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| Virgil |
 Posted: Fri Aug 25, 2006 3:25 pm |
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Guest
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In article <1156501377.098380.50700@m79g2000cwm.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:78ae53c564]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?
[/quote:78ae53c564]
The part that says a bijective image of the naturals bijects with a
bijective image of the power set of the naturals. |
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| MoeBlee |
| Posted: Fri Aug 25, 2006 3:54 pm |
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Tony Orlow wrote:
[quote:956ddf518e]MoeBlee wrote:
Tony Orlow wrote:
If the naturals are not a subset of the reals in ZFC and NBG, then those
theories are even more screwed up than they already seemed.
With either of the usual constructions of the reals (Dedekind cuts or
equivalence classes of Cauchy sequences), both the system of naturals
and the system of rationals are isomorphically embedded in the system
of reals. Mathematicians usually speak of the system of natural numbers
and the system of rational numbers as subsystems of the the system of
reals. Strictly speaking, that is incorrect, but it is harmless given
the isomorphism.
Harmless, even though incorrect? What makes it incorrect, if not
inconsistency? Doesn't inconsistency cause a problem?
[/quote:956ddf518e]
Now you're being captious. The informality is harmless, as I said,
because we are within ISOMORPHISM. It has nothing to do with the
consistency of the theory. It's only a matter of INFORMAL convenience
to not have to say each time "the system that is isomorphically
embedded" but instead speak directly of the system as if it were a
subsystem, since, to WITHIN ISOMORPHISM, it is a subsystem. This kind
of informality is common throughout mathematics and is harmless.
[quote:956ddf518e]I used y in the equation (or statement of inequality) and then specified
in parentheses at the end, with a space, which condition for y made that
statement true. I didn't think that was tooooo confusing.
[/quote:956ddf518e]
It was unclear enough that I had to check with you what you meant. It's
not a big deal, but you could acheive clarity by being more explicit.
[quote:956ddf518e]'number of any real sort' is not a defined predicate of set theory. w
is not in the standard ordering of the reals. That does not make w an
undefined object.
My experience is that asking amthemticians for a definition of "number"
results in.....nothing.
[/quote:956ddf518e]
Because it's more a philosophical issue or an issue of terminology
outside the system. The purpose of set theory is not to address the
question of what is and is not a number. Rather, among the purposes of
set theory is to axiomatize and construct the various number systems
that are of interest.
[quote:956ddf518e]you don't see scientists accepting transfinitology either.
Of course, you have a survey of scientists to support your claim.
Uh, yes, right here. Why don't you survey thise that object, regarding
what they do?
[/quote:956ddf518e]
That's quite an unscientific survey method of yours.
[quote:956ddf518e]You have no coherent system of definitions at all. It's not the job of
set theory to define objects that obey the whims of your informal
notions.
It's the job of mathematicians to work with numbers.
[/quote:956ddf518e]
Numbers are among the primary concern of mathematics. Set theory
axiomatizes and constructs number systems.
[quote:956ddf518e]The selection of any unit is done by simply choosing a point separate
from the origin. When it comes to division using infinite values, one
translates the geometric definition into symbolic form and applies
induction formulaically.
"Translates the geometric definition into symolic form and applies
induction formulaically" is no less doubletalk than "Coordinates the
numeric form into geometric postulates and applies the recursive
definition metrically."
Is that a question?
[/quote:956ddf518e]
Is that a rhetorical question?
MoeBlee |
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| Virgil |
| Posted: Fri Aug 25, 2006 3:57 pm |
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In article <44eef224@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
[quote:37dbe022e4]Virgil wrote:
In article <ecikqd$a1o$1@ruby.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
The Finlayson numbers
are nilpotent infinitesimal "degenerate" intervals, sequential yet
indistinguishable on the finite scale.
This garbage about scales again.
If a < b on any scale then a < b on every scale. A MATHEMATICAL
inequality does not become an equality when you put away your magnifying
glass.
