Choice function over finite sets
Peter Webb
anything in Google (presumably because I don't know the appropriate search
term).
I am reasonably confident that I can well order finite sets constructed in
ZF. These are basically just sets comprising the characters "{", "}", and
"," and syntactical rules (brackets must be balanced, elements of sets are
seperated by commas, etc). By finite sets, I means sets representable with a
finite number of these three characters.
However, these sets do not bear much resemblance to the sets discussed in
secondary school, which include sets like {Margaret Thatcher, the rules of
hockey, the colour blue}.
I now create a new set theory ZFP, which (for the sake of simplification)
drops the Axiom of infinity, and adds three new Axioms:
Their exists A = {Margaret Thatcher} (and by this I mean the set containing
the actual ex-Prime Minister of Britian of the same name)
Their exists B = {the rules of hockey}
A does not equal B (to make them distiinguishable within ZFP).
Note that I have not numbered these axioms, as the numbering of an axiom is
a conveneince for discussing it, and not part of ZF (and certainly not part
of ZFP). Axioms themselves are not ordered.
Can we well order (finite) sets in ZFP without AC?
This breaks down to establishing an order on "A"and "B", for example A < B <
{} so that A, and B are effectively -2, and -1, or similar.
However, to do this we must choose one of "A"and "B" to be the lowest. How
do we do this without an axiom of finite choice, allowing us to choose an
element of the set {A, B} to be the lowest? I cannot conceive of an explicit
choice function over {{Margaret Thatcher} , {the rules of hockey}}.
Certainly I could define A< B, but this still requires us to choose an A and
B for this definition to work. In the absence of some indicator in the
axioms of the explicit order, I can't see how to do this without just
choosing an order (creating a choice function).
It seems unfortunate and unlikely that adding three Axioms stating the
existence of different two sets {Margaret Thatcher} and {the rules of
hockey} requires the creation of either an explicit order function or an
explicit finite choice Axiom to order. Indeed, even adding a single
additional Axiom would seem to suffice, as the set it defines needs to be
ordered with respect to {}.
This really boils down to the following issue. I am thinking of an element
of {{Margaret Thatcher}, {the rules of hockey}}. You can't ask me a question
(a choice function) which identifies which element I am thinking of without
asking "Is it {Margaret Thatcher} or is it {the rules of hockey}", which
itself requires choice.
Wkipedia doesn't seem to think it is a problem; in
http://en.wikipedia.org/wiki/Axiom_of_choice they state:
"Not every situation requires the axiom of choice. For finite sets X, the
axiom of choice follows from the other axioms of set theory. In that case it
is equivalent to saying that if we have several (a finite number of) boxes,
each containing at least one item, then we can choose exactly one item from
each box. Clearly we can do this: We start at the first box, choose an item;
go to the second box, choose an item; and so on. There are only finitely
many boxes, so eventually our choice procedure comes to an end. "
This seems completely circular to me. How do they "choose" the first box
without a choice function?
Does adding a single Axiom adding a single additional set to ZF demand an
Axiom of Finite Choice for us to well order finite sets which are
constructed using this additional set?
Adding a countably infinite numer of axioms of the form: There exists {"x"},
there exists {"xx"}, there exists {"xxx"}, ... would also seem to require an
Axiom of Countable choice to allow well ordering.
Am I missing something here, or is this just how it is?
Rupert
Peter Webb wrote:
> I know this is a stupid question, but I can't get my head around it, or find
> anything in Google (presumably because I don't know the appropriate search
> term).
>
> I am reasonably confident that I can well order finite sets constructed in
> ZF. These are basically just sets comprising the characters "{", "}", and
> "," and syntactical rules (brackets must be balanced, elements of sets are
> seperated by commas, etc). By finite sets, I means sets representable with a
> finite number of these three characters.
>
> However, these sets do not bear much resemblance to the sets discussed in
> secondary school, which include sets like {Margaret Thatcher, the rules of
> hockey, the colour blue}.
>
> I now create a new set theory ZFP, which (for the sake of simplification)
> drops the Axiom of infinity, and adds three new Axioms:
>
> Their exists A = {Margaret Thatcher} (and by this I mean the set containing
> the actual ex-Prime Minister of Britian of the same name)
> Their exists B = {the rules of hockey}
> A does not equal B (to make them distiinguishable within ZFP).
>
To state these new axioms you would have to extend the language,
introducing constants for Margaret Thatcher and the rules of hockey.
Another issue would be that Margaret Thatcher and the rules of hockey
are distinct objects which have no members, so the axiom of
extensionality would need modification. The usual approach is to
introduce a predicate that says "x is an individual", and restrict the
axiom of extensionality to those objects that are not individuals.
> Note that I have not numbered these axioms, as the numbering of an axiom is
> a conveneince for discussing it, and not part of ZF (and certainly not part
> of ZFP). Axioms themselves are not ordered.
>
> Can we well order (finite) sets in ZFP without AC?
>
Yes, I would think so. It is pretty easy to prove in ZF that finite
sets are well-ordered, by induction on the number of elements. The
axiom of infinity is not necessary for the proof.
> This breaks down to establishing an order on "A"and "B", for example A < B <
> {} so that A, and B are effectively -2, and -1, or similar.
>
Oh, did you mean can we well-order the class of finite sets? We can
definitely well-order the class of hereditarily finite sets, but we
wouldn't be able to prove that the class of finite sets can be
well-ordered, because this would easily imply that every set can be
well-ordered.
> However, to do this we must choose one of "A"and "B" to be the lowest. How
> do we do this without an axiom of finite choice, allowing us to choose an
> element of the set {A, B} to be the lowest?
The axiom of choice is provable for finite families of sets.
> I cannot conceive of an explicit
> choice function over {{Margaret Thatcher} , {the rules of hockey}}.
{Margaret Thatcher} -> Margaret Thatcher, {the rules of hockey} -> the
rules of hockey.
> Certainly I could define A< B, but this still requires us to choose an A and
> B for this definition to work.
Well, if you've got different constant symbols referring to the A and B
this shouldn't be a problem...
> In the absence of some indicator in the
> axioms of the explicit order, I can't see how to do this without just
> choosing an order (creating a choice function).
>
There are two different possible orders and we can choose either one of
them.
> It seems unfortunate and unlikely that adding three Axioms stating the
> existence of different two sets {Margaret Thatcher} and {the rules of
> hockey} requires the creation of either an explicit order function or an
> explicit finite choice Axiom to order. Indeed, even adding a single
> additional Axiom would seem to suffice, as the set it defines needs to be
> ordered with respect to {}.
