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Sep 22, 1999, 3:00:00 AM9/22/99

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This is a crazy idea I had while reading some of the recent posts in

sci-logic.

sci-logic.

Assume that there is a universe where the integer 3 doesn't exist.

IOW, it was impossible to have 3 objects.

Even though 3 couldn't exist as an integer,

it might still exist as a real number.

How would mathematicians in that universe represent 3?

Is it possible that there are "integers" that don't exist in our universe?

In a universe that contained all of these "super-integers",

would all "super-real" numbers be "super-rational"?

I know that the irrational numbers are uncountable.

(Can't be put into a 1 to 1 relation with the integers.)

Is it possible that there is a set of irrational numbers

such that all other irrational numbers are expressible

as a ratio of two members from this set?

(Would such a set redefine "countable"?)

Could PI be a "super-integer"?

Russell

- integers are an illusion

Sep 22, 1999, 3:00:00 AM9/22/99

to

> This is a crazy idea I had while reading some of the recent posts in

> sci-logic.

> sci-logic.

Close your eyes and relax and it will go away :-)

> Assume that there is a universe where the integer 3 doesn't exist.

> IOW, it was impossible to have 3 objects.

It's not clear what such a universe would be like.

Since you mention you were in sci-logic, it's worth noting

that even if you begin with the empty set, you have to

be careful in order to construct a universe that doesn't

have "3", for example,

{} = null set

{ {} } = set containing one object, which is the null set.

{ {}, {{}} } = set containing two objects, one being the null

set, and the other being the set containing just

the null set

{ {}, {{}}, {{},{{}}} } = ....

The last thing sure looks like 3.

Of course this is a "mathematical" universe, not a "physical"

universe. It's even less clear how you could arrange

this is a physical universe. If you had two objects

and bought in a third, what would happen?

I suppose in either case you could insist that there be

only two objects, and nothing else.

> Is it possible that there are "integers" that don't exist in our universe?

> In a universe that contained all of these "super-integers",

> would all "super-real" numbers be "super-rational"?

People have played with various "extensions" of

the idea of numbers. Their is work on transfinite

cardinals, transfinite ordinals, reals with "infinitesimals",

sur-real numbers, not to mention algebraic things

like finite fields, etc. So there are many things

one can study which in one way or the other are

somewhat like numbers (and intgers) but not

quite.

> Is it possible that there is a set of irrational numbers

> such that all other irrational numbers are expressible

> as a ratio of two members from this set?

Well, you could just take the set of all irrationals, and

say that any irrational can be expressed as the

ratio of two irrationals, but this is a rather dull

theorm( eg, x = x^2 / x).

> (Would such a set redefine "countable"?)

Unless you find a fundamental flaw in logic, the

concept of "countable" will still be a valid

concept and everything we "know" about it

will still be true. Whether there might be

some different abstraction which in the future

we will discover and which will be more interesting

to study is a different matter, and of course

one which I can't answer until I've perfected

my time machine (any yesterday now . . .).

> Could PI be a "super-integer"?

Well, please feel free to define "super-integer".

Again, unless you muck with the underlying logical

system (which, in a sense, they are probably doing

in sci-logic) everything we currently know will

still be true, though one might find more interesting

things to study. For example, you divide the world

up into rationals and irrationals. Well, there

are other divisions. For example, some

numbers are the roots of polynomials with

integer coefficients (so sqrt(2) is a root

of x^2 - 2 = 0) but not all numbers all. The

ones which aren't are called "transcendental".

Similarly, one might be interesed in the

statistical properties of the distribution of

digits in decimal expansions, in which case

one has things like "normal" numbers, etc.

Well, there's always more to do (or at least

it seems that way). One can do even more

bizarre things if one tampers with the underlying

logical foundations. People have studied things

like three valued logics (true, false, huh?).

I am told that in a certain sense there is a

"countable" version of the real, in which

basically one considers the set of all reals

such that one can write down an algorithm for

computing the digits. This set is countable

because there are only a countable number of

algorithms one can write down. The usual

diagonalization argument fails because--I am

told--the number produced by the diagonalization

procedure is not given in a "constructive"

manner, so it "doesn't exist" in this "twisted"

logic (I'm told that this sort of constructive

logic is actually of interest to the computer

science crowd).