Nor does a MATHEMATICAL equality become an inequality when you look at
it under higher magnification.
Does TO also claim that while x is a member of A at one magnification,
it need not be so at another, or that A being or not being a subset of B
depends on the magnification used by the observer?
Silliness compounded!
No, it all rests on the notions of identity and equality.
[/quote:37dbe022e4]
TO would have one believe that equality is conditional on scale rather
than being, as it must be in mathematics, absolute, and independent of
scale.
Does 1 + 1 = 2 become false under a sufficiently strong magnifier, TO?
Does 2 < 3 become false if one moves far enough away? |
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| Virgil |
| Posted: Fri Aug 25, 2006 4:04 pm |
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Guest
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In article <44eef7f0@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
[quote:82964fb0ad]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)?
[/quote:82964fb0ad]
Depends. In ZF m not necessarily, but in ZFC or NBG, at least
theoretically yes.
[quote:82964fb0ad]
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.
[/quote:82964fb0ad]
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). |
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| Tony Orlow |
| Posted: Fri Aug 25, 2006 9:55 pm |
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MoeBlee wrote:
[quote:488d1a460f]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.
[/quote:488d1a460f]
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.
[quote:488d1a460f]
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.
[/quote:488d1a460f]
You prove two quantities equal by showing there is no difference, do you
not?
[quote:488d1a460f]
MoeBlee
[/quote:488d1a460f] |
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| Tony Orlow |
| Posted: Fri Aug 25, 2006 9:57 pm |
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MoeBlee wrote:
[quote:8990d02c2d]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.
[/quote:8990d02c2d]
The universe is governed by the properties of the elements within it,
which properties are statements true about those elements.
[quote:8990d02c2d]
Yes, personally I want concepts of infinity to be compatible with the
rest of math, as a logical extension. There is something amiss in set
theory in this respect.
"Logical extension". You have no idea what you're talking about.
[/quote:8990d02c2d]
But, I do.
[quote:8990d02c2d]
MoeBlee
[/quote:8990d02c2d]
Tony |
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| Virgil |
| Posted: Fri Aug 25, 2006 10:00 pm |
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Guest
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In article <44ef344b@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
[quote:68e6123fd1]For every two new edges, one new path is produced
[/quote:68e6123fd1]
But all such paths, by having to have the terminal edges which created
them, are finite, so irrelevant. |
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| Tony Orlow |
| Posted: Fri Aug 25, 2006 10:07 pm |
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MoeBlee wrote:
[quote:2b3e94290e]Tony Orlow wrote:
Indeed. Not step by step. And so? I have said this before. I am quite
sure others have said that before. You can not get the set of naturals
by adding the one by one. It is the axiom of infinity that asserts that
that set does exist.
And that it's infinite. Why do you need an axiom for that? WHy is it not
derivable logically?
What do you mean "why"? It is not derivable since there are finite
domains of discourse that satisfy Z-I (Z set theory without the axiom
of infinity).
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.
[/quote:2b3e94290e]
I am not convinced of that.
[quote:2b3e94290e]
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.
[/quote:2b3e94290e]
Logic is the primitive of argument. Logical truth is founded upon
quantity, and quantity upon geometry. Space is a priori. :)
[quote:2b3e94290e]
Yes, but in that case you are not using standard mathematical terminology.
And I have yet to see a *definition* of you about *infinite*.
Larger than any finite. The set of naturals is as large as, but no
larger than, every natural.
Define 'larger'.
[/quote:2b3e94290e]
m>n <=> A x (x>m ^ x>n) v (m>x ^ x>n) v (m>x ^ n>x)
Define 'finite'.
That is a little more difficult - it's relative.
Actually, nevermind, since you have
[quote:2b3e94290e]no logicistic system, primitives, or axioms from which to make
defintions. Moreover, you have no idea what are correct forms for
defintions, why such forms are correct and others are incorrect. You
even resort to circular definitions.
[/quote:2b3e94290e]
Comment on 'larger'.