>
> This really boils down to the following issue. I am thinking of an element
> of {{Margaret Thatcher}, {the rules of hockey}}. You can't ask me a question
> (a choice function) which identifies which element I am thinking of without
> asking "Is it {Margaret Thatcher} or is it {the rules of hockey}", which
> itself requires choice.
>
Why does it require choice? Both of those sets can be defined.
> Wkipedia doesn't seem to think it is a problem; in
> http://en.wikipedia.org/wiki/Axiom_of_choice they state:
>
> "Not every situation requires the axiom of choice. For finite sets X, the
> axiom of choice follows from the other axioms of set theory. In that case it
> is equivalent to saying that if we have several (a finite number of) boxes,
> each containing at least one item, then we can choose exactly one item from
> each box. Clearly we can do this: We start at the first box, choose an item;
> go to the second box, choose an item; and so on. There are only finitely
> many boxes, so eventually our choice procedure comes to an end. "
> This seems completely circular to me. How do they "choose" the first box
> without a choice function?
>
They're talking a bit loosely here. What you have to do is prove it by
induction on the number of boxes.
> Does adding a single Axiom adding a single additional set to ZF demand an
> Axiom of Finite Choice for us to well order finite sets which are
> constructed using this additional set?
>
Note that I'm assuming you actually have a constant symbol to refer to
the set. If this isn't what you had in mind then I'd have to ask you to
specify what you had in mind a bit more carefully.
> Adding a countably infinite numer of axioms of the form: There exists {"x"},
> there exists {"xx"}, there exists {"xxx"}, ... would also seem to require an
> Axiom of Countable choice to allow well ordering.
>
> Am I missing something here, or is this just how it is?
I think we need to be a bit more careful about what question you're
asking here. Specify in more detail the extensions of ZF that you're
considering.
smnew...@comcast.net
Hi,well if you have a non empty set E ,ie there exists x belonging to
E then the predicate calculus from logic allows you to say choose a
name,say "b" by saying choose b belonging to E.Thus the axiom of
choice for one set is a theorem of logic which is why its never
mentioned .For finite sets ,the existance of a choice function follows
by mathematical induction on the number of elements in your set which
you need and have in ZF when you develop the theory of finite sets
.Well ordering of finite sets is also by induction.When speking of the
language and creating sets with brackets you need to assume
mathematical induction to prove anything as logicians always do when
discussing metamathematical theorems.Regards,smn
Peter Webb
> I think we need to be a bit more careful about what question you're
> asking here. Specify in more detail the extensions of ZF that you're
> considering.
>
Three new axioms:
Exists A = {a}
Exists B={b}
x element A => x not element B
How are you going to order A and B without a finite choice Axiom (or an
explicit order as an Axiom)?
Rupert
Peter Webb wrote:
> >
> > I think we need to be a bit more careful about what question you're
> > asking here. Specify in more detail the extensions of ZF that you're
> > considering.
> >
>
> Three new axioms:
>
> Exists A = {a}
Is a a new constant which you're introducing into the language? Or are
you just saying there exists a singleton A? That can be proved in ZF
anyway.
> Exists B={b}
> x element A => x not element B
>
You could introduce two new constants a and b and introduce a new axiom
saying a!=b. Then the assertions that there are singletons {a} and {b}
could be proved. We can then prove the existence of an ordering of
{A,B} in which A<B, and also of an ordering in which B<A. There is no
problem.
If you don't want to introduce new constants, then all you have to do
is say there exist singletons A and B which are different. That can be
proved in ZF (without the axiom of infinity). Then we can prove in ZF
(without the axiom of infinity) that for all such A and B there exists
an ordering of {A,B} in which A<B and also an ordering in which B<A.
There is no problem.
It is easy to prove in ZF (without the axiom of infinity) that all
finite sets can be well-ordered. One does it by induction on the number
of elements.
Peter Webb
"Rupert" <rupertm...@yahoo.com> wrote in message
news:1153456507.8...@b28g2000cwb.googlegroups.com...
>
> Peter Webb wrote:
>> >
>> > I think we need to be a bit more careful about what question you're
>> > asking here. Specify in more detail the extensions of ZF that you're
>> > considering.
>> >
>>
>> Three new axioms:
>>
>> Exists A = {a}
>
> Is a a new constant which you're introducing into the language? Or are
> you just saying there exists a singleton A? That can be proved in ZF
> anyway.
>
>> Exists B={b}
>> x element A => x not element B
>>
>
> You could introduce two new constants a and b and introduce a new axiom
> saying a!=b. Then the assertions that there are singletons {a} and {b}
> could be proved.
That effectively exactly what I have done (I considered sets {a} and {b},
rather than constants a and b)
> It is easy to prove in ZF (without the axiom of infinity) that all
> finite sets can be well-ordered. One does it by induction on the number
> of elements.
>
This part I have trouble with. How do you choose the element corresponding
to n+1 ?
>
>> How are you going to order A and B without a finite choice Axiom (or an
>> explicit order as an Axiom)?
>
Let me get a little more specific.
I can easily provide an explicit choice function for a finite set of Natural
numbers, say choose the smallest of the set. If I think of a finite set of
numbers, you can always choose one in this manner. You have an explicit
choice function.
Now I give you ZFP, which is
ZFP = ZF + exists {a} + exists {b}
Now a and b are just symbols of course, for things like a specific
rhinoserous in the Adelaide zoo, through to things like the rules of hockey.
Lets say I know what a and b really are. Now, ask me for one of these in a
way that allows me to identify which one you want (a choice function). I
pick a=Chaitans number (the actual number), and b="My mental image of the
best way from the city to the airport". How do you make me choose between
these? On the basis of the minimum number of English language characters in
my description? I'm not sure how I would describe fully my route to the
airport, and if the listener doesn't know what set theory is, then the
description of Chaitans constant may take a little longer.
OK, but can't I just say is it a or b ? After all, a and b are defined to
exist as Axioms. Similar problem occurs.
The problem is best shown by considering two slightly different Axiom sets:
Here is a structure I will call ZFP'
Axioms 1 to 8: the same as ZF.
Axiom 9: Exists {a}
Axiom 10: Exists {b}
I can give you a finite choice function for this set. Scan down the Axioms
in order until you find an existence definition of the form "Exists
{symbol}" We find this on the ninth Axiom, Exists {a}. So you have what you
need for a choice function - you can say "Is it {a}"
Here is ZFP, as I construct it.
Axioms 1 to 8: the same as ZF.
