The strange set in R^2
bigli
point on each horisontal and each vertical line but A be dense in
R^2 .
José Carlos Santos
news:1186068245....@i38g2000prf.googlegroups.com...
> How do I construct a set A in R^2 such that A contains at most one
> point on each horisontal and each vertical line but A be dense in
> R^2 .
I did not check every detail, but it looks like this will work: take a
Hammel base of the reals
over the rational field. Take two elements _a_ and _b_ of that base and
consider the Q-linear
map _f_ from R into R such that f(a) = b, f(b) = a, and f(c) = c for any
element of the base
other than _a_ or _b_. Then the graph of _f_ will have the property that you
want.
Best regards,
Jose Carlos Santos
Arturo Magidin
<math...@hotmail.com.CUT> wrote:
Is there at most one point with rational coordinates on each
horizontal and each vertical line on the plane?
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
math...@hotmail.com.cut
Stephen J. Herschkorn
>In article <u514b31ljkocjidgr...@4ax.com>,
> <math...@hotmail.com.CUT> wrote:
>
>
>>On Thu, 02 Aug 2007 08:24:05 -0700, bigli <nimb...@yahoo.com> wrote:
>>
>>
>>
>>>How do I construct a set A in R^2 such that A contains at most one
>>>point on each horisontal and each vertical line but A be dense in
>>>R^2 .
>>>
>>>
>>How about the rational points?
>>
>>
>
>Is there at most one point with rational coordinates on each
>horizontal and each vertical line on the plane?
>
But I am wondering if we can construct (without choice) such an A
consisting entirely of rational points. For example, something like
this: Order all the dyadic rationals and all the (triadic?) rationals
of the form k / 3^n, and form ordered pairs by matching up the
orderings. Can we come up with such orderings so that A is dense in
the plane?
--
Stephen J. Herschkorn sjher...@netscape.net
Math Tutor on the Internet and in Central New Jersey and Manhattan
The World Wide Wade
bigli <nimb...@yahoo.com> wrote:
Choose pairwise disjoint countable subsets E_n of R with the property
that every open interval contains some E_n. Choose pairwise disjoint
countable *dense* subsets F_n of R. Map each E_n bijectively onto F_n.
Consider the resultant graph.
hagman
wrote:
> In article <1186068245.567878.74...@i38g2000prf.googlegroups.com>,
>
> bigli <nimbi...@yahoo.com> wrote:
> > How do I construct a set A in R^2 such that A contains at most one
> > point on each horisontal and each vertical line but A be dense in
> > R^2 .
>
> Choose pairwise disjoint countable subsets E_n of R with the property
> that every open interval contains some E_n. Choose pairwise disjoint
> countable *dense* subsets F_n of R. Map each E_n bijectively onto F_n.
> Consider the resultant graph.
Let E_n=Q+sqrt(p_{2n}), F_n=Q+sqrt(p_{2n+1}) where p_k is the kth
prime.
Mapping E_n->F_n, x|->x+sqrt(p_{2n+1})-sqrt(p_{2n}) will not produce a
dense graph.
hagman
Oops, I read "intersects" (which would have been redundant) where you
wrote "contains"
Hero
What about the diagonal
{ ( r, r ) | r element R}
and similar monoton advancing lines?
They are dense in the interpretation, that You can not cross from one
side to the other without intersecting a point of it.
With friendly greetings
Hero
Arturo Magidin
Never heard that "interpretation" of "dense" before.
"Dense" in this context means that the closure of the set is all of
R^2; that is, ->any<- open ball, ->anywhere<- in the plane, must
contain at least one point from the set.
Hero
> In article <1186075514.454816.299...@i13g2000prf.googlegroups.com>,
>
> Hero <Hero.van.Jind...@gmx.de> wrote:
> >bigli wrote:
> >> How do I construct a set A in R^2 such that A contains at most one
> >> point on each horisontal and each vertical line but A be dense in
> >> R^2 .
>
> >What about the diagonal
> >{ ( r, r ) | r element R}
> >and similar monoton advancing lines?
> >They are dense in the interpretation, that You can not cross from one
> >side to the other without intersecting a point of it.
>
> Never heard that "interpretation" of "dense" before.
