Bijection Between Complex Numbers and Real Numbers

[Michael Ejercito]

What bijective function exists such that every complex number maps
to a unique real number, and likewise every real number maps to a
unique complex number?


Michael

[William Elliot]

Let f:R -> R^2 be a bijection.
Define g:R^2 -> C, (x,y) -> x + iy.

gf:R -> C is a bijection.

[Virgil]

In article <Pine.NEB.4.64.12...@panix3.panix.com>,

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

> On Sat, 18 Feb 2012, Michael Ejercito wrote:
>
> > What bijective function exists such that every complex number maps
> > to a unique real number, and likewise every real number maps to a
> > unique complex number?

There is certainly no such bijective function which preserves any of the
arithmetical, order or topological properties between the two fields.
--

[Butch Malahide]

On Feb 19, 1:25 am, Virgil <vir...@ligriv.com> wrote:
> In article <Pine.NEB.4.64.1202182143080.27...@panix3.panix.com>,

That seems slightly overstated, in view of the fact that (assuming the
axiom of choice) the additive groups are isomorphic.

[Virgil]

In article
<607a757f-6f08-4e1e...@p13g2000yqd.googlegroups.com>,

But one need not assume the axiom of choice.
--

[Pubkeybreaker]

I expect that he wants an example for f.

[Pubkeybreaker]

Consider a bijective map from [0,1] to the unit square.

Let r \in [0,1]. Let (x,y) be a point in the unit square.
let r = .a1 a2 a3 .....

Let x = .a1 a3 a5 .....
y = .a2 a4 a6 .....

i.e. take x as being formed from every other digit in the decimal
representation of r. Similarly for y.

[Richard Tobin]

In article <cb39da8d-c390-4a10...@l14g2000vbe.googlegroups.com>,

Pubkeybreaker <pubkey...@aol.com> wrote:

>Consider a bijective map from [0,1] to the unit square.
>
>Let r \in [0,1]. Let (x,y) be a point in the unit square.
>let r = .a1 a2 a3 .....
>
>Let x = .a1 a3 a5 .....
> y = .a2 a4 a6 .....
>
>i.e. take x as being formed from every other digit in the decimal
>representation of r. Similarly for y.

Though this has the usual problem with multiple representations. For
example, both 0.1 and 0.00909090... map to (0.1, 0).

-- Richard

[Shmuel Metz]

In <virgil-D05694....@bignews.usenetmonster.com>, on
02/19/2012

at 01:50 AM, Virgil <vir...@ligriv.com> said:

>But one need not assume the axiom of choice.

Also, if one assumes the AOC then it is possible to preserve more than
just the additive structures. In fact, any group isomorphism between R
and C is also an isomorphism between R and C as vector spaces over Q.

--
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

[Michael Ejercito]

On Feb 19, 7:03 am, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:
> In article <cb39da8d-c390-4a10-a159-2fdaa5c7e...@l14g2000vbe.googlegroups.com>,

>
> Pubkeybreaker  <pubkeybrea...@aol.com> wrote:
> >Consider a bijective map from [0,1] to the unit square.
>
> >Let  r \in [0,1].   Let (x,y) be a point in the unit square.
> >let r = .a1 a2 a3 .....
>
> >Let x = .a1 a3 a5  .....
> >    y = .a2 a4 a6 .....
>
> >i.e. take x as being formed from every other digit in the decimal
> >representation of r.  Similarly for y.
>
> Though this has the usual problem with multiple representations.  For
> example, both 0.1 and 0.00909090... map to (0.1, 0).
>
> -- Richard

That is correct. This would imply that each complex number maps
onto MULTIPLE real numbers, and certainly the reals can not have a
GREATER cardinality than the complexes.

Then when you consider signs....


Michael

[Kaba]

Michael Ejercito wrote:
> > Though this has the usual problem with multiple representations.  For
> > example, both 0.1 and 0.00909090... map to (0.1, 0).
> >
> > -- Richard
> That is correct. This would imply that each complex number maps
> onto MULTIPLE real numbers, and certainly the reals can not have a
> GREATER cardinality than the complexes.

Just identify the representations which have the same value. No problem.
A real number exists independent of representation, just as a vector
exists independent of its coordinates in some basis.