Exists {a}
Exists {b}
The only difference is the two new Axioms are not numbered. Indeed, I don't
believe any of the Axioms in ZF are actually numbered; its just a notational
convenience. Now, you want to ask me a question of the form Exists{symbol}.
How are you going to choose which one is Axiom 9? Just choose one and make
it Axiom 9? How?
The problem completely goes away if the axioms are numbered, because
choosing one which is Exists {a} is simply finding the lowest Axiom number
in which it appears. But this is by explicitly encoding an order on {a} and
{b} in the Axioms. The Axioms of ZF are not actually numbered canonically;
the numbers are as I said before are a notational convenience in every
context I have seen them. Nor are my two new Axioms Exists{a} and Exists{b}.
Unless we have an Axiom of Finite choice, I cannot see how to choose between
them.
Jiri Lebl
> I am reasonably confident that I can well order finite sets constructed in
> ZF. These are basically just sets comprising the characters "{", "}", and
> "," and syntactical rules (brackets must be balanced, elements of sets are
> seperated by commas, etc). By finite sets, I means sets representable with a
> finite number of these three characters.
If you have a set that is constructed in this way, then it is obviously
well ordered. There are many ways to write down the set, but just take
ANY ONE of these. Then your set is already well ordered.
> However, to do this we must choose one of "A"and "B" to be the lowest. How
> do we do this without an axiom of finite choice, allowing us to choose an
> element of the set {A, B} to be the lowest? I cannot conceive of an explicit
> choice function over {{Margaret Thatcher} , {the rules of hockey}}.
> Certainly I could define A< B, but this still requires us to choose an A and
> B for this definition to work. In the absence of some indicator in the
> axioms of the explicit order, I can't see how to do this without just
> choosing an order (creating a choice function).
What is an order by a set of relations which you can most definately
construct by hand for the set {A,B}. So to make an ordered set you
just need to construct a relation A < B which is usually given as an
ordered pair, so just as the set {{A},{A,B}}. I didn't need any
"choice function" to do this. What I mean is that a well ordering of
A,B definately exists, I have just exhibited one. I can also exhibit
the other one easily.
Anything you can do in finite number of steps is fine. You can
explicitly choose finitely many elements to well order any given set.
So no axiom of choice is needed. You can just explicitly construct the
ordering from the axioms, that means this ordering must exist within
your model which is all we care about right? (that is the question:
does there exist a well ordering of a finite set).
Perhaps the name "Axiom of choice" is slightly misleading. It is not
telling you that you have this ONE particular choice. It is purely an
existence statement. I mean, if you apply the axiom of choice to well
order the reals, it doesn't mean that you have some particular well
ordering. All that the axiom of choice tells you that one exists. A
well ordering of R is a certain set (an ordered pair (R,S), where S is
a set of ordered pairs which give the order relation). Thus all that
the axiom of choice tells you is that such a set exists. You can
always "choose" a set (or element of a set) to work with in logic, you
just need to know it exists. That is, you don't need the axiom of
choice to "choose" to work with a certain ordering of the reals, that
just the set (R,S) as given above. What you need the axiom of choice
for, is for this (R,S) to exist in the first place.
Since for a finite set I can just exhibit such a set (a well order on
the original finite set) I don't need an axiom.
> It seems unfortunate and unlikely that adding three Axioms stating the
> existence of different two sets {Margaret Thatcher} and {the rules of
> hockey} requires the creation of either an explicit order function or an
> explicit finite choice Axiom to order. Indeed, even adding a single
> additional Axiom would seem to suffice, as the set it defines needs to be
> ordered with respect to {}.
I think you confuse order being something somehow universal. the set
{A,B} has two orderings. Just as the set {0,1,2} hsa one ordering
which is the common one 0 < 1 < 2, but I could just as well order it as
2 < 0 < 1.
> Does adding a single Axiom adding a single additional set to ZF demand an
> Axiom of Finite Choice for us to well order finite sets which are
> constructed using this additional set?
A priory your new fun elements "Margaret Thatcher" and "rules of
hockey" could a priory exist in a model of ZF without your additional
axiom. Thus, if "axiom of finite choice" is not required for well
ordering in ZF, it is not required in your larger system.
Here is probably another confusion on your part. Just because an axiom
system doesn't mandate a specific set exists, doesn't mean it doesn't
exist (wow that's a lot of negations in one sentence). Let me clarify
with an appropriate example. In ZF (without choice), it is still
concievable that the reals are well ordered, that is the set (R,S)
exists. That is, there is a model of ZF in which the reals are well
ordered and the set (R,S) exists. Of course there is also a model of
ZF in which this is not true. But if you are working within ZF you
cannot exclude the possibility that the reals are well ordered.
Similarly in your system, even if you don't have the axiom of infinity,
there exists a model for it which does in fact include infinite sets,
and there exists a model for it which even includes well ordered reals.
(Just take a model for ZFC, and add your two new elements "Margaret
Thatcher" and "rules of hockey").
> Adding a countably infinite numer of axioms of the form: There exists {"x"},
> there exists {"xx"}, there exists {"xxx"}, ... would also seem to require an
> Axiom of Countable choice to allow well ordering.
Well it would be tough to state the axiom system this way perhaps. But
for example ZFC really has infinitely many axioms in some sense (one of
the axioms is an axiom schema). But you don't need an axiom of choice
to "state" an axiom system. But to really state it you'd need some
consise way of stating the above axioms. But all it boils down to is
the axiom of infinity I think.
Jiri
Dave L. Renfro
> This seems completely circular to me. How do they
> "choose" the first box without a choice function?
This is a question that used bother me also, and I think
one of the biggest sources for this confusion is caused
by non-foundational textbooks that glibly throw around
comments about AC, such as measure theory texts when
nonmeasurable sets are "constructed".
Other people (Jiri Lebl being one of them) have
discussed your concern. However, I recently came
across a comment somewhere (in Fraenkel's book
"Abstract Set Theory", I think) which said that
Bernays had made the observation that the Axiom
of Choice for finite collections is essentially
due to the distributivity of "and" over "or" in
logic. I managed to track the precise reference
down in:
Paul Bernays, "Die Philosophie der Mathematik und
die Hilbertsche Beweistheorie" [The philosophy
of mathematics and Hilbert's proof theory],
Blätter für Deutsche Philosophie 4 (1930),
326-367. [JFM 56.0044.02]
http://www.emis.de/cgi-bin/JFM-item?56.0044.02
There is an English translation of Bernays'
1930 paper, and you can find this distributivity
connection for AC discussed on pp. 47-48 at:
http://www.phil.cmu.edu/projects/bernays/Pdf/bernays09_2002-07-26.pdf
Dave L. Renfro
Peter Webb
"Jiri Lebl" <ji...@5z.com> wrote in message
news:1153496570.8...@m79g2000cwm.googlegroups.com...