[quote:2b3e94290e]
You are using words without definition, using words from standard terminology
but with (apparently) not the standard definition. That is what makes you
on occasion not understandable.
I thought it was clear that I was using a notion of infinite, like WM,
from a quantitative standpoint, rather than set-theoretic.
Oh, yes, the "quantitative standpoint". Have you read
Zerbernieskoskywoskyozerlichmanosty's paper on the standard
quantitative geometric induction? It's right up your alley.
[/quote:2b3e94290e]
I'll have to google it. Thanks for the reference. :|
[quote:2b3e94290e]
MoeBlee
[/quote:2b3e94290e] |
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| Tony Orlow |
| Posted: Fri Aug 25, 2006 10:10 pm |
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MoeBlee wrote:
[quote:a8997367d4]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.
[/quote:a8997367d4]
Good guess, except I have no idea what your driving at.
[quote:a8997367d4]
However, the set of bit positions
required to list the naturals in binary defies classification in this
system.
Whatever you mean by that, based on remark that you know of no proof of
a contradiction in set theory, it is a problem of your own
misconception and not a contradiction in set theory. Fortunately, we
don't obligate ourselves to the burdens of your own misconceptions.
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.
First you need to define the set in set theory and prove that it exists
in set theory. Then, set theory would not be inconsistent for there not
being an algorithm to determine the ordinal index of the aleph that is
the cardinality of any given set.
[/quote:a8997367d4]
So, you're saying that identifying a set with no cardinality doesn't
make set theory inconsistent? I think I saw a different opinion yesterday.
[quote:a8997367d4]
Again, you're just yapping without regard for the specific definitions
of such things as 'consistent'.
[/quote:a8997367d4]
Consistent: Adj. Without contradiction.
[quote:a8997367d4]
MoeBlee
[/quote:a8997367d4] |
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| Tony Orlow |
| Posted: Fri Aug 25, 2006 10:16 pm |
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Guest
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Virgil wrote:
[quote:8a48d419d6]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.
[/quote:8a48d419d6]
But, you claim that a bijective image of the naturals bijects with a
bijective image of a set whose power set is the naturals, so who are you
to complain? The naturals in binary are the power set of the bit
positions in those strings. How can the set biject with its own power set? |
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| Tony Orlow |
| Posted: Fri Aug 25, 2006 10:23 pm |
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MoeBlee wrote:
[quote:0a503f61b3]Tony Orlow wrote:
MoeBlee wrote:
Tony Orlow wrote:
If the naturals are not a subset of the reals in ZFC and NBG, then those
theories are even more screwed up than they already seemed.
With either of the usual constructions of the reals (Dedekind cuts or
equivalence classes of Cauchy sequences), both the system of naturals
and the system of rationals are isomorphically embedded in the system
of reals. Mathematicians usually speak of the system of natural numbers
and the system of rational numbers as subsystems of the the system of
reals. Strictly speaking, that is incorrect, but it is harmless given
the isomorphism.
Harmless, even though incorrect? What makes it incorrect, if not
inconsistency? Doesn't inconsistency cause a problem?
Now you're being captious. The informality is harmless, as I said,
because we are within ISOMORPHISM. It has nothing to do with the
consistency of the theory. It's only a matter of INFORMAL convenience
to not have to say each time "the system that is isomorphically
embedded" but instead speak directly of the system as if it were a
subsystem, since, to WITHIN ISOMORPHISM, it is a subsystem. This kind
of informality is common throughout mathematics and is harmless.
[/quote:0a503f61b3]
So, is it technically incorrect to say 1 = 3/3 = 1.000... = 0.999...? I
never thought so.
[quote:0a503f61b3]
I used y in the equation (or statement of inequality) and then specified
in parentheses at the end, with a space, which condition for y made that
statement true. I didn't think that was tooooo confusing.
It was unclear enough that I had to check with you what you meant. It's
not a big deal, but you could acheive clarity by being more explicit.
'number of any real sort' is not a defined predicate of set theory. w
is not in the standard ordering of the reals. That does not make w an
undefined object.
