Choice!

[Zuhair]

If we add a primitive one place function symbol C to the language of
ZF and add the following axiom to the list of axioms of ZF:

Global Choice (GC): For all x. x =/= 0 -> C(x) e x

Informally this is taken to mean C(x) is chosen from a non empty set
x.

Another idea is to introduce a primitive two place function C(x,y) and
axiomtize

x =/= 0 -> C(x,y) e x

this informally means that C(x,y) is an element chosen from a non
empty set x by y.

we can actually extend this to any C(x,y1,...,yn), i.e. an n+1-ary
function symbol.

so for example if we add the axiom:

x =/= 0 -> C(x,y,z) e x

then this would stand for saying that C(x,y,z) is an element of a non
empty x that is chosen by y and z.

The >2-ary C functions represent what I call "Discussion", since the
choice of an element of x is determined by multiple objects.

We can further classify different situations occurring with those
discussions, those might be "unanimous" if C(x,y1,...,yn) = C(x,yi)
for all i=1,...,n Or non unanimous, and the later can be affirmative
of one decider i.e. there exist yi such that C(x,y1,...yn) = C(x,yi),
or it might not be affirmative. Also different "determination" rules
can be defined to determine the choice of y1,...yn from the individual
yi choices.

We can extend this method even to encounter infinitely many deciding
objects y1,y2,y3,...., by adopting a set notation in functions like
C(x,{yi| i e N}) , where N can be any set of any size.

Also all of those concepts can be restricted to sets of interest like
to countable sets (which enact countable choice) etc..

I don't know if something like that was done before or whether it is
useful either.

[Zuhair]

On Apr 8, 2:03 pm, Zuhair <zaljo...@gmail.com> wrote:
> If we add a primitive one place function symbol C to the language of
> ZF and add the following axiom to the list of axioms of ZF:
>
> Global Choice (GC): For all x. x =/= 0 -> C(x) e x

The ordinary axiom of choice is equivalent to

x =/= 0 & x e y -> C(x,y) e x

Zuhair

[Ross A. Finlayson]

You're not interested that I tell you the Dirac delta function, not a
real function, is built from approximative functions that you can vary
in the patch.

Heh, but, that's how it is.

If there's choice you built ordering or it is there.

Zuhair, here your elements satisfy these properties that then there is
order and structure throughout their array. Then to establish the
Axiom or Zorn's Lemma or the Well Ordering Principle, the Axiom of
Choice or here as you note "global" choice, here as we would expect
you would have in mind for this theory where each object is well-
behaved in that manner, then, where often that is generally true - how
would that build into the empty set?

I ask this question, how the element here the empty set would have
properties as a set, which one is it: is it none of the functions or
all of the functions?

Also, does, that make sense to you? (No, why would I ask that, it's
insulting to Zuhair Al Jofar to think that he doesn't make quite good
sense of it. Yet, I'll ask it as a courtesy.) What is the empty set
in the theory?

Thanks,

Ross Finlayson

[Zuhair]

On Apr 9, 10:50 am, "Ross A. Finlayson" <ross.finlay...@gmail.com>
wrote:

Any object in this theory is a set, and since the empty set can be
constructed from separation then it is a set. There is no choice map
for the empty set since it is empty of elements, yet that doesn't say
it is not a set.

Zuhair

[Graham Cooper]

There is no such thing as Axiom Of Choice.

It was tacked onto ZF to patch Cantor's theorem that hyperinfinite
sets exist. HOGWASH!

An Array of Arrays may have an *ALGORITHM OF CHOICE*

Say you project a 2 dimensional plane of a grid of digits
into infinite length in both directions.

+---------------------> PI
| 0 3 1 4 1 5 9 2 6 ...
| 1
| 4
| 1
| 2
| ...
v
_/2

LIST_ROW1 = PI
LIST_COL1 = _/2

let's populate the Infinite Plane with X+Y MOD 10
+---------------------> PI
| 0 3 1 4 1 5 9 2 6 ...
| 1 4 2 5 2 6 0 3 7 ...
| 4 7 5 8 5 9 3 6 0 ...
| 1 4 2 5 2 6 0 3 7 ...
| 2 5 3 6 3 7 1 4 8 ...
| ...
v
_/2

The smallest Algorithm Of Choice is LIST(n,n)
+---------------------> PI
| [0] 3 1 4 1 5 9 2 6 ...
| 1 [4] 2 5 2 6 0 3 7 ...
| 4 7 [5] 8 5 9 3 6 0 ...
| 1 4 2 [5] 2 6 0 3 7 ...
| 2 5 3 6 [3] 7 1 4 8 ...
| ...
v
_/2


Here's another Algorithm Of Choice!
+---------------------> PI
| 0 3 1 [4] 1 5 9 2 6 ...
| 1 [4] 2 5 2 6 0 3 7 ...
| 4 7 [5] 8 5 9 3 6 0 ...
| [1] 4 2 5 2 6 0 3 7 ...
| 2 5 3 6 [3] 7 1 4 8 ...
| ...
v
_/2

A DIFFERENT DIAGONAL!
A DIFFERENT ANTIDIAGONAL!