Whoops! I missed something there. How did you choose A to go on the left?
You say you construct the order "by hand". What does that mean?
In my other thread, I compare two Axiom systems as follows (and I
cut-and-paste here):
Here is a structure I will call ZFP'
Axioms 1 to 8: the same as ZF.
Axiom 9: Exists {a}
Axiom 10: Exists {b}
I can give you a finite choice function for ZFP'. Scan down the Axioms
in order until you find an existence definition of the form "Exists
{symbol}" We find this on the ninth Axiom, Exists {a}. So you have what you
need for a choice function - you can choose {a} < {b}.
Here is ZFP, as I construct it.
Axioms 1 to 8: the same as ZF.
Exists {a}
Exists {b}
The only difference is the two new Axioms are not numbered. Indeed, I don't
believe any of the Axioms in ZF are actually numbered; its just a notational
convenience. Now, you want to ask me a question of the form Exists{symbol}.
How are you going to choose which one is Axiom 9? Just choose one and make
it Axiom 9? How?
The problem completely goes away if the axioms are numbered, because
choosing one which is Exists {a} is simply finding the lowest Axiom number
in which it appears. But this is by explicitly encoding an order on {a} and
{b} in the Axioms. The Axioms of ZF are not actually numbered canonically;
the numbers are as I said before are a notational convenience in every
context I have seen them. Nor are my two new Axioms Exists{a} and Exists{b}
numbered. That
would be explicitly providing a Finite Choice function, which otherwise
doesn't exist. Unless we have an Axiom of Finite choice, I cannot see how to
choose between
{a} and {b}
> which is usually given as an
> ordered pair, so just as the set {{A},{A,B}}. I didn't need any
> "choice function" to do this. What I mean is that a well ordering of
> A,B definately exists, I have just exhibited one. I can also exhibit
> the other one easily.
>
> Anything you can do in finite number of steps is fine. You can
> explicitly choose finitely many elements
How?
>to well order any given set.
> So no axiom of choice is needed. You can just explicitly construct the
> ordering from the axioms, that means this ordering must exist within
> your model which is all we care about right? (that is the question:
> does there exist a well ordering of a finite set).
>
> Perhaps the name "Axiom of choice" is slightly misleading. It is not
> telling you that you have this ONE particular choice. It is purely an
> existence statement. I mean, if you apply the axiom of choice to well
> order the reals, it doesn't mean that you have some particular well
> ordering. All that the axiom of choice tells you that one exists. A
> well ordering of R is a certain set (an ordered pair (R,S), where S is
> a set of ordered pairs which give the order relation). Thus all that
> the axiom of choice tells you is that such a set exists. You can
> always "choose" a set (or element of a set) to work with in logic, you
> just need to know it exists. That is, you don't need the axiom of
> choice to "choose" to work with a certain ordering of the reals, that
> just the set (R,S) as given above. What you need the axiom of choice
> for, is for this (R,S) to exist in the first place.
With respect, I already know what you wrote above. Its an "in principal, an
element can be selected" sort of thing. So here's the challenge. I am
thinking of two arbitrary sets. Tell me how to choose one - give me an
explicit choice function. Don't say you can pick either; tell me how to
choose it explicitly. (Ofd couse, with the axiom of finite choice, we
wouldn't need to bother - we just use Choice to say in principle we can)
>
> Since for a finite set I can just exhibit such a set (a well order on
> the original finite set) I don't need an axiom.
>
>> It seems unfortunate and unlikely that adding three Axioms stating the
>> existence of different two sets {Margaret Thatcher} and {the rules of
>> hockey} requires the creation of either an explicit order function or an
>> explicit finite choice Axiom to order. Indeed, even adding a single
>> additional Axiom would seem to suffice, as the set it defines needs to be
>> ordered with respect to {}.
>
> I think you confuse order being something somehow universal. the set
> {A,B} has two orderings. Just as the set {0,1,2} hsa one ordering
> which is the common one 0 < 1 < 2, but I could just as well order it as
> 2 < 0 < 1.
>
Not at all; indeed exactly the opposite. Order is not universal, not even
over finite sets. Certainly {A,B} has two orderings; please tell me which
one to use and why. How do we choose the smaller element in the A < B
relation? (Without numbered Axioms).
>> Does adding a single Axiom adding a single additional set to ZF demand an
>> Axiom of Finite Choice for us to well order finite sets which are
>> constructed using this additional set?
>
> A priory your new fun elements "Margaret Thatcher" and "rules of
> hockey" could a priory exist in a model of ZF without your additional
> axiom. Thus, if "axiom of finite choice" is not required for well
> ordering in ZF, it is not required in your larger system.
>
Whoa here. Choice is required precisely because my system has additional
sets.
I have no problem ordering finite subsets of N, because I can always find a
smallest element.
What is the smallest elelement of { {the Crab Nebula explosion} , {my
recollection of food poisoning}}. How did you choose it?
> Here is probably another confusion on your part. Just because an axiom
> system doesn't mandate a specific set exists, doesn't mean it doesn't
> exist (wow that's a lot of negations in one sentence). Let me clarify
> with an appropriate example. In ZF (without choice), it is still
> concievable that the reals are well ordered, that is the set (R,S)
> exists. That is, there is a model of ZF in which the reals are well
> ordered and the set (R,S) exists. Of course there is also a model of
> ZF in which this is not true. But if you are working within ZF you
> cannot exclude the possibility that the reals are well ordered.
>
> Similarly in your system, even if you don't have the axiom of infinity,
> there exists a model for it which does in fact include infinite sets,
> and there exists a model for it which even includes well ordered reals.
> (Just take a model for ZFC, and add your two new elements "Margaret
> Thatcher" and "rules of hockey").
>
With ZFC, there is no problem. I can use the "C" to choose an element from
{{the smell of napalm in the morning} , {the best way to cut tiles}}. In
ZFP -
which is ZFC - C + Exists{a} + Exists {b}, I can't see how to create a C
explicitly, and
so I need finite choice as an Axiom.
>> Adding a countably infinite numer of axioms of the form: There exists
>> {"x"},
>> there exists {"xx"}, there exists {"xxx"}, ... would also seem to require
>> an
>> Axiom of Countable choice to allow well ordering.
>
> Well it would be tough to state the axiom system this way perhaps. But
> for example ZFC really has infinitely many axioms in some sense (one of
> the axioms is an axiom schema).
