MoeBlee wrote:
Tony Orlow wrote:
No, it all rests on the notions of identity and equality. As Leibniz
pointed out, when the properties of two objects are all exactly the
same, then they are the same object. So, when we say two numbers are
equal, that means all properties of the two are equal.

Ha! The fallacy of reversing implication right there! An example of
just about the most basic fallacy.

When you say "a=b", you say "A a A b P(a)=P(b)". The two are equivlent
statements, and therefore imply each other.

No, the indiscernibility of identicals does NOT imply the identity of
indiscernibles. You need both implications; you can't derive one from
the other. And, in first order logic, one direction can be posited only
in the semantics not in the axioms.

You prove two quantities equal by showing there is no difference, do you
not?

MoeBlee wrote:
Tony Orlow wrote:
Yes, and the universe is consistent by definition, so math should be
consistently overall as well.

Unless the universe is a set of sentences, the notion of consistency
does not even apply.

The universe is governed by the properties of the elements within it,
which properties are statements true about those elements.

There is NO system that will give you any kind of even minimal
mathematics without adopting axioms that are not true in all domains of
discourse.

I am not convinced of that.

If Ross and I each have our linearizations of the reals, then they
exist, independent of the axiom of choice (which might as well be called
the axiom of free will).

We prove existence of objects in a theory from axioms. You posit the
existence of objectw without any system of logic, primitives, or
axioms.

Logic is the primitive of argument. Logical truth is founded upon
quantity, and quantity upon geometry. Space is a priori.

Define 'finite'.

MoeBlee wrote:
Tony Orlow wrote:
Given a set x, can we always determine card(x)?

I'm guessing, but I'm pretty sure that there is not an algorithm that
will tell you the ordinal index of the aleph that is the cardinality of
any given set.

Good guess, except I have no idea what your driving at.

Virgil wrote:
In article <1156501377.098380.50700@m79g2000cwm.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

Virgil schrieb:


But as we just investigate consistency, you cannot presuppose it. With
your attitude it is impossible to find any inconsistency even in an
inconsistent theory. Deplorably you are too simple to recognize that.
If "Mueckenh" can deduce from any axiom system both a statement within
the system and its negation, "Mueckenh" will have found his
inconsistency.
Which part of my proof concerning the binary tree is not in accordance
with the ZFC axioms in your opinion?

The part that says a bijective image of the naturals bijects with a
bijective image of the power set of the naturals.

But, you claim that a bijective image of the naturals bijects with a
bijective image of a set whose power set is the naturals

Virgil wrote:
In article <44eef7f0@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:

MoeBlee wrote:
Tony Orlow wrote:
MoeBlee wrote:
Please say what sentence and its negation you believe are both theorems
of set theory.
I'm not sure how this would be proven in set theory (I don't think it
is),
So you don't know that set theory is inconsistent (by 'inconsistent' I
mean the usual definition).

but it appears to be a belief, anyway, that all sets can be
classified through cardinality.
With suitable axioms, we can define a cardinality operation 'card' so
that we get a theorem: card(x)=card(y) <-> x equinumerous with y.
Given a set x, can we always determine card(x)?

Depends. In ZF m not necessarily, but in ZFC or NBG, at least
theoretically yes.

No, but you are obligated to define a cardinality for this set which is
consistent, if you claim the theory is consistent. You can't.

For each index value there is a natural whose binary string requires
that index value.

Thus anything less than N is too small.

Thus N is required, with cardinality Card(N).

And yet, N is infinitely too large.

Virgil wrote:
In article <44eefa93@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:

On the contrary, that set of bit positions has cardinality equal to that
of N. The fact that the same set of bit positions is capable of more
does not mean that there need be any smaller set which is sufficient.

That depends on how many more it produces.

Then TO should be able to give the precise number of binary bits
required to represent every natural up through , say, ten, but not to
represent any larger naturals.


If he can not then his whole argument fails.



No, one needs to be able to specify to the nearest bit.

Unless TO can give the number of bits to count up to ten but no farther,
the issue of unused strings is a straw man.

To cover 10 but not 20: 4.

Unless TO can do every natural with a finite set of bits, an infinite
set is required.

Not that the same thing happens for the rationals and decimal
representations. Infinitely many decimal places are required.

But, which infinity?

Everybody but TO accepts Card(N) as necessary and (more than) sufficient.

Excessively sufficient, with no smaller sufficient alternative.

If TO claims to have a workable system, and wishes those claims to be
treated with anything but distain,then he must present that system in
its entirety with some better evidence than he has so far produced that
it actually does work.

I know.

So, TO, what is YOUR definition of TRUTH?
Truth is the value put on a statement, from 0 to 1. Truth is consistency
with reality. There are a few ways to talk about truth.

If truth is merely a value one puts on a statement, everyone's "truths"
will be different, which is not very useful.

In that sense, many truth values are false.

Any logical tautology derived only from a gIven set of axioms is TRUE in
that axiom system.

This is form of relative truth, and mathematics can be certain of no
other.

Logic determines truth.

Either every finite natural has a finite bit string and no infinite set
of positions is required
No inifinite set is requires fora any natural but for any finite number

Yes, given the current understanding, any unbounded set is infinite.

Find any dictionary anywhere that says otherwise, that allows TO's
self-contradictory unbounded but finite.
Dictionaries reflect the widest and most accepted usage of words.