>
> "Dense" in this context means that the closure of the set is all of
> R^2; that is, ->any<- open ball, ->anywhere<- in the plane, must
> contain at least one point from the set.
>
I see. And the rationals do not work. But may be, when You rotate
them around one point a certain angle?
( p, q ) ----> ( cos @ , sin @) times ( p, q)
With friendly greetings
Hero
Herman Rubin
Rotating the rational points by an angle with irrational
sine and cosine will work. Alternatively, one can get a
dense set in the open unit square by using the fractional
parts of n*sqrt(2), n*sqrt(3) and mapping the open square
onto the plane in any continuous manner.
--
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
The World Wide Wade
hru...@odds.stat.purdue.edu (Herman Rubin) wrote:
> In article <1186068245....@i38g2000prf.googlegroups.com>,
> bigli <nimb...@yahoo.com> wrote:
> >How do I construct a set A in R^2 such that A contains at most one
> >point on each horisontal and each vertical line but A be dense in
> >R^2 .
>
> Rotating the rational points by an angle with irrational
> sine and cosine will work.
No, e^(iPi/4)*(q+iq) is on the y-axis for every rational q.
> Alternatively, one can get a
> dense set in the open unit square by using the fractional
> parts of n*sqrt(2), n*sqrt(3) and mapping the open square
> onto the plane in any continuous manner.
??? You may have misread the problem.
Robert Sheskey
nimb...@yahoo.com says...
>
>
>How do I construct a set A in R^2 such that A contains at most one
>point on each horisontal and each vertical line but A be dense in
>R^2 .
>
For any irrational t, let A = { (p+qt, pt+q) : p,q rational }.
RS
Virgil
Robert Sheskey <rshe...@san.rr.com> wrote:
This set, being countable, projects vertically onto a countable subset
of the x axis, but the whole of the x axis is not countable, A so does
not contain a point on every vertical line.
If such an A can exist at all, it must at least be as uncountable as R.
Ben Rudiak-Gould
> hru...@odds.stat.purdue.edu (Herman Rubin) wrote:
>> Rotating the rational points by an angle with irrational
>> sine and cosine will work.
>
> No, e^(iPi/4)*(q+iq) is on the y-axis for every rational q.
Yes, replace "sine and cosine" with "tangent".
>> Alternatively, one can get a
>> dense set in the open unit square by using the fractional
>> parts of n*sqrt(2), n*sqrt(3) and mapping the open square
>> onto the plane in any continuous manner.
>
> ??? You may have misread the problem.
I think he means the points (f(frac(n*sqrt(2))), g(frac(n*sqrt(3)))), where
f and g are continuous bijections from (0,1) to R.
-- Ben
Ben Rudiak-Gould
> This set, being countable, projects vertically onto a countable subset
> of the x axis, but the whole of the x axis is not countable, A so does
> not contain a point on every vertical line.
Which is fine, since the original problem didn't say that it should. But I
would be interested to know if you can solve the "exactly one" variant
without AC.
-- Ben
quasi
Nice.
For brevity, that's hard to beat.
quasi
quasi
In the statement of the problem, you missed the words "at most one".
quasi
The World Wide Wade
Ben Rudiak-Gould <br276d...@cam.ac.uk> wrote:
Sure. Let D be any countable dense subset of the plane that contains
at most one point on each horizontal and each vertical line, e.g. one
of the several posted in this thread. Let D_x, D_y be the projections
of D into the x and y axes resp. D is then the graph of a bijection f
: D_x -> D_y. Because D_x, D_y are countable, R \ D_x, R \ D_y have
the cardinality of R. Let g be any bijection of R \ D_x onto R \ D_y.
If we define h = f on D_x, h = g on R \ D_x, then h is a bijection of
R onto R. The graph of h has the "exactly one" property as well as
being dense in the plane.
David Bernier
> Ben Rudiak-Gould <br276d...@cam.ac.uk> wrote:
>
>> Virgil wrote:
>>> This set, being countable, projects vertically onto a countable subset
>>> of the x axis, but the whole of the x axis is not countable, A so does
>>> not contain a point on every vertical line.
>> Which is fine, since the original problem didn't say that it should. But I
>> would be interested to know if you can solve the "exactly one" variant
>> without AC.