Induction only works if you can choose a next element. This is the problem I
have with the inductive proof of Finite Choice. Pick an element from the
first set and call this the lowest. How, exactly, without a Finite Choice
function?
But you don't need an axiom of choice
> to "state" an axiom system. But to really state it you'd need some
> consise way of stating the above axioms. But all it boils down to is
> the axiom of infinity I think.
>
Unless the Axiom system provides an explicit order, or you have an Axiom of
finite choice, how do you pick an element of { {pi} , {the contents of my
navel}} ???
> Jiri
>
Rupert
Peter Webb wrote:
> "Rupert" <rupertm...@yahoo.com> wrote in message
> news:1153456507.8...@b28g2000cwb.googlegroups.com...
> >
> > Peter Webb wrote:
> >> >
> >> > I think we need to be a bit more careful about what question you're
> >> > asking here. Specify in more detail the extensions of ZF that you're
> >> > considering.
> >> >
> >>
> >> Three new axioms:
> >>
> >> Exists A = {a}
> >
> > Is a a new constant which you're introducing into the language? Or are
> > you just saying there exists a singleton A? That can be proved in ZF
> > anyway.
> >
> >> Exists B={b}
> >> x element A => x not element B
> >>
> >
> > You could introduce two new constants a and b and introduce a new axiom
> > saying a!=b. Then the assertions that there are singletons {a} and {b}
> > could be proved.
>
> That effectively exactly what I have done (I considered sets {a} and {b},
> rather than constants a and b)
>
> > It is easy to prove in ZF (without the axiom of infinity) that all
> > finite sets can be well-ordered. One does it by induction on the number
> > of elements.
> >
>
> This part I have trouble with. How do you choose the element corresponding
> to n+1 ?
>
It's actually even more trivial than I thought. The usual definition of
a finite set is a set that's equipollent to a finite ordinal. Are you
familiar with the proof that ordinals can be well-ordered? Or maybe
you'd like to use some other definition of "finite". Which one?
The idea that I vaguely had in mind was that we can easily prove in ZF
that a set S with n+1 elements is nonempty, so there exists an element
in it, x say. Then well-order the complement of {x} in S by the
induction hypothesis and now you've got that there exists a
well-ordering. But, as I say, this is really unnecessary. It's even
more trivial than I thought.
> >
> >> How are you going to order A and B without a finite choice Axiom (or an
> >> explicit order as an Axiom)?
> >
>
> Let me get a little more specific.
>
> I can easily provide an explicit choice function for a finite set of Natural
> numbers, say choose the smallest of the set. If I think of a finite set of
> numbers, you can always choose one in this manner. You have an explicit
> choice function.
>
> Now I give you ZFP, which is
> ZFP = ZF + exists {a} + exists {b}
>
> Now a and b are just symbols of course, for things like a specific
> rhinoserous in the Adelaide zoo, through to things like the rules of hockey.
>
> Lets say I know what a and b really are. Now, ask me for one of these in a
> way that allows me to identify which one you want (a choice function). I
> pick a=Chaitans number (the actual number), and b="My mental image of the
> best way from the city to the airport". How do you make me choose between
> these? On the basis of the minimum number of English language characters in
> my description? I'm not sure how I would describe fully my route to the
> airport, and if the listener doesn't know what set theory is, then the
> description of Chaitans constant may take a little longer.
>
Well, I think we need to clarify what langage we're working in. If our
language has the constant symbols a and b, why can't I just say "I want
you to choose a"? I considered two cases, one where we've got the
constant symbols a and b in our language and one where we don't, and I
told you how I'd approach the problem in each case.
> OK, but can't I just say is it a or b ? After all, a and b are defined to
> exist as Axioms. Similar problem occurs.
>
> The problem is best shown by considering two slightly different Axiom sets:
>
> Here is a structure I will call ZFP'
>
> Axioms 1 to 8: the same as ZF.
> Axiom 9: Exists {a}
> Axiom 10: Exists {b}
>
> I can give you a finite choice function for this set. Scan down the Axioms
> in order until you find an existence definition of the form "Exists
> {symbol}" We find this on the ninth Axiom, Exists {a}. So you have what you
> need for a choice function - you can say "Is it {a}"
>
I really don't see how this can be translated into a definition in the
first-order language of set theory, or a definition in the first-order
language of set theory with the constant symbols a and b added. If we
were working with the latter language, I would just use the constant
symbols a and b to specify the choice function I wanted. If we were
working in the former language, I guess I wouldn't actually be able to
define the choice function, but I could still prove it existed.
> Here is ZFP, as I construct it.
>
> Axioms 1 to 8: the same as ZF.
> Exists {a}
> Exists {b}
>
> The only difference is the two new Axioms are not numbered. Indeed, I don't
> believe any of the Axioms in ZF are actually numbered; its just a notational
> convenience. Now, you want to ask me a question of the form Exists{symbol}.
> How are you going to choose which one is Axiom 9? Just choose one and make
> it Axiom 9? How?
>
Well, you're right that we are not given any particular order for the
axioms, but I think this is a red herring. If we have constant symbols
a and b in our language, then we can use these to define a choice
function. If we don't actually have the constant symbols a and b, we
can't actually define the choice function, but we can still prove it
exists. Say S={a,b} and we know that there exist x and y in the union
set of S such that x!=y. (Here x and y are variables rather than
constant symbols, and we don't know which of them is a and which is b).
Since there exist such x and y, we can easily argue that there exists
an ordering of S in which x<y and an ordering in which y>x. We can't
actually define the ordering, since we don't know the identity of x and
y, but we do know that there exist at least two orderings.
> The problem completely goes away if the axioms are numbered, because
> choosing one which is Exists {a} is simply finding the lowest Axiom number
> in which it appears. But this is by explicitly encoding an order on {a} and
> {b} in the Axioms. The Axioms of ZF are not actually numbered canonically;
> the numbers are as I said before are a notational convenience in every
> context I have seen them. Nor are my two new Axioms Exists{a} and Exists{b}.
> Unless we have an Axiom of Finite choice, I cannot see how to choose between
> them.
The numbering of the axioms is a red herring. Obviously our theory
doesn't know anything about the numbering of the axioms.
Peter Webb
The normal definition will do fine.
Here is a finite set: {{a} , {b}}
Here is an ordinal: { {}, {{}} }
Show me how they are equillpotent without an axiom of finite choice.
Specifically, I would expect your mapping will rely upon choosing an element
from the set { {a} , {b} }. Tell me how you will pick the "first" element
from this set.