They also, if they are any good, include most, if not all, special
meanings. So if it cannot be found in any of them, even the OED, it is
because it ain't so.

Oh.

If it is a theorem that every set has a cardinality, then it would be
inconsistent if it were also a theorem that there exists a set without
a cardinality. But the fact that we don't have a procedure to always
announce what SPECIFIC cardinality a set has does not contradict any
theorem of set theory.

However, if we have a set which we can prove does NOT have any
cardinality within the system, then there's a hole, yes?

So, is "every set has a cardinality, either finite or an indexed aleph"
a theorem or not?

You suggest that the axiom of choice makes it so. I
say that the set of bit positions required by the binary naturals does
not have such a cardinality.

mueckenh@rz.fh-augsburg.de wrote:
In real mathematics (as opposed to matheology)
we have the valid foundation: What cannot be addressed, that does not
exist.

Do you have axioms for this "real mathematics"?

Logic
dictates that every set must have a cardinality, either zero, a
finite cardinality, or an infinite cardinality.

I have never said I think there is a largest natural. I have said that
some of your assumptions lead to that conclusion.

You assume that the axioms of set theory lead to that conclusion,
but you've never proved it.

And similar holds for the width. So either you state that nor the height,
nor the width are 1, or you state that both are 1. That depends entirely
on your viewpoint. But as the height and width of my staircase at step n
is 1 - 1/2^n, I would say that the height and width of the completed
staircase is the limit of that number. Apparently you disagree.

There is no asymptote. Hence your
comparison of 1 and oo fails.

Yes, I wanted to simplify the picture to show that what holds for the
height also holds for the width.

And that is just the deviation from the natural numbers. Here the 1 in

As long as we are in the naturals we know: Each natural is the sum of
all preceding differences. Infinitely many differences require an
infinite sum.

Wrong. In mathematics the concept of "infinite sum" is not defined,

as I
have told you already *many* times. We can talk losely and state that
the infinite sum of all those numbers is indeed aleph-0, but note that
that is talking losely. All infinite sums as in sum{n = 1 .. oo}
used in mathematics are defined by limits.

If you insist that there are infinitely many naturals and
if infinity is a number aleph_0, then there is also a magnitude
aleph_0.

Makes no sense, again. What do you mean with the term "magnitude"?

The only connection I know of is where it is used in connection
with real or complex numbers. And it can also denote some norm
(in general Euclidean norm) in vector spaces. And as the finite
cardinals equate to the non-negative integers, that equate to their
magnitude, you might consider a cardinal number to represent its own
magnitude. Do you mean that?

1/3 is that path which turns left, right, left, right, left, right, ...
and never ceases to change its direction.
Which edge do you miss? Which dge could not be enumerated?

As in your tree, by definition of the tree, each edge terminates at a node,
what is the number of the edge that terminates at 1/3? What is the number of
the edge that terminates at 1/5?

As the set of all edges is countable, you
should be able to state that. The problem with your counting method of
edges (count first level 1, next count level 2, etc.) you will already
have exhausted all natural numbers by counting all edges that are at the
end of finite paths.

Do you believe that the set of edges of 1/3 is uncountable?

That is not the set of all edges in the tree.

Again, definition: K is the number with K[p] = 1 for all p in N and no
other digits.
Do you agree that is a valid definition? If not, why not?

K can be completely indexed because each digit position is a natural number.
Do you agree with this? If not, why not?

K can not be covered because there is no natural number n such that all
digit positions are less than or equal to n.
Do you agree with this? If not, why not?

No, don't tell us that this was
true or that you defined that, but show us *how* you manage this trick
which, in my eyes, is impossible.

With the axiom of infinity it is dead easy, see above. And I thought we
were arguing with the axiom of infinity in mind.

Give us at least one example how you
index a number without covering all the preceding numbers.

I have no idea what you are stating here.

It is your lack of a proper proof that if each digit of a number can
be indexed that number can be covered. And such a proof does not
exist.

I do not see how I could avoid my conclusion. But if you are so sure
then give us at least one example how you completely index a number
without covering all the preceding numbers.

I have done so many times, and am doing it here again.

In article <1156364184.155913.12090@74g2000cwt.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
Dik T. Winter schrieb:
What is the sum of infinitely many finite numbers? What is the sum of
all finite numbers?

I state that it holds for infinitely many 'n'. Not that it holds for
infinite 'n' (whatever that may be).

Infinitely many have infinitely many differences. If you sum them up
then you get an infinite sum.

Well, if you give me a proper definition of the sum of infinitely many
numbers, I may have some idea about your meaning. In mathematics such
a thing is not defined. Your statement above is not a definition.

But I am doing so in order to show that his arguing concerns
impredicative definitions and is inconclusive.f is the mapping, n is a
natural number, M_f(n) is a set which contains all nongenerators,
including n if not including n which is mapped on M.

Yes. Such triples do not exist. And that precisely shows why Hessenberg's
proof was right.

That it is false. An argument which makes use of the last digit of pi
is void, because the last digit of pi does not exist.

Again avoiding the issue by going on with completely different things.

If there is a surjective mapping f from N to P(N) it is
a requirement (of surjectivity) that such a triple *does* exist.

I gave an example that this set cannot exist, independent of the
surjectivity, independent of the cardinalities of the sets involved in
the mapping.

But it is trivial that such a triple can not exist.

But if there is
a surjective mapping from N to P(N), such a triple *must* exist.