Well, always something new under the sun :-)

Best wishes,

Mike

Sep 23, 1999, 3:00:00 AM9/23/99

to

"Russell Easterly" <logi...@wolfenet.com> writes:

>Is it possible that there are "integers" that don't exist in our universe?

Since integers are defined in our universe, this question doesn't

even make sense.

>Could PI be a "super-integer"?

In base PI, yes.

Doug

----------------------------------------------------------------------------

Douglas Todd Norris (norr...@euclid.colorado.edu) "The Mad Kobold"

Hockey Goaltender Home Page:http://ucsu.colorado.edu/~norrisdt/goalie.html

----------------------------------------------------------------------------

"Maybe in order to understand mankind, we have to look at the word itself.

Mankind. Basically, it's made up of two separate words---"mank" and "ind".

What do these words mean? It's a mystery, and that's why so is mankind."

- Deep Thought, Jack Handey

Sep 23, 1999, 3:00:00 AM9/23/99

to

In article <7sc0or$835$1...@sparky.wolfe.net>,

"Russell Easterly" <logi...@wolfenet.com> wrote:

> This is a crazy idea I had while reading some of the recent posts in

> sci-logic.

>

it too seriously.

"Russell Easterly" <logi...@wolfenet.com> wrote:

> This is a crazy idea I had while reading some of the recent posts in

> sci-logic.

>

> Assume that there is a universe where the integer 3 doesn't exist.

> IOW, it was impossible to have 3 objects.

> IOW, it was impossible to have 3 objects.

> Even though 3 couldn't exist as an integer,

> it might still exist as a real number.

> How would mathematicians in that universe represent 3?

>

> it might still exist as a real number.

> How would mathematicians in that universe represent 3?

>

> Is it possible that there are "integers" that don't exist in our

universe?

universe?

> In a universe that contained all of these "super-integers",

> would all "super-real" numbers be "super-rational"?

>

> would all "super-real" numbers be "super-rational"?

>

> I know that the irrational numbers are uncountable.

> (Can't be put into a 1 to 1 relation with the integers.)

>

> (Can't be put into a 1 to 1 relation with the integers.)

>

> Is it possible that there is a set of irrational numbers

> such that all other irrational numbers are expressible

> as a ratio of two members from this set?

> such that all other irrational numbers are expressible

> as a ratio of two members from this set?

> (Would such a set redefine "countable"?)

>

>

> Could PI be a "super-integer"?

>

The following is meant for entertainment purposes only. Don't take>

it too seriously.

In the time-honored tradition, since I can't answer your questions I

will answer some that I can, or at least give a couple hints.

>

> Is it possible that there is a set of irrational numbers

> such that all other irrational numbers are expressible

> as a ratio of two members from this set?

The real numbers form a vector space over the rationals. They have

a basis (known as a Hamel Basis), necessarily uncountable. One element

is rational -- throw it out since you only asked about irrationals.

So, for any rational number r there are finitely many elements of

the Hamel Basis r_1,...,r_n and rational numbers q_1,...,q_n not

all 0 so that r = r_1*q_1 + r_2*q_2 + ... + r_n*q_n

Multiply through by the least common multiple of the denominators

of the q_i and you have a relationship with only integer coefficients.

Of course, it is not actually possible to construct a Hamel Basis --

one

can show, using the axiom of choice, that they exist. Their usual

use is to show the existence of a set which is not Lebesgue measurable.

> Could PI be a "super-integer"?

Consider the lattice of points in the cartesian plane with only

integer components. Suppose only these points existed. Further suppose

for any pair of neighboring points either horizontally or vertically

it takes the same amount of energy to "move form one to the other"

no matter which pair you pick, i.e. for x,y integers it takes "the

same amount of energy to move from" (x,y) to (x+1,y) or (x,y+1) no

matter what x any y are chosen. Further suppose that the only way to

get to "diagonal points" is to first go horizontally and then

vertically (or vice-versa), i.e. to get from (x,y) to (x+1,y+1)

one must first go from (x,y) to (x+1,y) and then to (x+1,y+1) (or

else go vertically first.