>
> The idea that I vaguely had in mind was that we can easily prove in ZF
> that a set S with n+1 elements is nonempty, so there exists an element
> in it, x say.
This step uses the Axiom of Finite Choice. You are defining x to be the
result of the Choice function.
You can of course say "I want you to choose a". You can also say "I want you
to choose b". But I can't see how you decide which question you are going to
ask without an axiom of finite choice.
The situation in ZFP is fundamentally different to ZF. In ZF, we don't need
an axiom of finite choice, because we don't have to ask "Is it {x}", we can
ask "what is the smallest number in the set?". Obviously if we have a well
ordering (as we do for finite sets in ZF), we have a choice function.
However, we cannot create an ordering of {a} and {b} without a choice
function.
>I considered two cases, one where we've got the
> constant symbols a and b in our language and one where we don't, and I
> told you how I'd approach the problem in each case.
>
>> OK, but can't I just say is it a or b ? After all, a and b are defined to
>> exist as Axioms. Similar problem occurs.
>>
>> The problem is best shown by considering two slightly different Axiom
>> sets:
>>
>> Here is a structure I will call ZFP'
>>
>> Axioms 1 to 8: the same as ZF.
>> Axiom 9: Exists {a}
>> Axiom 10: Exists {b}
>>
>> I can give you a finite choice function for this set. Scan down the
>> Axioms
>> in order until you find an existence definition of the form "Exists
>> {symbol}" We find this on the ninth Axiom, Exists {a}. So you have what
>> you
>> need for a choice function - you can say "Is it {a}"
>>
>
> I really don't see how this can be translated into a definition in the
> first-order language of set theory, or a definition in the first-order
> language of set theory with the constant symbols a and b added. If we
> were working with the latter language, I would just use the constant
> symbols a and b to specify the choice function I wanted.
How would you do this without being able to choose an element of the set
{a,b} ? If choice is required, then where does the choice function come
from, unless we define that one exists in th Axioms?
Beautiful. We agree that there is no way of explitly ordering {a} and {b}.
You also agree that there might be orderings based upon x<y and y<x.
You also agree that we have no way of choosing which ordering to use.
It appears that you agree that there is no explicit way of choosing an order
on {a} and {b}. There are two ways it can be done, but if we want an
explicit order, we have to choose one of these two possibilities. Tell me
how to do this without an Axiom of Finite Choice?
>
>> The problem completely goes away if the axioms are numbered, because
>> choosing one which is Exists {a} is simply finding the lowest Axiom
>> number
>> in which it appears. But this is by explicitly encoding an order on {a}
>> and
>> {b} in the Axioms. The Axioms of ZF are not actually numbered
>> canonically;
>> the numbers are as I said before are a notational convenience in every
>> context I have seen them. Nor are my two new Axioms Exists{a} and
>> Exists{b}.
>> Unless we have an Axiom of Finite choice, I cannot see how to choose
>> between
>> them.
>
> The numbering of the axioms is a red herring. Obviously our theory
> doesn't know anything about the numbering of the axioms.
>
I agree that it is a red herring. I was trying to get across the concept
that the Axiom scheme does not impose a natural ordering on {a} and {b}, and
an Axiom of Finite Choice is required to impose an ordering.
Rupert
So, can I assume that there exists an x and a y such that x!=y, and an
object is a member of the set if and only if it is either the singleton
of x or the singleton of y? Then it is easy to deduce from that that it
is equipollent to your ordinal.
> Specifically, I would expect your mapping will rely upon choosing an element
> from the set { {a} , {b} }. Tell me how you will pick the "first" element
> from this set.
>
I don't have to pick one. I just have to be allowed to assume that
there are an x and a y with the requisite properties. I don't have to
decide whether x is a and y is b or the other way around.
If a and b are constant symbols which are being introduced into the
language, then I could explicitly construct the bijection: a->{},
b->{{}}. But if I'm just assuming that there exist an x and a y with
the requisite properties exist, without knowing which is which, then I
can prove the bijection exists, without actually being able to define
it.
> >
> > The idea that I vaguely had in mind was that we can easily prove in ZF
> > that a set S with n+1 elements is nonempty, so there exists an element
> > in it, x say.
>
> This step uses the Axiom of Finite Choice. You are defining x to be the
> result of the Choice function.
>
No, no choice is needed to prove that there exists an x in a nonempty
set. That's just the definition of a nonempty set. I don't have to pick
an x, I just have to observe that at least one exists. That's enough.
But this is all really beside the point. I can't think of a definition
of "finite" on which it isn't absolutely trivial that finite sets can
be well-ordered. Maybe you can suggest one, then there will actually be
something for my induction argument to prove and we can argue about it
on that basis.
Which question I'm going to ask? Do you mean, which answer I'm going to
give?
There's no problem. I can write down whatever definition I like. I
don't need to appeal to any axioms to justify me in doing this. As long
as the symbols a and b are actually in the language, there's no
problem. Just pick one. You don't need to invoke an axiom to justify
this. Mathematicians choose definitions all the time. No axioms are
required for this.
> The situation in ZFP is fundamentally different to ZF. In ZF, we don't need
> an axiom of finite choice, because we don't have to ask "Is it {x}", we can
> ask "what is the smallest number in the set?". Obviously if we have a well
> ordering (as we do for finite sets in ZF), we have a choice function.
> However, we cannot create an ordering of {a} and {b} without a choice
> function.
>
I don't follow this. If it's a set of natural numbers, then yes, we can
ask what the smallest natural number in the set is. We can also prove
in ZF or ZFP that finite sets can be well-ordered. I don't understand
what distinction you are drawing between ZF and ZFP.
> >I considered two cases, one where we've got the
> > constant symbols a and b in our language and one where we don't, and I
> > told you how I'd approach the problem in each case.
> >
> >> OK, but can't I just say is it a or b ? After all, a and b are defined to
> >> exist as Axioms. Similar problem occurs.
> >>
> >> The problem is best shown by considering two slightly different Axiom
> >> sets:
> >>
> >> Here is a structure I will call ZFP'
> >>
> >> Axioms 1 to 8: the same as ZF.
> >> Axiom 9: Exists {a}
> >> Axiom 10: Exists {b}
> >>
> >> I can give you a finite choice function for this set. Scan down the
> >> Axioms
> >> in order until you find an existence definition of the form "Exists
> >> {symbol}" We find this on the ninth Axiom, Exists {a}. So you have what
> >> you
> >> need for a choice function - you can say "Is it {a}"
> >>
> >
> > I really don't see how this can be translated into a definition in the
> > first-order language of set theory, or a definition in the first-order
> > language of set theory with the constant symbols a and b added. If we
> > were working with the latter language, I would just use the constant
> > symbols a and b to specify the choice function I wanted.