--
http://kaba.hilvi.org

[Virgil]

In article
<cb39da8d-c390-4a10...@l14g2000vbe.googlegroups.com>,
Pubkeybreaker <pubkey...@aol.com> wrote:

That maps the single point r = 0.1000... = 0.0999...
to both the point x = 0.1000..., y = 0.000...
and to the point x = 0.0999... = 0.01 , y = 0.999... = 1, and maps those
two complex points back to a single real point.

And similarly for all the infinitely many reals in [0,1] having dual
representation.
--

[David R Tribble]

Michael Ejercito wrote:
>> What bijective function exists such that every complex number maps
>> to a unique real number, and likewise every real number maps to a
>> unique complex number?
>

Pubkeybreaker wrote:
> Consider a bijective map from [0,1] to the unit square.
> Let r \in [0,1]. Let (x,y) be a point in the unit square.
> let r = .a1 a2 a3 .....
> Let x = .a1 a3 a5 .....
> y = .a2 a4 a6 .....
> i.e. take x as being formed from every other digit in the decimal
> representation of r. Similarly for y.

Yep. This is Cantor's mapping of R to R^2, i.e., mapping
the points on a line to the points on a 2D plane. I was
going to suggest exactly the same.

But you still have to handle problematic equivalent decimal
fractions, such as .4999... (from <.4999..., .9999...>) versus
.5000... (from <.5000..., .0000...>).

[Peter Webb]

"Kaba" <ka...@nowhere.com> wrote in message
news:MPG.29ab7f9ab...@news.cc.tut.fi...

Michael Ejercito wrote:
> > Though this has the usual problem with multiple representations. For
> > example, both 0.1 and 0.00909090... map to (0.1, 0).
> >
> > -- Richard
> That is correct. This would imply that each complex number maps
> onto MULTIPLE real numbers, and certainly the reals can not have a
> GREATER cardinality than the complexes.

Just identify the representations which have the same value. No problem.

_____________________________________________
Problem still exists.

A real number exists independent of representation, just as a vector
exists independent of its coordinates in some basis.

_____________________________________
The bijection depends upon the representation, so its not independent of the
representaion.


--
http://kaba.hilvi.org

[Peter Webb]

"David R Tribble" <da...@tribble.com> wrote in message
news:368d7bf4-7b35-44c0...@f14g2000yqe.googlegroups.com...

There is a proper proof of this on page 99 of "Proofs from the Book" (Aigner
and Zielger). It explicitly produces the bijection using a slightly
different way of combining the decimal digits of x and y. Couldn't find an
explicit bijection on-line, which is strange considering what an important
result this is.

[Tim Little]

On 2012-02-19, Kaba <ka...@nowhere.com> wrote:
> Michael Ejercito wrote:
>> > Though this has the usual problem with multiple representations.  For
>> > example, both 0.1 and 0.00909090... map to (0.1, 0).

[...]

> Just identify the representations which have the same value. No problem.

Then the function fails to be a bijection.


--
Tim

[Kaba]

You are misunderstanding what I mean. No matter how you construct the
set of real numbers, it is a set, denoted by RR. We would like to
construct a bijective function f: RR --> RR^2. To do this, we first
construct a bijective function g : RR --> P subset ZZ^ZZ mapping each
real number to a sequence of digits, such that if we allow us think that
ZZ subset RR, then

g^{-1}((..., x_0, x_1, x_2, ...)) = sum_{i in ZZ} x_i b^i.

Now, let h+ : P --> P such that

h+((x_i)) = (x_(2i)),

and let h-: P --> P such that

h-((x_i)) = (x_(2i + 1)).

Then the claim is that if f is defined as

f(x) = ((g^{-1} of h+ of g)(x), (g^{-1} of h- of g)(x))

then f is a bijection between RR and RR^2.

This is not a proof, just a description of the approach.

--
http://kaba.hilvi.org

[Kaba]

Kaba wrote:
> Now, let h+ : P --> P such that
>
> h+((x_i)) = (x_(2i)),
>
> and let h-: P --> P such that
>
> h-((x_i)) = (x_(2i + 1)).
>

It is not clear though whether the results of h+ and h- are actually in
P...