>>
>> -- Ben
>
> Sure. Let D be any countable dense subset of the plane that contains
> at most one point on each horizontal and each vertical line, e.g. one
> of the several posted in this thread. Let D_x, D_y be the projections
> of D into the x and y axes resp. D is then the graph of a bijection f
> : D_x -> D_y. Because D_x, D_y are countable, R \ D_x, R \ D_y have
> the cardinality of R. Let g be any bijection of R \ D_x onto R \ D_y.
If we are working in ZF, and not using the Axiom of Choice, AC,
how do we prove the existence of g: R\ D_x -> R\ D_y (bijective)?
As I see it, the g above is used to pair up vertical lines that don't
meet D with horizontal lines that don't meet D in a
one-to-one fashion. So maybe we can assume D is of the type
constructed by R. Sheskey ...
David Bernier
David Libert
To guarantee that a constructed set is dense R^2, it suffices that
ti has a member in each non-empty open rectangle having rational
coordinates for the horizontal position of the vertical sides and for
the vertical postion of the horizontal sides. Theere are only
countably many of these (indexed by suitable quadruplres of
rationals).
So (without choice) order these in an omega sequence.
Now define a set A in omega stages by inductively at stage n adding
to A an element of the n'th such rectangle. Only finitely many points
were previously added to A, so only finitely many veritcal and
horizontal coordinates have been rules out for further use (to
maintain the at most one condition), so pick from the open rectangle
to avoid those finitely many values. To avoid AC, pick the least
rational coordinate (both coordinates) point according to an omega
ordering of Q.
So produce a countable A as required.
--
David Libert
manin...@googlemail.com
> "bigli" <nimbi...@yahoo.com> escreveu na mensagemnews:1186068245....@i38g2000prf.googlegroups.com...
How about this (I'm busking this so let's see if it works):
First I thought:
\{ x \in \R^2 : |x| \in \Rationals, (\arcsin ( \frac{x_1}{x_2} } + |
X| ) \in \Rationals \}
or if you prefer:
{X=(x,y) in R^2 : |X| in Q, (arcsin(x/y} + |X| ) in Q}
Assuming Pi is irrational (nor a square root of a rational - perhaps)
I think this may work. If not perhaps just one more layer of
obfuscation is required, i.e. multiplying the second |X| in the
formula above by another irrational number which is related to Pi by
an irrational factor.
CAsper Clemence
manin...@googlemail.com
discussion thread already, just saw the original post.
Casper Clemence
David Libert
> > Ben Rudiak-Gould <br276delet...@cam.ac.uk> wrote:
>
> >> Virgil wrote:
> >>> This set, being countable, projects vertically onto a countable subset
> >>> of the x axis, but the whole of the x axis is not countable, A so does
> >>> not contain a point on every vertical line.
> >> Which is fine, since the original problem didn't say that it should. But I
> >> would be interested to know if you can solve the "exactly one" variant
> >> without AC.
>
> >> -- Ben
>
> > Sure. Let D be any countable dense subset of the plane that contains
> > at most one point on each horizontal and each vertical line, e.g. one
> > of the several posted in this thread. Let D_x, D_y be the projections
> > of D into the x and y axes resp. D is then the graph of a bijection f
> > : D_x -> D_y. Because D_x, D_y are countable, R \ D_x, R \ D_y have
> > the cardinality of R. Let g be any bijection of R \ D_x onto R \ D_y.
>
> If we are working in ZF, and not using the Axiom of Choice, AC,
> how do we prove the existence of g: R\ D_x -> R\ D_y (bijective)?
The articles below give two proofs in ZF only of:
2-partioning Claim: there are no A,B partitioning the reals
with A countable and |B| < c.
[1] David Libert May 1 2001 "Re: Strategy vs Tactics. was:
Famous Quote"
http://groups.google.com/group/sci.math/msg/d25c654ad77ba21d
[2] Fred Galvin May 1 2001 "Re: Strategy vs Tactics. was:
Famous Quote"
http://groups.google.com/group/sci.math/msg/f6144b80e5066778
Since D_x and D_y are both countable, by the claim R\D_x and R
\D_y are both c-sized and hence bijective.
--
David Libert
David Libert
So the above settles the "exactly one" variant without AC, as
noted.