Now there is a "natural" metric function on all this, called the

"taxi-cab metric" which is d((a,b),(c,d)) = |a-c| + |b-d|.

Now consider the "circle centered at (0,0) with radius 4", the set

of points (x,y) such that d((0,0),(x,y)) = 4. It is easy to make a

list of the points: {(4,0), (3,1), (2,2),... (3,-1)} . Plot the

points and see what it looks like.

Now fill in more points between the lattice points but keep the rules

about how to move between them. So you keep the taxi-cab metric.

Now, if you have cared enough to follow all this, and you look at

"circles" with this taxi-cab metric you should see that they "look

like" diagonal squares. Furthermore the ratio of the "circumference" of

these "circles" to their radii is 4. Now, it seems to me, that if we

lived in a universe which had a "natural law" that it took the same

amount of energy to go diagonally as to first go horizontally

and then vertically the taxi-cab metric would be the "natural" way

of thinking of distances. Further if there were only discreet, evenly-

spaced points, or only points corresponding to those with rational

components, rational numbers would be enough to do all measurements.

In that case the pythagorean dilemma might never be discovered and

irrational numbers never thought of. "Pi" would be 4.

> Assume that there is a universe where the integer 3 doesn't exist.

> IOW, it was impossible to have 3 objects.

> Even though 3 couldn't exist as an integer,

> it might still exist as a real number.

> How would mathematicians in that universe represent 3?

>

There are (at least) two different kinds of "numbers" arising from

two different human activities -- counting and measuring. It is one

of the beauties of mathematics that there is a very natural way

to embed the "counting numbers" into the "measuring numbers" by

picking a fixed (arbitrary) length and calling it 1, ...

(see the relevant part of Euclid's Elements).

This embedding has been done at least since the time of the

pythagoreans and probably in all cultures with mathematics.

It works because adding lengths of segments "works just like"

adding pebbles.

Consider, then, a universe where this embedding did not work -- where

you needed different rules for adding pebbles than for adding

lengths. So 2 + 1 = 3 for pebbles but 2 + 1 = 2.756398 for line

segments.

Sent via Deja.com http://www.deja.com/

Share what you know. Learn what you don't.

Sep 23, 1999, 3:00:00 AM9/23/99

to

Russell Easterly wrote in message <7sc0or$835$1...@sparky.wolfe.net>...

>This is a crazy idea I had while reading some of the recent posts in

>sci-logic.

>

>Assume that there is a universe where the integer 3 doesn't exist.

>IOW, it was impossible to have 3 objects.

>Even though 3 couldn't exist as an integer,

>it might still exist as a real number.

>How would mathematicians in that universe represent 3?

>

As the love child of Santa Claus and the Easter Bunny.

That's as defensible as anything anyone else is going to be able to say.

Sep 23, 1999, 3:00:00 AM9/23/99

to

Dr. Michael Albert wrote (in part)...

> { {}, {{}}, {{},{{}}} } = ....

>

>The last thing sure looks like 3.

>

>Of course this is a "mathematical" universe, not a "physical"

>universe. It's even less clear how you could arrange

>this is a physical universe. If you had two objects

>and bought in a third, what would happen?

>

>I suppose in either case you could insist that there be

>only two objects, and nothing else.

In Edwin Abbott Abbott's "Flatland"

http://www.ulib.org/webRoot/Books/_Gutenberg_Etext_Books/etext94/flat11.txt

http://www.ofcn.org/cyber.serv/resource/bookshelf/flat10/

there is a fine description of Pointland: a zero-dimensional universe which

contains exactly one object. This object is a god-like creature: he _is_ his

own thoughts, his own words... he _is_ the universe. He is his own past and

his own future. He is at one with every conceivable phenomenon in his

universe. He has no conception of the number two.

Abbott was, for 25 years, the headmaster of the City of London School, where

he had been a student. I went to that school, and was in Abbott House (one

of six houses).

