math. Are they all wrong? Can you have finite S and L and infinite
S^L?
>
> 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.

--
Smiles,

Tony
.

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