Any random sequence of digits will fit the DIAG or ANTIDIAG by
selecting the AOC.

_____________________________________


The *constructable* mathematics Universe U is devoid of backtracking.

There is no way to specify a set of consistent objects.

DEFINE: x e Y :: (a e Y) ^ (a <-/-> x) -> !(x e Y)

or CONTRADICTION --> x ~e Y

ZFC is just an over-elaborate *Subset* Specification system to push
all definitions 1 subset lower than Russell's Paradox, which is easily
contravened in much simpler ways.

Y = { x | P(x) } <-> !PROOF(!(EXIST(Y))

PROOF(C) <-> C v (PROOF(A) ^ PROOF(B) ^ A^B->C)

Herc

[Tonico]

> Herc-



Idiot

[William Elliot]

On Sun, 8 Apr 2012, Zuhair wrote:

> > If we add a primitive one place function symbol C to the language of
> > ZF and add the following axiom to the list of axioms of ZF:
> >
> > Global Choice (GC): For all x. x =/= 0 -> C(x) e x
>
> The ordinary axiom of choice is equivalent to
>
> x =/= 0 & x e y -> C(x,y) e x
>

In ZF + GC there should be a well ordering of all sets.

In New Foundations + GC there should be a well ordering of all sets.
I doubt that well ordering can be extended to include classes.

If fact, New Foundations + choice is inconsistent.
Thus GC shouldn't be added to New Foundations or can it?

[Zuhair]

No you can't do that with NF but you can do it with fragments of NF
like NFU for example.
So NF+GC is inconsistent, but NFU + GC is consistent, actually since
we have a universal
set in NFU then choice would imply global choice as well.

Zuhair

[Ross A. Finlayson]

On Apr 9, 12:50 am, "Ross A. Finlayson" <ross.finlay...@gmail.com>
wrote:

What is Zorn proving for his lemma?

You know, I did read it, but maybe just to say it later.

"Zorn's Lemma": Most people assume that is true for anything or not
true for anything.

Thanks,

Ross Finlayson

PS Well order the reals.

[Aatu Koskensilta]

Zuhair <zalj...@gmail.com> writes:

> I don't know if something like that was done before or whether it is
> useful either.

It's known that adding to ZFC a global choice function, i.e. a
function symbol c together with the axiom

If x is a non-empty set, c(x) is an element of x.

results in a conservative extension of ZFC. The proof, using
compactness, is not difficult. Just observe that in any proof involving
c it can only be applied to finitely many sets so in any given proof we
can take it to be a suitable function (depending on the sets involved)
that provably exists in ZFC. It's also known that NBG with global choice
is conservative over ZFC, but the proof (using class forcing) is
slightly more involved. I don't offhand see any use for your multiplace
choice functions; in any case it would seem they can easily be handled
using similar devices.

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, darüber muss man schweigen."
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

[Aatu Koskensilta]

William Elliot <ma...@panix.com> writes:

> In ZF + GC there should be a well ordering of all sets.

Should there, now? How so?

[Michael Stemper]

In article <87obqq7...@uta.fi>, Aatu Koskensilta <aatu.kos...@uta.fi> writes:
>Zuhair <zalj...@gmail.com> writes:

>> I don't know if something like that was done before or whether it is
>> useful either.
>
> It's known that adding to ZFC a global choice function, i.e. a
>function symbol c together with the axiom
>
> If x is a non-empty set, c(x) is an element of x.

I can see that this isn't the same as the standard formulation of Choice,
but I'm having trouble figuring out whether it's weaker or stronger.

I'm leaning in the "weaker" direction because this version just says
that you can choose an element from any set. However, it states this
in terms of a single function (called "c" here). That makes it sound
as if it's stronger, since the standard statement just says that there's
a choice function for any non-empty collection of non-empty sets.

This gets me to wondering what the domain of "c" is, since the domain
of a function is, under ordinary circumstances, a set. Or, is this
version simply saying that there is *a* "c" for each "x"? I'm guessing
not, based on the following remarks.

>compactness, is not difficult. Just observe that in any proof involving
>c it can only be applied to finitely many sets so in any given proof we
>can take it to be a suitable function (depending on the sets involved)
>that provably exists in ZFC.

Now, I'm guessing that there's a different "c" for different families
of sets, which seems to get back to the standard statement of Choice.

Could somebody give a layman-accessible description of how this
differs from the standard statement of Choice (if it does)?