This is from Section 20 of the book:

http://ofcn.org/cyber.serv/resource/bookshelf/flat10/chapter20.html

"Look yonder," said my Guide, "in Flatland thou hast lived; of Lineland thou

hast received a vision; thou hast soared with me to the heights of

Spaceland; now, in order to complete the range of thy experience, I conduct

thee downward to the lowest depth of existence, even to the realm of

Pointland, the Abyss of No dimensions. Behold yon miserable creature. That

Point is a Being like ourselves, but confined to the non-dimensional Gulf.

He is himself his own World, his own Universe; of any other than himself he

can form no conception; he knows not Length, nor Breadth, nor Height, for he

has had no experience of them; he has no cognizance even of the number Two;

nor has he a thought of Plurality; for he is himself his One and All, being

really Nothing. Yet mark his perfect self-contentment, and hence learn his

lesson, that to be self-contented is to be vile and ignorant, and that to

aspire is better than to be blindly and impotently happy. Now listen."

He ceased; and there arose from the little buzzing creature a tiny, low,

monotonous, but distinct tinkling, as from one of your Spaceland

phonographs, from which I caught these words, "Infinite beatitude of

existence! It is; and there is nothing else beside It."

"What," said I, "does the puny creature mean by 'it'?"

"He means himself," said the Sphere: "have you not noticed before now, that

babies and babyish people who cannot distinguish themselves from the world,

speak of themselves in the Third Person? But hush!"

"It fills all Space," continued the little soliloquizing Creature, "and what

It fills, It is. What It thinks, that It utters; and what It utters, that It

hears; and It itself is Thinker, Utterer, Hearer, Thought, Word, Audition;

it is the One, and yet the All in All. Ah, the happiness, ah, the happiness

of Being!"

"Can you not startle the little thing out of its complacency?" said I. "Tell

it what it really is, as you told me; reveal to it the narrow limitations of

Pointland, and lead it up to something higher."

"That is no easy task," said my Master; "try you."

Hereon, raising my voice to the uttermost, I addressed the Point as follows:

"Silence, silence, contemptible Creature. You call yourself the All in All,

but you are the Nothing: your so-called Universe is a mere speck in a Line,

and a Line is a mere shadow as compared with--"

"Hush, hush, you have said enough," interrupted the Sphere, "now listen, and

mark the effect of your harangue on the King of Pointland."

The lustre of the Monarch, who beamed more brightly than ever upon hearing

my words, shewed clearly that he retained his complacency; and I had hardly

ceased when he took up his strain again. "Ah, the joy, ah, the joy of

Thought. What can It not achieve by thinking! Its own Thought coming to

Itself, suggestive of its disparagement, thereby to enhance Its happiness!

Sweet rebellion stirred up to result in triumph! Ah, the divine creative

power of the All in One! Ah, the joy, the joy of Being!"

"You see," said my Teacher, "how little your words have done. So far as the

Monarch understands them at all, he accepts them as his own -- for he cannot

conceive of any other except himself --and plumes himself upon the variety

of 'Its Thought' as an instance of creative Power. Let us leave this God of

Pointland to the ignorant fruition of his omnipresence and omniscience:

nothing that you or I can do can rescue him from his self-satisfaction."

--

Clive Tooth

http://www.pisquaredoversix.force9.co.uk/

End of document

Sep 23, 1999, 3:00:00 AM9/23/99

to

On Wed, 22 Sep 1999 18:52:35 -0700,

Russell Easterly <logi...@wolfenet.com> wrote:

>This is a crazy idea I had while reading some of the recent posts in

>sci-logic.

>

>Assume that there is a universe where the integer 3 doesn't exist.

>IOW, it was impossible to have 3 objects.

>Even though 3 couldn't exist as an integer,

>it might still exist as a real number.

>How would mathematicians in that universe represent 3?

Russell Easterly <logi...@wolfenet.com> wrote:

>This is a crazy idea I had while reading some of the recent posts in

>sci-logic.

>

>Assume that there is a universe where the integer 3 doesn't exist.

>IOW, it was impossible to have 3 objects.