As in the comments above, the "exactly one" variant can be seen from
its statement, not thinking directly about actual constructed
solutions but just inferring from the statement of the question, that
any such set as required must be the graph of a bijection of R >->> R.
In
[1] Robert Sheskey Aug 2 '07 "Re: The strange set in R^2"
http://groups.google.com/group/sci.math/msg/bfa4927136081278
there is the elegant solution
For any irrational t, let A = { (p+qt, pt+q) : p,q rational }.
Not having read [1] carefully after it was posted I wrote
[2] David Libert Aug 4 '07 "Re: The strange set in R^2"
http://groups.google.com/group/sci.math/msg/d505e638a313e80e
giving a longer solution.
On the other hand, [2] gets, without AC, all the points in AS to
have both coordinates rational, answered a question of Stephen
Herschkorn.
Also, we can modify [2] in combination with Wade's solution from he
parent article above. Make an omega list of the rationals, and
interleave with the steps in [2] steps to add a pairing, avoiding
previously used values as in [2], which in turn throw the n'th
rational (according tho the omega listing) as first a vertical value
and then separately a horizontal value, if that n'th rational was not
previously so used, else do nothing for that step.
This has the effect of making the contructed A be not merely a
bijection between 2 subsets of Q = the rationals, but a full
bijection of Q to itself, ie all of Q got thrown in.
Next take the identity function on R-Q, and union that with A the
graph of the bijection of Q. This is like Wade's solution, but
knowing Range A = Domain A, we can use the identity function on the
other part.
So we end up in ZF with a graph dense in R^2, identiy function off
Q. So on a comeager set and on a full Lebesgue measure set the
function is just the identity.
We can also replace Q by any other countable dense set, for example
"sparse" subsets of Q.
--
David Libert
µ
> How do I construct a set A in R^2 such that A contains at most one
> point on each horisontal and each vertical line but A be dense in
> R^2 .
Using AC, I think the following works: consider R as a Q-vector space,
let H be a supplementary of
Q in R and f be the symetry about Q parallelly to H. Then the graph of
f answers the problem.
--
Mū
David Bernier
> On Aug 2, 10:36 pm, David Bernier <david...@videotron.ca> wrote:
>> The World Wide Wade wrote:
>>
>>
>>
>>> In article <f8th9b$m3...@gemini.csx.cam.ac.uk>,
>>> Ben Rudiak-Gould <br276delet...@cam.ac.uk> wrote:
>>>> Which is fine, since the original problem didn't say that it should. But I
>>>> would be interested to know if you can solve the "exactly one" variant
>>>> without AC.
>>>> -- Ben
>>> Sure. Let D be any countable dense subset of the plane that contains
>>> at most one point on each horizontal and each vertical line, e.g. one
>>> of the several posted in this thread. Let D_x, D_y be the projections
>>> of D into the x and y axes resp. D is then the graph of a bijection f
>>> : D_x -> D_y. Because D_x, D_y are countable, R \ D_x, R \ D_y have
>>> the cardinality of R. Let g be any bijection of R \ D_x onto R \ D_y.
>> If we are working in ZF, and not using the Axiom of Choice, AC,
>> how do we prove the existence of g: R\ D_x -> R\ D_y (bijective)?
>
> The articles below give two proofs in ZF only of:
>
> 2-partioning Claim: there are no A,B partitioning the reals
> with A countable and |B| < c.
>
> [1] David Libert May 1 2001 "Re: Strategy vs Tactics. was:
> Famous Quote"
> http://groups.google.com/group/sci.math/msg/d25c654ad77ba21d
>
> [2] Fred Galvin May 1 2001 "Re: Strategy vs Tactics. was:
> Famous Quote"
> http://groups.google.com/group/sci.math/msg/f6144b80e5066778
>
> Since D_x and D_y are both countable, by the claim R\D_x and R
> \D_y are both c-sized and hence bijective.
Thanks for those references to archived articles.
In the message
< http://groups.google.com/group/sci.math/msg/d25c654ad77ba21d>
David Libert wrote in part:
"A difficult case which we can't rule out happening over ZF is the
reals partioning into sets A, B each of cardinality strictly less
than c = 2^Aleph_0. " (***)
[BTW: I was very intrigued by your proof of 2-partitioning Claim.]