>
> How would you do this without being able to choose an element of the set
> {a,b} ?
I can choose. Watch me: a. There, you see, I chose. I don't need to use
any axioms to justify my choice of which definition I write down. If I
needed to use an axiom every time I was about to write something down
I'd never get started. I can write down what I like, so long as I
validly argue from the axioms of ZF.
If a and b are symbols in the language, then I can define an ordering.
I don't think that there is any way to do this without using the
symbols a and b. But I can still prove existence.
>
> >
> >> The problem completely goes away if the axioms are numbered, because
> >> choosing one which is Exists {a} is simply finding the lowest Axiom
> >> number
> >> in which it appears. But this is by explicitly encoding an order on {a}
> >> and
> >> {b} in the Axioms. The Axioms of ZF are not actually numbered
> >> canonically;
> >> the numbers are as I said before are a notational convenience in every
> >> context I have seen them. Nor are my two new Axioms Exists{a} and
> >> Exists{b}.
> >> Unless we have an Axiom of Finite choice, I cannot see how to choose
> >> between
> >> them.
> >
> > The numbering of the axioms is a red herring. Obviously our theory
> > doesn't know anything about the numbering of the axioms.
> >
>
> I agree that it is a red herring. I was trying to get across the concept
> that the Axiom scheme does not impose a natural ordering on {a} and {b}, and
> an Axiom of Finite Choice is required to impose an ordering.
Well, the axiom of finite choice can be proved. There's no problem
about proving the existence of an ordering. What you're really talking
about is the problem of *defining* an ordering, and the axiom (actually
theorem) of finite choice won't help you there.
Similarly, we can prove in ZFC that the real numbers can be
well-ordered, but we can't actually define an ordering.
MoeBlee
As I understand, you've extended the language of set theory by adding
two primitive constant symbols, 'a' and 'b'. And you you've added the
axioms:
Ex x={a}
Ex x={b}
But we don't need to add those as axioms to Z set theory, since they're
theorems anyway of Z in the extended language..
And I understand that you've (implicity?) added the axiom:
~a=b
Okay, that is not a theorem of Z with the added constants, so we do
need it as an axiom if you want it in the theory. And then it follows
that ~{a}={b}.
As to finite choice, it's a theorem of Z. Here's a proof:
Show:
(x is finite & Ay(yex -> ~y=0)) -> Ef(f is a function & dom(f)=x &
Ay(yex -> f(y) e y))
Let n be the finite cardinality of x. Induction on n.
n = 0. Then f = 0 suffices, since 0 is a function on 0 and Ay(ye0 ->
f(y) e y)) holds vacuously.
Suppose holds for n, show holds for n+1. So the finite cardinality of x
is n+1.
So, for some z that is a member of x, the finite cardinality of x/{z}
is n.
So, by the inductive hypothesis, let:
g is a function & dom(g)=x/{z} & Ay(yex/{z} -> g(y) e y).
Since ~z=0, for some v, we have v e z.
Let f = g u {<z v>}.
f is a function & dom(f)=x & Ay(yex -> f(y) e y)
Peter Webb wrote:
> Rupert wrote:
> Here is a finite set: {{a} , {b}}
>
> Here is an ordinal: { {}, {{}} }
> Show me how they are equillpotent without an axiom of finite choice. As to your question, above,
Again, I am supposing that we have ~a=b.
Then let h = {<0 {a}> <1 {b}>, which is a bijection between {0 1} and
{{a} {b}}.
> Specifically, I would expect your mapping will rely upon choosing an element
> from the set { {a} , {b} }. Tell me how you will pick the "first" element
> from this set.
My reason for choosing {a} first was not mathematical. The proof
doesn't depend on my giving any reason at all for preferring {a} before
{b}.
> > The idea that I vaguely had in mind was that we can easily prove in ZF
> > that a set S with n+1 elements is nonempty, so there exists an element
> > in it, x say.
>
> This step uses the Axiom of Finite Choice. You are defining x to be the
> result of the Choice function.
No, he's using existential instantiation, a rule of predicate logic.
> The situation in ZFP is fundamentally different to ZF. In ZF, we don't need
> an axiom of finite choice, because we don't have to ask "Is it {x}", we can
> ask "what is the smallest number in the set?". Obviously if we have a well
> ordering (as we do for finite sets in ZF), we have a choice function.
> However, we cannot create an ordering of {a} and {b} without a choice
> function.
It seems to me that you misconstrue what a choice function is. I can
prove the existence of two different well orderings of {{a} {b}}
without using a choice function. The two well orderings are:
{<{a} {b}>} is a well ordering of {{a} {b}}
and
{<{b} {a}>}.is a well ordering of {{a} {b}}.
Therefore, there exist two well orderings of {{a} {b}}.
All we need to do is prove the existence of the well orderings, not how
we discovered them. And the existence of those two well orderings
follows from previous theorems that we proved as to the existence of
ordered pairs and singletons, since each well ordering in this instance
is a singleton of an ordered pair.
Also, your notion of numbering the axioms is not relevent to any of
this, as well as, unless you do some Godel numbering or something like
that (so that you've specified how the theory talks about its own
formulas), the numbering is in the meta-theory, not the object theory,
and is thus not applicable to a proof in the object theory.
And, as I showed, finite choice is provable from the Z axioms, so
finite choice doesn't need to be called an 'axiom'.
MoeBlee
Peter Webb
>> This step uses the Axiom of Finite Choice. You are defining x to be the
>> result of the Choice function.
>
> No, he's using existential instantiation, a rule of predicate logic.
>
That observation pretty much kills my argument.
Existential instantiation is essentially an Axiom of finite choice, because
it lets us name an element we know exists, and with a name we can order. It
doesn't tell us how to choose an element to associate with a given name, but
then it doesn't need to - its an Axiom.
MoeBlee
> >> This step uses the Axiom of Finite Choice. You are defining x to be the
> >> result of the Choice function.
> >
> > No, he's using existential instantiation, a rule of predicate logic.
> >
>
> That observation pretty much kills my argument.
>
> Existential instantiation is essentially an Axiom of finite choice, because
> it lets us name an element we know exists,
I guess you could call it the 'rule of individual choice', since it
lets you give a temporary name to an object. Then it allows 'finite
choice' by a finite number of iterations. But don't forget that it's
not specific to set theory, but rather is a rule of predicate logic in
general (and in many formulations of predicate logic is not even a
primitive rule but rather is a derived rule, so that it is reducible to
rules and/or axioms of universal quantification).