--
http://kaba.hilvi.org

[Richard Tobin]

In article <513f89be-feb4-4e36...@x19g2000yqh.googlegroups.com>,

Michael Ejercito <meje...@hotmail.com> wrote:

>> Though this has the usual problem with multiple representations.  For
>> example, both 0.1 and 0.00909090... map to (0.1, 0).

> That is correct. This would imply that each complex number maps
>onto MULTIPLE real numbers, and certainly the reals can not have a
>GREATER cardinality than the complexes.

Given a surjection both ways, the dual Schroder Bernstein theorem
shows that there is a bijection. This however relies on the axiom
of choice.

If we take the interleaved digits as a base-11 number we have an injection
both ways, and standard Schroder Bernstein gives us a bijection.

-- Richard

[Josh]

[David C. Ullrich]

On Sun, 19 Feb 2012 01:50:02 -0700, Virgil <vir...@ligriv.com> wrote:

>In article
><607a757f-6f08-4e1e...@p13g2000yqd.googlegroups.com>,
> Butch Malahide <fred....@gmail.com> wrote:
>
>> On Feb 19, 1:25 am, Virgil <vir...@ligriv.com> wrote:
>> > In article <Pine.NEB.4.64.1202182143080.27...@panix3.panix.com>,
>> >  William Elliot <ma...@panix.com> wrote:
>> >
>> > > On Sat, 18 Feb 2012, Michael Ejercito wrote:
>> >
>> > > >    What bijective function exists such that every complex number maps
>> > > > to a unique real number, and likewise every real number maps to a
>> > > > unique complex number?
>> >
>> > There is certainly no such bijective function which preserves any of the
>> > arithmetical, order or topological properties between the two fields.
>>
>> That seems slightly overstated, in view of the fact that (assuming the
>> axiom of choice) the additive groups are isomorphic.
>
>But one need not assume the axiom of choice.

Doesn't matter. You made an assertion. AC implies that your assertion
is false. _Hence_ you cannot _prove_ your assertion, because
a proof of your assertion would then prove "not AC", and that's
impossible.

(Ok, it's impossible if ZF is consistent. When you prove ZF is
inconsistent let us know.)

[David C. Ullrich]

Even without AC it's not hard to show that there _does_ exist a
bijection that preserves Lebesgue measure.

[Michael Ejercito]

It does disturb me. I mean, there are several non-constructive
proofs demonstrating a bijection exists, and yet...


Michael

[jbriggs444]

>  Michael- Hide quoted text -
>
> - Show quoted text -

It is important and obvious that a bijection exists.

That a particular bijection can be explicitly given is
less so.

One way to proceed would be to establish an explicit
bijection between the reals and the infinite binary
digit strings/

Google is able to find one such.

http://mathoverflow.net/questions/56633/simple-bijection-between-reals-and-sets-of-natural-numbers

With that in hand, an explicit bijection between the reals
and the complex numbers would be trivial to construct using
the digit-interleaving approach.

[Virgil]

In article <27h4k7l303sssvd42...@4ax.com>,

So I overstated the case. If I hadn't mentioned topological properties,
would I have still been so wrong?
--

[Peter Webb]

"jbriggs444" <jbrig...@gmail.com> wrote in message
news:1874ebfd-7f04-48bc...@z31g2000vbt.googlegroups.com...

__________________________________________
If its trivial, produce the bijection.

[Peter Webb]

"Richard Tobin" <ric...@cogsci.ed.ac.uk> wrote in message
news:jhtdqe$163e$1...@matchbox.inf.ed.ac.uk...

> In article
> <513f89be-feb4-4e36...@x19g2000yqh.googlegroups.com>,
> Michael Ejercito <meje...@hotmail.com> wrote:
>
>>> Though this has the usual problem with multiple representations. For
>>> example, both 0.1 and 0.00909090... map to (0.1, 0).
>
>> That is correct. This would imply that each complex number maps
>>onto MULTIPLE real numbers, and certainly the reals can not have a
>>GREATER cardinality than the complexes.
>
> Given a surjection both ways, the dual Schroder Bernstein theorem
> shows that there is a bijection.

But doesn't provide an obvious bijection. And this is using a comparatively
advanced result to prove that a solution to a simple problem exists.

> This however relies on the axiom
> of choice.
>

No. Cantor Bernstein Schroder does not require AoC. It is constructive.