My experience is that asking amthemticians for a definition of "number"
results in.....nothing.
Because it's more a philosophical issue or an issue of terminology
outside the system. The purpose of set theory is not to address the
question of what is and is not a number. Rather, among the purposes of
set theory is to axiomatize and construct the various number systems
that are of interest.
[/quote:0a503f61b3]
So, how do you know when you're doing math, and when you've strayed into
other territory?
[quote:0a503f61b3]
you don't see scientists accepting transfinitology either.
Of course, you have a survey of scientists to support your claim.
Uh, yes, right here. Why don't you survey thise that object, regarding
what they do?
That's quite an unscientific survey method of yours.
[/quote:0a503f61b3]
Yo, man, it's empirical evidence. Ask around. Life's an experiment.
[quote:0a503f61b3]
You have no coherent system of definitions at all. It's not the job of
set theory to define objects that obey the whims of your informal
notions.
It's the job of mathematicians to work with numbers.
Numbers are among the primary concern of mathematics. Set theory
axiomatizes and constructs number systems.
[/quote:0a503f61b3]
Which are the other primary concerns, if any? (how did we define
"number" again?)
[quote:0a503f61b3]
The selection of any unit is done by simply choosing a point separate
from the origin. When it comes to division using infinite values, one
translates the geometric definition into symbolic form and applies
induction formulaically.
"Translates the geometric definition into symolic form and applies
induction formulaically" is no less doubletalk than "Coordinates the
numeric form into geometric postulates and applies the recursive
definition metrically."
Is that a question?
Is that a rhetorical question?
[/quote:0a503f61b3]
I'll take as an "I dunno".
[quote:0a503f61b3]
MoeBlee
[/quote:0a503f61b3] |
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| Tony Orlow |
| Posted: Fri Aug 25, 2006 10:31 pm |
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Virgil wrote:
[quote:2bec6952d3]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).
[/quote:2bec6952d3]
And yet, N is infinitely too large. |
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| MoeBlee |
| Posted: Fri Aug 25, 2006 10:47 pm |
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Guest
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Tony Orlow wrote:
[quote:35885c0925]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.
[/quote:35885c0925]
Of course you don't, because you are ignorant of even of the basics of
your own question. With the axiom of choice (which is one method),
every set has a cardinality. That cardinality is a cardinal number. And
that cardinal number is an aleph indexed by an ordinal. So I answered
your quesion: No, I don't think there is a procedure to determine which
cardinal is the cardinality of a given set. But in my first answer I
gave an even sharper formulation by pining the question down to the
ordinal index. Of course, for some sets, we can prove that a certain
cardinal is the cardinality that set; but I don't think there can be a
general procedure to produce an answer for any set that might be
submitted to test.
[quote:35885c0925]First you need to define the set in set theory and prove that it exists
in set theory. Then, set theory would not be inconsistent for there not
being an algorithm to determine the ordinal index of the aleph that is
the cardinality of any given set.
So, you're saying that identifying a set with no cardinality doesn't
make set theory inconsistent? I think I saw a different opinion yesterday.
[/quote:35885c0925]
You INSIST on twisting just about every answer given you.
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.
[quote:35885c0925]Again, you're just yapping without regard for the specific definitions
of such things as 'consistent'.
Consistent: Adj. Without contradiction.
[/quote:35885c0925]
Yes, and your remarks were without sense of that definition, as
SPECIFICALLY you've been told about a million times that a theory is
inconsistent iff there is a contradiction in the theory, which means
that there is sentence and its negation that are both members of the
theory.
MoeBlee |
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| MoeBlee |
| Posted: Fri Aug 25, 2006 10:51 pm |
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Tony Orlow wrote:
[quote:753bd24775]So, is it technically incorrect to say 1 = 3/3 = 1.000... = 0.999...? I
never thought so.