If I remember correctly what someone wrote quite recently,
the Cantor–Bernstein–Schroeder theorem is a ZF-theorem.
I'd be interested to know if the following is a
correct rendering in ZF of
"the subset A of the reals has cardinality
strictly less than 2^Aleph_0."
[Proposed rendering]
"There is no injective mapping f: R -> A."
I'm wondering if the problem (***) is equivalent to:
"Is it a theorem in ZF that
``If A and B are disjoint subsets of
the reals whose union is the reals and if
there is no injection
i: R ->A, then there will always be
an injective map j: R ->B.''
?."
===
For Robert Sheskey's solution,
"For any irrational t, let A = { (p+qt, pt+q) : p,q rational }."
with t=sqrt(2), isn't it the case that, (following Wade's ideas),
the projections of A onto the x and y-axes are the same?
The plane set I'm thinking of is then
A \union ( R - {p+qt: p,q rational}) x ( R - {p+qt: p,q rational}).
David Bernier
David Libert
You're welcome. I thought that thread was interesting, the other
articles too than the two I gave.
> In the message
> <http://groups.google.com/group/sci.math/msg/d25c654ad77ba21d>
> David Libert wrote in part:
>
> "A difficult case which we can't rule out happening over ZF is the
> reals partioning into sets A, B each of cardinality strictly less
> than c = 2^Aleph_0. " (***)
>
> [BTW: I was very intrigued by your proof of 2-partitioning Claim.]
Thanks. The second article by Fred Galvin had a nice proof too,
shorter than mine. And it also proved the stronger result, with A
countable weakened to A well-orderable.
> If I remember correctly what someone wrote quite recently,
> the Cantor-Bernstein-Schroeder theorem is a ZF-theorem.
Yes it is. The usual zig-zag proof doesn't use any AC. And as I
noted in this thread, it actually defines the bijection from
parameters for the two injections.
> I'd be interested to know if the following is a
> correct rendering in ZF of
> "the subset A of the reals has cardinality
> strictly less than 2^Aleph_0."
>
> [Proposed rendering]
> "There is no injective mapping f: R -> A."
Yes.
> I'm wondering if the problem (***) is equivalent to:
>
> "Is it a theorem in ZF that
> ``If A and B are disjoint subsets of
> the reals whose union is the reals and if
> there is no injection
> i: R ->A, then there will always be
> an injective map j: R ->B.''
> ?."
Yes.
> ===
>
> For Robert Sheskey's solution,
> "For any irrational t, let A = { (p+qt, pt+q) : p,q rational }."
>
> with t=sqrt(2), isn't it the case that, (following Wade's ideas),
> the projections of A onto the x and y-axes are the same?
Yes! Doh! I never thought to check this. So all my comments about
ny solution apply to Sheskey's as well. That is my commments about
using the identity function off A. My other comments about making A
being a designated countable dense set remain.
> The plane set I'm thinking of is then
> A \union ( R - {p+qt: p,q rational}) x ( R - {p+qt: p,q rational}).
Yep, as I was writing the identity function off A.
> David Bernier
By the way, I will mention an additional side point. I made a
staged argument to build a dense A. Later I proposed a variant of it,
putting in side steps to make A be the graph of a full bijection of
the rationals Q.
I later realized the original construction actually aleady had the
property without the side clauses, so I could have just used the
original construction without the modification.
Namely, considering the A from the original conmstruction, and since
it is the graph of a bijection between subsets of Q, how could a
specific rational number r ever get out of
dom A ?
There are infinitely many rational rectangles containing r in their
horizontal sides. So in the omega seauence of stages there are
infniitely many stages at which r had the potential to be used as the
horizontal coordinate.
What could stop r from being picked at such stages. If it had been
used as a horizontal coordinate at some earlier stage, but then it is
in dom A already. Or if other rationals listed earlier than r in
the omega sequence also qualify so got used instead. But there are
only finitely many rationals preceding r in the omega listing. Once
one of those rationals ruins a step for r, it has been used and so
will never be picked again and so never block r again. So in the
infinite sequence of stages which could have considered r according
to its interval membership, only finitely many sidestepped it for an
earlier rational.
So eventually we get a stage which used r and puts it into dom A.
Similarly for ran A.