[Peter Webb]

"Michael Ejercito" <meje...@hotmail.com> wrote in message
news:6224d05f-746d-4151...@r20g2000yqn.googlegroups.com...

__________________________________________
Represent each Real with its infinite decimal expansion, eg 0.5 is
represented as 0.499....

Consider

x = 0.301200708 ..
y = 0.0092051008 ..

break each into groups by always going to the next non-zero digit inclusive:

x = 0.3 01 2 007 08 ....
y = 0.009 2 05 1 0008 ...

Interleave the groups:

z = 0.3 0009 01 2 2 05 007 1 08 0008 ...

This forms an explicit bijection between R and R^2.




Michael

[jbriggs444]

On Feb 20, 9:31 pm, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> If its trivial, produce the bijection.

Why?

[Peter Webb]

"jbriggs444" <jbrig...@gmail.com> wrote in message

news:0b798015-4105-4415...@hk10g2000vbb.googlegroups.com...

On Feb 20, 9:31 pm, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> If its trivial, produce the bijection.

Why?

____________________________
Because if its trivial, then it should be easy to do so. And I don't think
it is.

Here is his post in full, which you snipped:

"One way to proceed would be to establish an explicit
bijection between the reals and the infinite binary
digit strings/

Google is able to find one such.

http://mathoverflow.net/questions/56633/simple-bijection-between-reals-and-sets-of-natural-numbers

With that in hand, an explicit bijection between the reals
and the complex numbers would be trivial to construct using
the digit-interleaving approach."

The bit that was explained with a web page link is in that a bijection
exists between R and infinite binary strings. Well duh.

The next bit is certainly not trivial. A simple digit interleaving approach
won't work, as 1 = 0.1111... and I have never seen a way around this problem
that actually works when you try all combinations to ensure its truly a
bijection.

There is a digit interleaving algorithm that will work, it is the same as
the base 10 version I described elsewhere, but its hardly trivial to find.

Hence my suggestion that he produces the algorithm, if he thinks it trivial,
so I can see if it works.

Now, do *you* have any more questions?

[G. A. Edgar]

In article <jhv0ah$dhf$1...@news.albasani.net>, Peter Webb

<r.peter...@gmail.com> wrote:

> Represent each Real with its infinite decimal expansion, eg 0.5 is
> represented as 0.499....
>
> Consider
>
> x = 0.301200708 ..
> y = 0.0092051008 ..
>
> break each into groups by always going to the next non-zero digit inclusive:
>
> x = 0.3 01 2 007 08 ....
> y = 0.009 2 05 1 0008 ...
>
> Interleave the groups:
>
> z = 0.3 0009 01 2 2 05 007 1 08 0008 ...
>
> This forms an explicit bijection between R and R^2.
>
>
>
>
> Michael
>

This looks nice. I haven't seen it before.
Presumably the claim is that it is a bijection
from (0,1]^2 to (0,1].

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

[Peter Webb]

"G. A. Edgar" <ed...@math.ohio-state.edu.invalid> wrote in message
news:210220120626456049%ed...@math.ohio-state.edu.invalid...

Yes.

Judging from the responses in this thread, the proof is not particularly
well known. I have only ever seen it in the book I cited. I remembered it
because I have seen so many explicit bijections that don't quite work (and
rely on hand waving or Cantor Bernstein to complete), it was interesting to
see that a minor change to the digit interleaving produces something that
actually does work and is completely self-contained.

[Tim Little]

On 2012-02-21, Peter Webb <r.peter...@gmail.com> wrote:
> "jbriggs444" <jbrig...@gmail.com> wrote in message

> http://mathoverflow.net/questions/56633/simple-bijection-between-reals-and-sets-of-natural-numbers
>
> With that in hand, an explicit bijection between the reals
> and the complex numbers would be trivial to construct using
> the digit-interleaving approach.
> __________________________________________
> If its trivial, produce the bijection.

Let the function defined in the linked article be denoted f:R->P(N).
Then the bijection is g:C->R given by g(a+bi) = f^-1((2f(a)+1) u (2f(b))).

The arithmetic operations on the sets of naturals are defined to be
the obvious element-wise ones.

Is that trivial enough for you?