[/quote:753bd24775]
Those are notations that can be understood precisely only in context of
the particular treatment in which they occur. I said that IN A SYSTEM
that constructs the reals as either Dedekind cuts or as equivalence
classes of Cauchy sequences, then the naturals and rationals are
isomorphically embedded but are not actual subsets of the reals. (That
is BASIC undergraduate mathematics. I mean, even I know that, and I
don't even have an education in mathematics! It's appalling how
ignorant you are WHILE you so categorically opine.) If the system and
construction are of a different sort, then we'd have to evaluate upon
the specifics of that system and construction. But, again, as I said,
for contexts that do not need to be so pedantic, such notation is
understood well enough that the equations you mentioned do of course
hold.
By the way, a while ago I posted a very rigorous explication of decimal
notation in a post to you.
[quote:753bd24775]Because it's more a philosophical issue or an issue of terminology
outside the system. The purpose of set theory is not to address the
question of what is and is not a number. Rather, among the purposes of
set theory is to axiomatize and construct the various number systems
that are of interest.
So, how do you know when you're doing math, and when you've strayed into
other territory?
[/quote:753bd24775]
That's a question for philosophy. I have my own views on what
mathematics is, but I don't hold that my own concept of mathematics is
definitive.
[quote:753bd24775]you don't see scientists accepting transfinitology either.
Of course, you have a survey of scientists to support your claim.
Uh, yes, right here. Why don't you survey thise that object, regarding
what they do?
That's quite an unscientific survey method of yours.
Yo, man, it's empirical evidence. Ask around. Life's an experiment.
[/quote:753bd24775]
It's anecdotal evidence only. Check it out. A scientific experiment is
not just any life experience.
[quote:753bd24775]Numbers are among the primary concern of mathematics. Set theory
axiomatizes and constructs number systems.
Which are the other primary concerns, if any? (how did we define
"number" again?)
[/quote:753bd24775]
I don't define 'number' in a formal theory. But in informal discussion
about mathematics I don't demur from using the word 'number' in its
ordinary dictionary senses.
Other concerns of mathematics are sets, relations, spaces, geometries,
topologies, algebras, for example, and more.
[quote:753bd24775]I'll take as an "I dunno".
[/quote:753bd24775]
Then you'll, AS USUAL, take incorrectly.
MoeBlee |
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| Tony Orlow |
| Posted: Fri Aug 25, 2006 10:51 pm |
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MoeBlee wrote:
[quote:6fe319101e]Tony Orlow wrote:
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.
Of course you don't, because you are ignorant of even of the basics of
your own question. With the axiom of choice (which is one method),
every set has a cardinality. That cardinality is a cardinal number. And
that cardinal number is an aleph indexed by an ordinal. So I answered
your quesion: No, I don't think there is a procedure to determine which
cardinal is the cardinality of a given set. But in my first answer I
gave an even sharper formulation by pining the question down to the
ordinal index. Of course, for some sets, we can prove that a certain
cardinal is the cardinality that set; but I don't think there can be a
general procedure to produce an answer for any set that might be
submitted to test.
First you need to define the set in set theory and prove that it exists
in set theory. Then, set theory would not be inconsistent for there not
being an algorithm to determine the ordinal index of the aleph that is
the cardinality of any given set.
So, you're saying that identifying a set with no cardinality doesn't
make set theory inconsistent? I think I saw a different opinion yesterday.
You INSIST on twisting just about every answer given you.
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.
[/quote:6fe319101e]
However, if we have a set which we can prove does NOT have any
cardinality within the system, then there's a hole, yes?
[quote:6fe319101e]
Again, you're just yapping without regard for the specific definitions
of such things as 'consistent'.
Consistent: Adj. Without contradiction.
Yes, and your remarks were without sense of that definition, as
SPECIFICALLY you've been told about a million times that a theory is
inconsistent iff there is a contradiction in the theory, which means
that there is sentence and its negation that are both members of the
theory.
[/quote:6fe319101e]
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.
[quote:6fe319101e]
MoeBlee
[/quote:6fe319101e] |
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