> and with a name we can order.
I guess you can put it that way, but seems odd. We make such orderings
by pretty much explicitly stipulating them as we iterate existential
instantiation a finite number of times. I guess you could say that the
axiom of choice is a kind of "infinite existential instantiation". We
can't existentially instantiate an infinite number of times within any
given proof. But I guess you couuld say, roughly speaking, the axiom of
choice allows us to "do an infinite number of existential instantions"
in one single step.
MoeBlee
Dave L. Renfro
> [...] I recently came across a comment somewhere (in Fraenkel's
> book "Abstract Set Theory", I think) which said that Bernays
> had made the observation that the Axiom of Choice for finite
> collections is essentially due to the distributivity of "and"
> over "or" in logic.
I just double-checked this (I'm home now, where my books are),
and the place I came across this isn't where I thought it was.
It's actually footnote 3 on p. 49 (Chapter II.4.2) of "Foundations
of Set Theory" by Abraham A. Fraenkel and Yehoshua Bar-Hillel,
North-Holland Publishing Company, 1958:
"3) Cf. *Littlewood I, p. 14. As has been pointed out
by Bernays (e.g., in 30, pp. 359f; cf. Hilbert-Bernays
34, p. 41), the assertion of the multiplicative axiom
for finite sets t is nothing but an application of one
of the distributive laws connecting logical conjunction
and disjunction. Hence in the general case one may
conceive the axiom as a generalization of this elementary
logical law to infinite sets; in other words, as a
supplement to the logical rules governing general and
existential statements. Cf. Collins 54. (Cf. the
intuitionistic attitude to the principle of the
excluded middle; see Chapter IV, [Section] 3.)"
As I posted yesterday, there's an English translation of
the relevant Bernays paper on the internet, and you can find
Bernays' observation about AC and distributivity on pp. 47-48:
Russell Easterly
"MoeBlee" <jazz...@hotmail.com> wrote in message
news:1153564871....@m73g2000cwd.googlegroups.com...
> Peter Webb:
>> Here is a finite set: {{a} , {b}}
>>
>> Here is an ordinal: { {}, {{}} }
>> Show me how they are equillpotent without an axiom of finite choice. As
>> to your question, above,
>
> Again, I am supposing that we have ~a=b.
> Then let h = {<0 {a}> <1 {b}>, which is a bijection between {0 1} and
> {{a} {b}}.
>
>> Specifically, I would expect your mapping will rely upon choosing an
>> element
>> from the set { {a} , {b} }. Tell me how you will pick the "first" element
>> from this set.
>
> My reason for choosing {a} first was not mathematical. The proof
> doesn't depend on my giving any reason at all for preferring {a} before
> {b}.
The fact you have to make a choice is mathematical.
You can not choose not to choose.
<snip>
> It seems to me that you misconstrue what a choice function is. I can
> prove the existence of two different well orderings of {{a} {b}}
> without using a choice function. The two well orderings are:
>
> {<{a} {b}>} is a well ordering of {{a} {b}}
> and
> {<{b} {a}>}.is a well ordering of {{a} {b}}.
> Therefore, there exist two well orderings of {{a} {b}}.
Sets are not natural numbers.
One must choose to define some set to some natural number.
Choosing a<b instead of b<a is arbitrary.
Choosing {}<{{}} instead of {{}}<{} is also arbitrary.
This is easier to see with Boolean functions.
I want to assign a 2-bit string to the numbers 0,1,2, and 3.
zero = 00
one = 01
two = 10
three =11
I can define the successor function for this mapping.
Let a = the low order bit and b = the high order bit.
Succ(ba) = b'a' where
a' = ~a
b' = a&~b + ~a&b
What is the successor of one?
a' = ~1 = 0
b' = 1&~0 + ~1&0 = 1
Succ(01) = (10).
I can choose any mapping I want and still define successor.
zero = 00
one = 10
two = 01
three =11
succ(ba) = b'a' where
a' = a&~b + ~a&b
b' = ~a
There are at least 4! ways I can map symbols to 0, 1, 2, and 3.
No one mapping is any more "natural" than any other.
I can define successor, addition, or any other mathematical
operation for any mapping.
Assigning sets to natural numbers in ZF is no different.
There must exist a mapping of sets to natural numbers,
and no one mapping is any more natural than any other.
Russell
- 2 many 2 count
Virgil
"Russell Easterly" <logi...@comcast.net> wrote:
> Assigning sets to natural numbers in ZF is no different.
> There must exist a mapping of sets to natural numbers,
> and no one mapping is any more natural than any other.
>
Once one has seen the von Neumann model of the finite ordinal numbers,
one can quite confidently say that those finite ordinals make much
better natural numbers than anything else anyone else has ever come up
with.
So that in that sense, there is one set of naturals that is "more
natural" than any others.
MoeBlee
> > doesn't depend on my giving any reason at all for preferring {a} before
> > {b}.
>
> The fact you have to make a choice is mathematical.
> You can not choose not to choose.
Your remark reflects that you didn't get the point of my remark. That's
not surprising, since you refuse to read a single book on the subject
of mathematical logic or set theory.
> > {<{a} {b}>} is a well ordering of {{a} {b}}
> > and
> > {<{b} {a}>}.is a well ordering of {{a} {b}}.
> > Therefore, there exist two well orderings of {{a} {b}}.
>
> Sets are not natural numbers.
Not all sets are natural numbers. That has no bearing on what I said.
> One must choose to define some set to some natural number.
No, there is no "must" to make such "definition".
> Choosing a<b instead of b<a is arbitrary.
Yes, that is what I said.
> Choosing {}<{{}} instead of {{}}<{} is also arbitrary.
Yes. And it doesn't weaken anything I said.
I omitted quoting the rest of your post, since it is not needed to
answer the question of this thread.
MoeBlee
george
Peter Webb wrote:
> Beautiful. We agree that there is no way of explitly ordering {a} and {b}.
> You also agree that there might be orderings based upon x<y and y<x.
> You also agree that we have no way of choosing which ordering to use.
Of course we have a way. In fact, Vee Haff VAYS.
If the finite set has n elements then we have n! DIFFERENT ways.
The fact that there is no "predetermined" or "axiomatic" way to
choose one of the n! ways, or that there is no way to choose
which of the n elements is first, IS IRRELEVANT. The point is that
YOU are free to choose WHICHEVER way you like. Thus, if the
set has n elements, RATHER than having "no" way to choose which
of the n is first, we in fact have n ways.