--
Tim

[Richard Tobin]

In article <jhv0ah$dhf$1...@news.albasani.net>,
Peter Webb <r.peter...@gmail.com> wrote:

>Represent each Real with its infinite decimal expansion, eg 0.5 is
>represented as 0.499....

What about zero?

>This forms an explicit bijection between R and R^2.

If you considered (0,1) instead of R then zero wouldn't be a problem.

-- Richard

[Peter Webb]

"Richard Tobin" <ric...@cogsci.ed.ac.uk> wrote in message

news:ji2jac$guv$1...@matchbox.inf.ed.ac.uk...

atan function blah blah. There are lots of bijections between intervals of R
and R itself.

[Peter Webb]

"Tim Little" <t...@little-possums.net> wrote in message
news:slrnjk9gb...@soprano.little-possums.net...

> On 2012-02-21, Peter Webb <r.peter...@gmail.com> wrote:
>> "jbriggs444" <jbrig...@gmail.com> wrote in message
>> http://mathoverflow.net/questions/56633/simple-bijection-between-reals-and-sets-of-natural-numbers
>>
>> With that in hand, an explicit bijection between the reals
>> and the complex numbers would be trivial to construct using
>> the digit-interleaving approach.
>> __________________________________________
>> If its trivial, produce the bijection.
>
> Let the function defined in the linked article be denoted f:R->P(N).
> Then the bijection is g:C->R given by g(a+bi) = f^-1((2f(a)+1) u (2f(b))).
>

> I assume you use "u" to mean

> The arithmetic operations on the sets of naturals are defined to be
> the obvious element-wise ones.
>
> Is that trivial enough for you?
>

No.

A worked example would be nice.

To start with, I need to know how the mapping handles 0.49999... = 0.5; for
Reals with more than one decimal representation it is not clear how the
function (and more importantly its inverse) are uniquely defined.

And then how this supposedly uses digit interleaving.

You will note that I did give a simple algortithm (though hardly trivial to
find) which does correctly use digit interleaving to produce a bijection,
and I even gave a worked example.

Could you do the same?


>
> --
> Tim

[jbriggs444]

On Feb 21, 8:23 am, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> "jbriggs444" <jbriggs...@gmail.com> wrote in message

>
> news:0b798015-4105-4415...@hk10g2000vbb.googlegroups.com...
> On Feb 20, 9:31 pm, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
>
> > If its trivial, produce the bijection.
>
> Why?
> ____________________________
> Because if its trivial, then it should be easy to do so. And I don't think
> it is.

But you are the one who sees value in the result so you
are the one who should be working to produce the bijection.

> Here is his post in full, which you snipped:
>
> "One way to proceed would be to establish an explicit
> bijection between the reals and the infinite binary
> digit strings/
>
> Google is able to find one such.
>

> http://mathoverflow.net/questions/56633/simple-bijection-between-real...

>
> With that in hand, an explicit bijection between the reals
> and the complex numbers would be trivial to construct using
> the digit-interleaving approach."
>
> The bit that was explained with a web page link is in that a bijection
> exists between R and infinite binary strings. Well duh.

An explicit bijection. You are the one who thinks that
explicit bijections are hard to produce. You should be
going "holy cow", not "duh".

> The next bit is certainly not trivial. A simple digit interleaving approach
> won't work, as 1 = 0.1111... and I have never seen a way around this problem
> that actually works when you try all combinations to ensure its truly a
> bijection.

You are being a stupid idiot.

The explicit bijection between digit strings and real numbers gets
you completely past the dual-representation problem.

> There is a digit interleaving algorithm that will work, it is the same as
> the base 10 version I described elsewhere, but its hardly trivial to find.

You are being a stupid idiot.

> Hence my suggestion that he produces the algorithm, if he thinks it trivial,
> so I can see if it works.

You are being a stupid idiot.

> Now, do *you* have any more questions?

Why are you being a stupid idiot?

[Peter Webb]

"jbriggs444" <jbrig...@gmail.com> wrote in message

news:83681164-ab4b-4cba...@9g2000vbq.googlegroups.com...

On Feb 21, 8:23 am, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> "jbriggs444" <jbriggs...@gmail.com> wrote in message
>
> news:0b798015-4105-4415...@hk10g2000vbb.googlegroups.com...
> On Feb 20, 9:31 pm, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
>
> > If its trivial, produce the bijection.
>
> Why?
> ____________________________
> Because if its trivial, then it should be easy to do so. And I don't think
> it is.