>Even though 3 couldn't exist as an integer,

>it might still exist as a real number.

>How would mathematicians in that universe represent 3?

It is possible that there aren't 2^2^2^100 objects in our universe.

Whether or not this presents a difficulty is up to you.

--

Anatoly Vorobey,

mel...@pobox.com http://pobox.com/~mellon/

"Angels can fly because they take themselves lightly" - G.K.Chesterton

Sep 23, 1999, 3:00:00 AM9/23/99

to

> Is it possible that there is a set of irrational numbers

> such that all other irrational numbers are expressible

> as a ratio of two members from this set?

> such that all other irrational numbers are expressible

> as a ratio of two members from this set?

If there is such a set it would have to be uncountable. Otherwise the set

of ratios would be countable, and the reals would be the union of two

countable sets.

Dan

Sep 23, 1999, 3:00:00 AM9/23/99

to

Dr. Michael Albert <alb...@esther.rad.tju.edu> wrote in message

news:Pine.GSO.3.95.99092...@esther.rad.tju.edu...

> > Assume that there is a universe where the integer 3 doesn't exist.

> > IOW, it was impossible to have 3 objects.

>

> It's not clear what such a universe would be like.

> (...)> Of course this is a "mathematical" universe, not a "physical"

> universe. It's even less clear how you could arrange

> this is a physical universe. If you had two objects

> and bought in a third, what would happen?

>

> I suppose in either case you could insist that there be

> only two objects, and nothing else.

This is called pre-DETERMINED -- a SUBJECTIVE property.

> > Is it possible that there are "integers" that don't exist in our

universe?

> > In a universe that contained all of these "super-integers",

> > would all "super-real" numbers be "super-rational"?

> (...)

> > Is it possible that there is a set of irrational numbers

> > such that all other irrational numbers are expressible

> > as a ratio of two members from this set?

>

> Well, you could just take the set of all irrationals, and

> say that any irrational can be expressed as the

> ratio of two irrationals, but this is a rather dull

> theorm( eg, x = x^2 / x).

No, you cannot create a SINGULARITY like this AND your previous examples of

NULL objects.

>

> > (Would such a set redefine "countable"?)

>

> Unless you find a fundamental flaw in logic, the

> concept of "countable" will still be a valid

> concept and everything we "know" about it

> will still be true.

Sorry, but all the kids have already been taught that the SUCCESSOR function

covers the entire RANGE from the complete DOMAIN.

(...)

> Again, unless you muck with the underlying logical

> system (which, in a sense, they are probably doing

> in sci-logic) everything we currently know will

> still be true, though one might find more interesting

> things to study. For example, you divide the world

> up into rationals and irrationals.

Sorry, but that's not allowed up here. You take BOTH the rational AND the

irrational regardless of momentary incoherence NOR illegality.

> Well, there

> are other divisions. For example, some

> numbers are the roots of polynomials with

> integer coefficients (so sqrt(2) is a root

> of x^2 - 2 = 0) but not all numbers all. The

> ones which aren't are called "transcendental".

> Similarly, one might be interesed in the

> statistical properties of the distribution of

> digits in decimal expansions, in which case

> one has things like "normal" numbers, etc.

No such concept.

> Well, there's always more to do (or at least

> it seems that way). One can do even more

> bizarre things if one tampers with the underlying

> logical foundations. People have studied things

> like three valued logics (true, false, huh?).

> I am told that in a certain sense there is a

> "countable" version of the real, in which

> basically one considers the set of all reals

> such that one can write down an algorithm for

> computing the digits.

Broaden this part out. You ENUMERATE and EVALUATE both.

> This set is countable

> because there are only a countable number of

> algorithms one can write down. The usual

> diagonalization argument fails because--I am

> told--the number produced by the diagonalization

> procedure is not given in a "constructive"

> manner, so it "doesn't exist" in this "twisted"

> logic (I'm told that this sort of constructive

> logic is actually of interest to the computer

> science crowd).

(...) No, there's NOTHING NEW under this sun. Karl M

Sep 23, 1999, 3:00:00 AM9/23/99

to

[...]