--
Michael F. Stemper
#include <Standard_Disclaimer>
If we aren't supposed to eat animals, why are they made from meat?

[Herman Rubin]

This cannot be done in ZF, as in ZF, with or without choice,
the domain of a function is a set. It can be done in NBG,
which has proper classes which are not sets, and hence
cannot be the arguments of functions. In that case, having
such a function is one of the forms of the Axiom of Choice.

I suggest you read the book by my late wife and myself,
_Equivalents of the Axiom of Choice, II_. It is self-contained,
and should help you to understand the various forms. If you
want more, the book by Jean Rubin and Paul Howard,
_Consequences of the Axiom of Choice_, should handle
most of your concerns about it. It uses ZF or NBG, as
appropriate.


--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558

[Zuhair]

On Apr 18, 8:55 pm, mstem...@walkabout.empros.com (Michael Stemper)
wrote:
> In article <87obqq78ws....@uta.fi>, Aatu Koskensilta <aatu.koskensi...@uta.fi> writes:

Yes there is a difference here it is stronger than choice. Choice is
equivalent to having a two place primitive choice function c(x,y),
where for every y where x e y we have c(x,y) e x for a non empty x.
The formulation I was speaking about (the one that uses one place
primitive function c whose domain of course is the domain of discourse
itself) is equivalent to *Global choice*. Countable choice is
equivalent to saying that for every countable non empty set x every y
where x e y we have c(x,y) e x. We can restrict this notion to any
desired set of interest to get weaker forms. However choice with more
than two arguments is interpretable from the two place function
c(x,y).

Zuhair

[Frederick Williams]

Michael Stemper wrote:
>
> In article <87obqq7...@uta.fi>, Aatu Koskensilta <aatu.kos...@uta.fi> writes:

> > It's known that adding to ZFC a global choice function, i.e. a
> >function symbol c together with the axiom
> >
> > If x is a non-empty set, c(x) is an element of x.

[...]

>
> This gets me to wondering what the domain of "c" is, since the domain
> of a function is, under ordinary circumstances, a set.

No function, just a function symbol. (As I learned myself somewhere or
other.)

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

[Shmuel Metz]

In <jmmv5t$7hi$1...@dont-email.me>, on 04/18/2012

at 05:55 PM, mste...@walkabout.empros.com (Michael Stemper) said:

>I'm leaning in the "weaker" direction because this version just says
>that you can choose an element from any set. However, it states
>this in terms of a single function (called "c" here).

Not quite; "c" is not actually a function. However, for any non-empty
set S of non-empty sets, {(x,c(x)), x \in S} is a function with domain
S. Or if you're using a definition of function that includes domain
and range, ({(x,c(x)), x \in S}, S, U(S)) is a function.

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to spam...@library.lspace.org

[Aatu Koskensilta]

mste...@walkabout.empros.com (Michael Stemper) writes:

> Could somebody give a layman-accessible description of how this
> differs from the standard statement of Choice (if it does)?

Well, let me try. In set theory we can and do introduce various
operations on sets. These include the powerset operation P that takes a
set x to the set P(x) of all its subsets, the union operation taking a
set to its union, and so on. These operations do not correspond to
functions in set theoretic sense, i.e. they're not sets of ordered pairs
stipulated to exist by this axiom or that. When formalizing set
theoretic reasoning we don't usually take these operations as
primitives, that is, we don't introduce symbols for them in the official
formal language. Rather, whenever we want to formally express some
statement about unions, powersets, cardinality, what not, of sets, we
simply spell out the defining conditions for these operations. In more
detail, if we want to formally express e.g. that every set is an element
of its powerset, which in terms of the powerset operation P could be
phrased as

For all sets x, x in P(x).

we use instead a rephrasing not involving P, e.g.

For all sets x, for all sets y, if y is such that a is in y iff a is
a subset of x, then x is in y.

Now, the reason we can do this for an operation O such as union,
powerset, cardinality, and so on, is that they are definable, in the
sense that there's a condition C expressible in terms of set membership
alone, such that

O(x1,...,xn) = y iff C(x1,...,xn,y).

For instance, for the powerset operation P we have that

P(x) = y iff for all z, z in y iff z is a subset of x

where of course "is a subset of" is to be spelled out in full as usual.

There's nothing to stop us from introducing a choice operation c by
the stipulation that for all non-empty sets x, c(x) is an element of
x. But there's no immediately apparent way of specifying such an
operation explicitly in terms of set membership alone. (And indeed from
results in logic we know there is no such explicit definition.) So on
the face of it, it's not at all clear whether formulating the axiom of
choice in its usual form, saying for any set of non-empty sets there's a
choice function, is equivalent to positing a choice operation. Indeed,
one might naturally expect this posit to be stronger, since as noted
there's no way of getting choice accross the board, so to speak, from
the piecemeal choice functions we get from the usual axiom of choice. As
it turns out, although there's no way to eliminate the choice operation
by means of an explicit definition, in any given proof we can use in
place of c various functions, in the ordinary set theoretic sense, that
provably exist. So the choice operation doesn't allow us to prove
anything about sets we couldn't prove without it.

