Re: infinity

In article <MPG.1d7d39827adcb13298a17c@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:

> Virgil said:
> > In article <MPG.1d75587e488473a98a151@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> >
> > > Virgil said:
> > > > In article
> > > > <MPG.1d73c56abbd0d05498a11a@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > > > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> > > >
> > > > > Virgil said:
> > > > > > In article
> > > > > > <MPG.1d6d6d461dbe3aa498a0fc@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > > > > > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> > > > > >
> > > > > > > > What, nothing but numbers can be elements of an
> > > > > > > > infinite set?
> > > > > >
> > > > > > > When working with infinite sets, there is really no way
> > > > > > > to compare the sizes of sets without resorting to
> > > > > > > properties of their elements.
> > > > > >
> > > > > > There is, but TO doesn't like to admit it. Bijections and
> > > > > > injections work without reference to any properties except
> > > > > > those already needed to determine the sets themselves.
> > > >
> > > > > Yes, the properties of the elements used in defining the set.
> > > >
> > > > Then, according to TO, the set {1,2,3} and the set {a,b,c}
> > > > cannot be compared by looking merely at positions in the
> > > > listings of members, but depend in some way on the actual
> > > > nature of the members themselves?

> > > Of course finite sets are easily compared. Infinite sets require
> > > some mapping between members to be compared. But, you already
> > > knew that.
> >
> > Finite sets also require such bijective mappings, any two order
> > isomprphic sets are naturally of the same "size", and with listings
> > of members there is a natural mapping between sets which is either
> > an injection of a smaller into a larger or a bijection between two
> > of the same size.

> Yeah well that's trivial.

Then why is TO having so much trouble with it?
> >
> > > > > >
> > > > > > > Numeric sets are one of the most common, but discussions
> > > > > > > of infinite sets also involve structures such as trees,
> > > > > > > processors such as Turing machines, and infinite systems
> > > > > > > such as symbolic languages, which include symbolic number
> > > > > > > systems.
> > > > > >
> > > > > > > When drawing a bijection using an arithmetic formula, we
> > > > > > > are dealing with sets of quantities, and inverse
> > > > > > > functions are the way to compare those kinds of sets.
> > > > > >
> > > > > > If the functions are bijections, no inverse functions are
> > > > > > needed.
> > > >
> > > > > The inverse functions are what cause them to be bijections.
> > > >
> > > > That they are bijections is what causes them to have inverses.
> >
> > > No kidding. Same thing.
> >
> > Not quite. There are lots of functions which are not bijections, so
> > that bijection comes first.

> That would make bijections a subset of functions.

Precisely correct, for a wonder. TO is so seldom corRect, that we
should take special not when he is!
> > > >
> > > > > The inverse of the function which generates a set from
> > > > > another describes it's size relative to the other.
> > > >
> > > > If there is an inverse, they are of the same size
> > > > (cardinality).
> >
> > > If there is a bijection, then they are the same cardinality. If
> > > there is a bijection from the naturals to a set, then the inverse
> > > of the function that forms the bijection determines the size of
> > > that set relative to N.
> >
> > TO can't even count finite sets!
> >
> > If there is a bijection from N to any image set then the image set
> > is of the same size as N. If there is a surjection from N that is
> > bijective on some initial segment of N, only then can the image set
> > be smaller, and then it is always finite of cardinality equal to
> > that initial segment of N.

> You obviously misunderstood.

I may easily have misunderstood what TO meant, but I understood quite
correctly what he said, and what he said was wrong.
> >

> > If either the number of characters in the alphabet or the string
> > lengh of tha allowed words is without finite limit, then the number
> > of words in the language is also without finite limit.


> If you either allow an infinite alphabet or infinite strings? Which
> of those do you allow?

An infinite set of numbers may have no finite upper bound, at least in
standard set theory, without containing infinite members. TO's
confusion on this issue is the basis of his confusion on many other
issues, too.

> >
> > ONLY if both the number of characters in the alphabet AND the
> > string lengh of tha allowed words have finite limits, will the
> > number of words in the language is also have finite limit.

> Did you not say that all the strings are finite in length, and do we
> not have a finite alphabet?

That each string in a set of strings is finite in length does not
require that there be any finite upper bound on the lengths of the sey
as a whole.


TO is repeating by implication here his demonstrably false assumption
that the set of (finite) naturals must be finite, even though it is
provably infinite, at least in the Cantor sense.

> >
> > If TO wishes to require a specific finite limit to the size of the
> > character set AND a specific finite limit to the allowable word
> > length, only then will he convince anyone that the corresponding
> > word list is finite. But that is only for that limited word list,
> > and not others.

> Only then can we specifically know the finite maximum size of the
> set, but if the alphabet and all strings constructed from it in the
> language are finite, then the language is also finite.


WRONG! AGAIN! In any system in which the Peano postulates are valid, TO
is wrong, persistently wrong, and stupidly wrong!
>
> Now, I asked for a formula, O Virgil. Where is your formula?

My formula is that Peano posutlates which gualrantees that the set of
(finite) naturals is, by Cantor's criterion, infinite.

Thus the set of allowable finite string lengths, unless it can be shown
to have some finite maximum member, is infinite, and thus the language
is also infinite.
> > > > > >
> > > > > > Only for computer languages for which there is always some
> > > > > > limit on both S and L. For non-computer languages, there is
> > > > > > no inherent limit on either S or L, so no inherent limit on
> > > > > > N. That TO limits himself does not mean that everyone is
> > > > > > similarly limited.
> > > >
> > > > > If you limit your words to a finite length while using a
> > > > > finite alphabet, then you have limited the language as a
> > > > > whole to a finite maximum size.
> > > >
> > > > See, TO is doing it again. WRONG AGAIN, TO!
> >
> > > Actually I am exactly right, and you offer no alternative formula
> > > or interpretation.