But you are the one who sees value in the result so you
are the one who should be working to produce the bijection.

____________________________
I did produce a bijection. And gave a fully worked example.

> Here is his post in full, which you snipped:
>
> "One way to proceed would be to establish an explicit
> bijection between the reals and the infinite binary
> digit strings/
>
> Google is able to find one such.
>
> http://mathoverflow.net/questions/56633/simple-bijection-between-real...
>
> With that in hand, an explicit bijection between the reals
> and the complex numbers would be trivial to construct using
> the digit-interleaving approach."
>
> The bit that was explained with a web page link is in that a bijection
> exists between R and infinite binary strings. Well duh.

An explicit bijection. You are the one who thinks that
explicit bijections are hard to produce. You should be
going "holy cow", not "duh".

__________________________________________
No. Lots of bijections are easy to find. A bijection between R and R^2 using
digit interleaving is not.

> The next bit is certainly not trivial. A simple digit interleaving
> approach
> won't work, as 1 = 0.1111... and I have never seen a way around this
> problem
> that actually works when you try all combinations to ensure its truly a
> bijection.

You are being a stupid idiot.

The explicit bijection between digit strings and real numbers gets
you completely past the dual-representation problem.

__________________________________________________
I don't see how. Could you do as I have done and produce an explicit
bijection? I have posted one, but it was hardly "trivial" to find.

> There is a digit interleaving algorithm that will work, it is the same as
> the base 10 version I described elsewhere, but its hardly trivial to find.

You are being a stupid idiot.

________________________________________
Huh? How?

> Hence my suggestion that he produces the algorithm, if he thinks it
> trivial,
> so I can see if it works.

You are being a stupid idiot.

_____________________________________
Huh? How?

> Now, do *you* have any more questions?

Why are you being a stupid idiot

__________________________________________
I'm not. If you think I have said something incorrect in this thread, you
should cut-and-paste it in context and we can discuss. Its almost certainly
some lack of mathematical knowledge on your part.

In the mean time, it would be great if you provided a "trivial" digit
interleaving scheme which provides a bijection between R and R^2. I provided
a worked example of one elsewhere in this thread; its clearly not trivial to
find because not one single poster has managed to find it.

[jbriggs444]

On Feb 24, 1:00 am, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> The explicit bijection between digit strings and real numbers gets
> you completely past the dual-representation problem.
> __________________________________________________
> I don't see how.

Right there. That is you being a stupid idiot.

If you have a bijection between the digit strings and the
real numbers then, by definition of bijection, you do
not have a single real number represented by two distinct
digit strings.

That takes you past the dual-representation problem.

> If you think I have said something incorrect in this thread, you
> should cut-and-paste it in context and we can discuss. Its almost certainly
> some lack of mathematical knowledge on your part.

Challenge met above. The lack is yours.

[Peter Webb]

"jbriggs444" <jbrig...@gmail.com> wrote in message

news:6d5c038b-55ec-4b34...@9g2000vbq.googlegroups.com...

On Feb 24, 1:00 am, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> The explicit bijection between digit strings and real numbers gets
> you completely past the dual-representation problem.
> __________________________________________________
> I don't see how.

Right there. That is you being a stupid idiot.

If you have a bijection between the digit strings and the
real numbers then, by definition of bijection, you do
not have a single real number represented by two distinct
digit strings.

That takes you past the dual-representation problem.

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Certainly, but it does not allow for a trivial digit interleaving to form a
bijection between R and R^2, which is the step I asked to be shown and which
you snipped.

> If you think I have said something incorrect in this thread, you
> should cut-and-paste it in context and we can discuss. Its almost
> certainly
> some lack of mathematical knowledge on your part.

Challenge met above. The lack is yours.

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I did in fact post a correct bijection based on digit interleaving elsewhere
in this thread. We have seen nothing from you. If you think that using the
mapping R -> P(N) to form a bijection between R and R^2 using digit
interleaving is trivial, show us how it is done. You will note that with the
digit interleaving bijection I provided earlier I provided a working
example, if you claim that a more trivial digit interleaving bijection
exists please show us how it works. Put up or shut up.