> No, you cannot create a SINGULARITY like this AND your previous examples of

> NULL objects.

[...]

> Sorry, but all the kids have already been taught that the SUCCESSOR function

> covers the entire RANGE from the complete DOMAIN.

> No, you cannot create a SINGULARITY like this AND your previous examples of

> NULL objects.

> Sorry, but all the kids have already been taught that the SUCCESSOR function

> covers the entire RANGE from the complete DOMAIN.

Is there any reason for using caps like that? It looks like yet

another Usenet kook posting style, but I noticed an acm.org email

address during killfiling, so probably there's another

explanation...

--

Remco Gerlich, scar...@pino.selwerd.cx

"This gubblick contains many nonsklarkish English flutzpahs, but the

overall pluggandisp can be glorked from context" (David Moser)

Sep 23, 1999, 3:00:00 AM9/23/99

to

In article <slrn7uktmv.f2n.s...@flits104-37.flits.rug.nl>,

Remco Gerlich <scarb...@pino.selwerd.cx> wrote:

>karl malbrain <kar...@acm.org> wrote:

>> This is called pre-DETERMINED -- a SUBJECTIVE property.

>[...]

>> No, you cannot create a SINGULARITY like this AND your previous examples of

>> NULL objects.

>[...]

>> Sorry, but all the kids have already been taught that the SUCCESSOR function

>> covers the entire RANGE from the complete DOMAIN.

>

>Is there any reason for using caps like that? It looks like yet

>another Usenet kook posting style, but I noticed an acm.org email

>address during killfiling, so probably there's another

>explanation...

Remco Gerlich <scarb...@pino.selwerd.cx> wrote:

>karl malbrain <kar...@acm.org> wrote:

>> This is called pre-DETERMINED -- a SUBJECTIVE property.

>[...]

>> No, you cannot create a SINGULARITY like this AND your previous examples of

>> NULL objects.

>[...]

>> Sorry, but all the kids have already been taught that the SUCCESSOR function

>> covers the entire RANGE from the complete DOMAIN.

>

>Is there any reason for using caps like that? It looks like yet

>another Usenet kook posting style, but I noticed an acm.org email

>address during killfiling, so probably there's another

>explanation...

<shrug> I've read karl's other posts, and he always does this. Combined with

the often illogical statements the capitalized words are in, I get a distinct

"Zippy the Math Geek" feel from them--very surreal.

I have a VISION! It's a RANCID double-FISHWICH on an ENRICHED BUN!!

+--First Church of Briantology--Order of the Holy Quaternion--+

| A mathematician is a device for turning coffee into |

| theorems. -Paul Erdos |

+-------------------------------------------------------------+

| Jake Wildstrom |

+-------------------------------------------------------------+

Sep 23, 1999, 3:00:00 AM9/23/99

to

Jake Wildstrom <wil...@mit.edu> wrote in message

news:7se5e4$g...@senator-bedfellow.MIT.EDU...

(...)

> <shrug> I've read karl's other posts, and he always does this. Combined

with

> the often illogical statements the capitalized words are in, I get a

distinct

> "Zippy the Math Geek" feel from them--very surreal.

Perhaps you missed the post where I declared the label NOMINATIVE. And,

what exactly do YOU do for a living??? Karl M

Sep 23, 1999, 3:00:00 AM9/23/99

to

bo...@my-deja.com wrote in message <7sdh0d$aa9$1...@nnrp1.deja.com>...

[snip]

[snip]

Frankly, at this point I'd rather calculate Pi to the 3 billionth place in

base 37 using only an abacus than get into another discussion about it.

--

Kevin Allegood ribotr...@mindspring.pants.com

Remove the pants from my email address to reply

"I am convinced people who worry about spelling and punctuation

on the net fold their underwear before they have sex. It's just got to be."

- Bunboy

Sep 24, 1999, 3:00:00 AM9/24/99

to

>This is a crazy idea I had while reading some of the recent posts in

>sci-logic.

>

>sci-logic.

>

>Assume that there is a universe where the integer 3 doesn't exist.