TO will only be right when he can PROVE, without any of his many
outside asumptions, that the sets guaranteed by the Peano postulates,
whose members each involve only finitely many successor operations, are
finite (do not allow any injection into proper subsets).

Since this can never happen, TO is not right now, and bodes well never
to be.




> > > You can way "wrong" till your teeth fall out, but your
> > > declarations don't convince anyone of anything.
> >
> > They convince everyone but TO that TO is wrong, though by now few
> > need much in the way of convincing evidence.

> Those declarations don't convince even your best friends of anything
> they don't already think. In fact, I can see in the months since I
> started talking here, that there is more and more discussion of this
> topic, and more and more discontent with the standard theory.

There have always been those who have had reservations about the set
theory of ZF, but they are mostly too smart to say that once one
assumes ZF, the conclusions that follow are wrong.

Since mathematics, regardless of which version of set theory one works
with, has nothing to do with physical reality, one can choose any set
of axioms one wants.

Mainstream mathematics chooses ZF of ZFC or some similar variation.

TO doesn't want to play the game by those rules, and that's OK.

But TO also doesn't want to let anyone else play by those rules, and
that is definitely not OK.


> > One maximal binary tree works for both. That TO sees it as wrong is
> > his error.
> No, it is exactly how you tried to prove your lie.

That TO does not understand it, and he clearly does not wish to
understand it, is insufficient to demonstrate its falsehood.

Those with the brains to understand it have raised no objections, only
those who have misrepresented it, either deliberately or out of
ignorance as to what it actually says.
>

> > I mapped bijectively each branch to a finite string of binary bits
> > representing a bijection between the set of branches and N.
>
> No, you mapped each branch to a bit in a path denoting a string of
> bits, which you claimed to be finite, even though you said the tree
> was infinite. What would a path represent on this tree? Let's
> see.....
>
> > I mapped paths from the same tree to sets of naturals so as to
> > establish a bijection between them and the power set of N.
>
> Now you are trying to change things, but fine, let's examine what you
> are now proposing. You claim to have branches mapped to all the
> naturals in a 1-1 correspondence, and therefore countable. Then you
> claim that the set of paths is the powerset of that set, and
> therefore uncountable? You really should draw pictures before writing
> words. Then you won't waste 1,000 of them sounding stupid. Here's
> your tree, with each number being a branch/node:
>
> .
> 1 2
> 3 4 5 6
> 7 8 9 10 11 12 13 14
>
> Which branch contains the set {1,2,8,9}? None? How can this be? Is
> there really a bijection between the paths of this tree and the
> powerset of the naturals? Absolutely not. Each path will start with
> either 1 or 2, so one of those will ALWAYS be a member of any path.
> For any given subset defined by a path in this tree, the next largest
> element to any n in the subset is either 2n+1 or 2n+2, not any
> arbitrary m>n. This is obviously not corresponding to the power set
> of N, but of a set of subsets which, given proper attention, can
> undoubtedly be shown to be exactly N/2, if there are N branches,
> since this is the correct answer.
> >
> > > Using the first, you "proved" that the branches are "countable",
> > > and using the second you "proved" that the paths are
> > > "uncountable".
> >
> > Precisely!

> Ummmm.....didn't you just deny that you used two different trees, and
> claim they were the same, above?

No! I used one maximal binary tree, which has both branches and paths. I
showed that the branches biject with the naturals and the paths bijecte
with the power set of the naturals,


> > As no legitimate refutation is possible, TO is wise not to attempt
> > one.
> Look at my refutation of your nonsense above, and correct me if I'm
> wrong.

You have made no refutation!

You are not even wrong!
> > > >
> > > >
> > > > > which tells me you have proven nothing with it.
> > > >
> > > > TO's "understanding" tells him only what he wants to hear,
> > > > which never includes the fact that he is so often wrong.
> >
> > > If you say so, Bergil.
> >
> > While I do say so, I am not by any means the only one to do so.

> I know, Mommy agrees with whatever Virgie says.

I note that very few corrections or objections appear in this thread to
my posts, other than by TO, but the objections and corrections to TO's
posts are manifold.
.

Relevant Pages

  • Re: infinity
    ... >>> members, but depend in some way on the actual nature of the members ... > injection of a smaller into a larger or a bijection between two of the ... > If either the number of characters in the alphabet or the string lengh ... One single maximal binary tree for both bijections. ...
    (sci.math)
  • Re: infinity
    ... >>> that bijection comes first. ... >>> If either the number of characters in the alphabet or the string ... >> If you either allow an infinite alphabet or infinite strings? ... >>> One maximal binary tree works for both. ...
    (sci.math)
  • Re: An uncountable countable set
    ... which infinite whole numbers are a subset. ... "infinite naturals" are members of N? ... the string up to and including that bit ...
    (sci.math)
  • Re: infinity
    ... > Did you have a specific objection to the mapping through the infinite bit ... in his "construction" of that bijection that for each ... infinite string of binary digits there is another such string with ... >>> Maybe as far as cardinality goes. ...
    (sci.math)
  • Re: abundance of irrationals!)
    ... >> If a bijection between bit numbers and a partial set of them is ... then we see that the set is infinite. ... between an unlimited string of digits and an infinite string of digits. ...
    (sci.math)
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