Anyway, given any other A which is the graph of a countable subset
of Q to some countable subset of Q, you could always union the
original dom A and ran A, and if necessary throw in countably many
more points, and extend A to a larger bijection on this bigger set
B, by mapping B - dom A to B - ran A, both made countabyl
infinite. And obtain that way a new bijection on coubntable B to
itself, then extend to R by identity on R - B.
So your comments above about Shesky's solution already got dom =
ran. And as I just said, my first solution did too.
But if there were any other solution A not doing this, after the
fact it can make a B as I just said, without tinkering with any
details of the construction of A.
--
David Libert
David Libert
[...]
>
> In the message
> <http://groups.google.com/group/sci.math/msg/d25c654ad77ba21d>
> David Libert wrote in part:
>
> "A difficult case which we can't rule out happening over ZF is the
> reals partioning into sets A, B each of cardinality strictly less
> than c = 2^Aleph_0. " (***)
I still don't know the status of this. Though in another article
of the archived thread containing the above I wrote:
> Possibly relevant is the 1982 thesis listed in the Math Genealogy Project
>at
>
> http://hcoonce.math.mankato.msus.edu/html/id.phtml?id=31782
>
>of Mark van Liere supervised by Leo Harrington and titled:
>Splitting the Reals into Two Small Pieces .
But I just realized the following about this general property of
partitions like this in ~AC models.
ZF proves AC <-> all cardinalities are comparable, where we
make a version of cardinality to cover the cases of non-well-orderable
sets. The last condition says that for any sets A,B either A can
inject into B or B can inject into A.
To see this <->, for -> every set is well-orderable by AC, so all
infinite cardinals are alephs and all alephs are comparable. For <-,
for each infiniite set A there is the alpha the Hartog's number of
A, which is an aleph which cannot inject into A. If all sets are
comparable A must therefore inject into alpha, so the well ordering on
alpha can be pulled back to A, so an arbitrary infinite A can be well-
ordered, hence AC.
So in any ~AC model, there are sets A, B incomparable to each
other. By disjointifying if necessary, for example replace B by {A}
x B, we can assume without loss of generality these are disjoint.
Let C = A union B. If C could inject into A, then B injects
into C (= A union B) and C injects into A, so composing B injects
into A contrary to the incomparability. Hence C doesn't inject into
A.
Similarly C doesn't inject into B.
So C has a partitioning (A,B) into two pieces each of strictly
smaller cardinality than #C, as the question asked about for R.
So I don't know what can happen with R. But just the basic sort of
partitioning is inevitable for some sets, just from ~AC.
Also, AC proves there is no such partitioning (with at least one
argument infinite, under AC binary cardinal + just returns the maximum
of its two cardinal arguments, which easily shows the larger of A,B
is bijectrive with A union B).
--
David Libert
David Bernier
Thanks for your two replies. I can't follow everything, but
I can follow some of it.
I don't know if I've got this right, but it would seem likely
true that "strange partitions" , or the "difficult case"
in (***) , cannot occur with ZF + CH...
On the other hand, ZF + ~CH implies the existence of
an infinite subset S of the reals which is uncountable but
not equipollent to the reals. Then the reals can
be partitioned as S \union R\S . So there is
no injection j: R->S (else, |R| = |S|). The same question
arises: is there then an injection i: R -> (R\S) ?
We have S uncountable, but no AC. This is as far
as I can go.
David Bernier
David Libert
>
> Thanks for your two replies. I can't follow everything, but
> I can follow some of it.
>
> I don't know if I've got this right, but it would seem likely
> true that "strange partitions" , or the "difficult case"
> in (***) , cannot occur with ZF + CH...
Yes, that's right. CH says c = Aleph_1, in particular R is well-
orderable. But ZF proves that for for any infinite C which is well-
orderable (not necessarily C = R, but including that case), all
subsets A and B of C are well-orderable, so given a partitioning into
infinite subsets A,B of C, #A, #B, #C are all alephs, so cardinal
addition just returns maximum, so #A, #B can't both be < #C. So
there are no examples like this for well-orderable C, in particular
none for R if CH.
This is like my comments 2nd last article from AC, but all I really
needed was C well-orderable.