Earlier, we noted that operations such as the powerset operation P,
the union operation U, the cardinality operation |.|, and so on, do not
correspond to functions in the set theoretic sense of a set of ordered
pairs satisfying certain conditions. But naturally we can introduce
objects that represent such operations, i.e. collections of sets that
are not themselves sets (aka proper classes). Indeed, we meet in
ordinary set theoretic reasoning in addition to P, U, etc. also the
class of all ordinals, the class of all cardinals, the rank function,
and so on. Instead of eliminating this aspect of our reasoning when
going formal, as we do in ZFC, we can mirror it in our tedious
formalities. That is, we can add to ZFC a new type of object, or in
technicalese a sort, for collections of sets, and add an axiom (schema)
saying that to any predicate P(x) expressible in terms of the set
membership relation, there is an object C, the class of all sets x such
that P(x), with the property that

x is an element of C iff P(x).

The end result is a formal theory known as von Neumann-Bernays-Gödel set
theory or NBG for short. What about the choice operation? It does not
correspond to any predicate expressible in terms of set membership
alone, so we don't get such an operation from the axiom (schema) we
introduced for classes. But we can just stipulate the choice operation
into existence, i.e. add an axiom, known as the axiom of global choice,
that says

There is a class C of ordered pairs that's a function, i.e.

if <a,b> and <a,c> are in C, then b = c,

and such that for any non-empty set x there's an element y of x such
that <x,y> is in C.

Once again, it's not clear whether this new axiom allows us to prove
more things about sets. Unlike the case with a primitive operation added
to ZFC, we can't in any straightforward way eliminate the choice class C
in a given proof. This is because we can do more with C than just apply
it to sets. We can build new classes, operations, etc. from C, by
recursion, by all manner of set theoretic machinations. Conceivably,
these classical contortions might have implications in the lower realm
of sets. It turns out, however, that global choice doesn't give us
anything new (at least in so far as we're only interested in sets), but
proving this requires the use of forcing. Somewhat impressionistically,
we show that given any model of NBG we can extend it to a new model,
differing only in the class portion, having exactly the same sets, that
contains a global choice function, cleverly cobbled together from
partial (set) choice functions in the original model.

[Michael Stemper]

In article <4F8F23DF...@btinternet.com>, Frederick Williams <freddyw...@btinternet.com> writes:
>Michael Stemper wrote:
>> In article <87obqq7...@uta.fi>, Aatu Koskensilta <aatu.kos...@uta.fi> writes:

>> > It's known that adding to ZFC a global choice function, i.e. a
>> >function symbol c together with the axiom
>> >
>> > If x is a non-empty set, c(x) is an element of x.
>
>[...]
>>
>> This gets me to wondering what the domain of "c" is, since the domain
>> of a function is, under ordinary circumstances, a set.
>
>No function, just a function symbol.

A subtlety that I missed. Thanks.

--
Michael F. Stemper
#include <Standard_Disclaimer>

This sentence no verb.

[Michael Stemper]

In article <87aa27y...@uta.fi>, Aatu Koskensilta <aatu.kos...@uta.fi> writes:
>mste...@walkabout.empros.com (Michael Stemper) writes:

>> Could somebody give a layman-accessible description of how this
>> differs from the standard statement of Choice (if it does)?
>
> Well, let me try. In set theory we can and do introduce various
>operations on sets. These include the powerset operation P that takes a
>set x to the set P(x) of all its subsets, the union operation taking a
>set to its union, and so on. These operations do not correspond to
>functions in set theoretic sense, i.e. they're not sets of ordered pairs
>stipulated to exist by this axiom or that.

[snip lucid explanation]

That was very helpful. I allowed me to better understand the particular
formulation being presented. It also gave me a better understanding of
why Choice is considered problematic by some. [1]

It especially helped me with some concerns that I had with fundamental
things such as union, intersection, subset, and power set. For instance,
"is a subset of" looks so much like a relation that I had been thinking
that it was defined as one. Similarly "powerset of" looked a lot like a
function, and union and intersection looked a lot like binary operations.

But, in each case the issue of domain and range came up. You've addressed
that quite well.

Thanks for the time taken to write that up.


[1] I knew it couldn't be Banach-Tarski that was the hang-up.

--
Michael F. Stemper
#include <Standard_Disclaimer>

2 + 2 = 5, for sufficiently large values of 2