> On the other hand, ZF + ~CH implies the existence of
> an infinite subset S of the reals which is uncountable but
> not equipollent to the reals. Then the reals can
> be partitioned as S \union R\S . So there is
> no injection j: R->S (else, |R| = |S|). The same question
> arises: is there then an injection i: R -> (R\S) ?
> We have S uncountable, but no AC. This is as far
> as I can go.
Yes. And that's a good point you raise. S is uncountable. Because
if instead S were denumerable, then the 2 partitioning claim earlier
would rule out such a partitioning.
But with S uncountable that doesn't say. And like you say, with no
AC, it sure doesn't seem easy to see how to construct i : R >-> R
\S . We know if instead R injected into S, then we could take
examples with S big and R\S small so there is no i, so somehow a
proof of such i must use that there are no injections R >-> S. How
to use the absence of those injections to construct one into R\S ?
I still don't know about the A,B partitioning like this of C for C =
R, as you were jjust discussing.
I will mention something I thought of since my last post about the
general C case. Earlier I wrote that in a ~AC model, if A and B
are incomparable disjoint, then C = A union B has A,B as such a
partitioning.
I also noted that with AC, there are no infinite C with such an
A,B partitioning.
(Hey an amusing side note: if C is finite and #C > 1 , then any
partitioning of C has the required properties :) ).
Anyway, that contrast earlier, incomparables make examples, and no
examples with AC (which AC also implies : no incomparables) led me
to wonder: are all example partitions of the form I first
constructed?
Ie: does ZF prove: if A,B are comparable cardinality and
disjoint, is A,B NOT an example partition of C = A union B ?
Ie, yes to that would say all examples look like the one I gave.
Anyway, it turns out the answer the the last is no. Paul Cohen's
first ~AC model in _Set Theory and the Continuum Hypothesis_
constructs V a set of real numbers with symmetry properties. Or
there is also a contruction of a type level up, of a subset of P(R)
which is amorphous (as opposed to V, a subset of R).
Anyway, using V as either of those constructions, I think you can
show in the ~AC models constructed the only functions V -> V differ
at finitely many places from either identity_V or else soem constant
function on V. Ie, to make a function V -> V, you can use the
identity function, or you can pick some V member and use the constant
function going to that member. You can also modify those examples so
far by changing one of those starting functions at finitely many
domain values to be an arbitrary finite pairing of V values. And
these are all the functions you can have in the model.
Anyway, with that, take A = V X {0}, B = v x {1} to disjointify
them. Let C = A union B.
So not only are A,B comparable, they are actually bijective. But
there is no injection of C into either A or B, because there isn't
enough room to squeeze 2 copies of V back into one copy, otherwise the
copoes would induce bad functions V -> V.
So apparently in some ~AC models we can have A,B partitioning C
even with #A = #B yet still an example as wanted.
--
David Libert
David Libert
[...]
>
> Ie: does ZF prove: if A,B are comparable cardinality and
> disjoint, is A,B NOT an example partition of C = A union B ?
>
> Ie, yes to that would say all examples look like the one I gave.
>
> Anyway, it turns out the answer the the last is no. Paul Cohen's
> first ~AC model in _Set Theory and the Continuum Hypothesis_
> constructs V a set of real numbers with symmetry properties. Or
> there is also a contruction of a type level up, of a subset of P(R)
> which is amorphous (as opposed to V, a subset of R).
>
> Anyway, using V as either of those constructions, I think you can
> show in the ~AC models constructed the only functions V -> V differ
> at finitely many places from either identity_V or else soem constant
> function on V. Ie, to make a function V -> V, you can use the
> identity function, or you can pick some V member and use the constant
> function going to that member. You can also modify those examples so
> far by changing one of those starting functions at finitely many
> domain values to be an arbitrary finite pairing of V values. And
> these are all the functions you can have in the model.
No, that's wrong for Cohen's V subset of R. The linear ordering
of R induces partitions of V by intervals, so let's you combine
different definitions by cases according to interval membership. This
V is dense in R. (Actually Cohen phrases it for "R" = P(w) but we
can pass back and forth to mathematician's R by bijections.)
So you could define V - > V as a specific constant fo negative V
elements and identity for non-negative, against what I said.
So use the other construction of V amorphous and then I think the
above and its uses later are ok.
--
David Libert