>IOW, it was impossible to have 3 objects.

>IOW, it was impossible to have 3 objects.

>Even though 3 couldn't exist as an integer,

>it might still exist as a real number.

>How would mathematicians in that universe represent 3?

>it might still exist as a real number.

>How would mathematicians in that universe represent 3?

Well, Russell, I think mathematicians would call it:

"The number that cannot be the square root of nine, nor the half of six,

because it does not exist".

They would not be able to arrange nine apples for example in a three rows by

three columns square pattern. Some mysterious force would always prevent it

being done.

Like the way you cannot put together two powerful magnets, which repel each

other.

Then, if they had six apples in a line, they could not partition that group

in all the ways which we can in our universe. They would encounter some

force again. If they wanted to place either of the middle apples to either

end, it would not go (like the repelling magnets again) only to one end, to

form groups of two and four apples, but not three plus three.

This unique phenomenon would fascinate and occupy physicist, though, in that

strange universe. I can only envisage the non-existence of three as integer

through assuming such a repelling force that prevents anything to be grouped

in threes. This would be a fundamental physics issue, rather than a

mathematical one.

Also, there would be no shamrock in that world...

Thomas

Sep 24, 1999, 3:00:00 AM9/24/99

to karl malbrain

On Thu, 23 Sep 1999, karl malbrain wrote:

[snip]

> Perhaps you missed the post where I declared the label NOMINATIVE. And,

> what exactly do YOU do for a living??? Karl M

Turnabout being fair play, what do you do for a living?

Sep 24, 1999, 3:00:00 AM9/24/99

to scarb...@pino.selwerd.cx

On 23 Sep 1999, Remco Gerlich wrote:

> karl malbrain <kar...@acm.org> wrote:

> > This is called pre-DETERMINED -- a SUBJECTIVE property.

[snip]

> Is there any reason for using caps like that? It looks like yet

> another Usenet kook posting style, but I noticed an acm.org email

> address during killfiling, so probably there's another

> explanation...

Well, as far as I can tell from their website all you have to do is send

them some money and you get an acm.org email forwarding address. Not

exactly a reliable sign of credibility.

Sep 25, 1999, 3:00:00 AM9/25/99

to

> I know that the irrational numbers are uncountable.

> (Can't be put into a 1 to 1 relation with the integers.)

The Irrationals are countable, Transcendentals are not, yet.

Ulie

Sep 25, 1999, 3:00:00 AM9/25/99

to

>>>>> "Ulie" == Ulrich Sondermann <usond...@earthlink.net> writes:

>> I know that the irrational numbers are uncountable. (Can't be put

>> into a 1 to 1 relation with the integers.)

Ulie> The Irrationals are countable, Transcendentals are not, yet.

Transcendentals are irrational. Perhaps you mean "algebraic numbers."

"Yet"?

Ulie> Ulie

Rory

Sep 25, 1999, 3:00:00 AM9/25/99

to

dlrenfro [that's me, folks!]

wrote

wrote

[[in sci.logic, sci.math, and (gasp!) even

rec.arts.sf.science (whatever that is) on

Sat, 25 Sep 1999 22:27:29 -0500]]

[snip]

> measure zero. Moreover, no Hamel basis is Borel (same

> paper by Sierpinski).

>

> Actually, I think there exists a closed Hamel basis,

[snip]

Actually, I think not. I don't want Sierpinski turning

over in his grave!

I think what is true is that any Hamel basis for R^n

(R = reals, n > 1) over the rationals is not Borel

[this seems to be what Kormes proves in his dissertation

(cited in my previous post), but the archaic terminology

and notation makes it difficult to tell for sure],

while there exist perfect (necessarily nowhere dense)

Hamel bases for R over the rationals. In fact, I think

the following is true (note my use of *think*):

(a) There exists a perfect set in R (necessarily nowhere

dense and of measure zero) that is a transcendence basis

for R over the rationals. [John von Neumann, Math. Ann.

99 (1928), 134-141]

(b) Given any perfect set P in R, there exists a transcendence

basis for R over the rationals that is perfect and is

contained in P. [theorem 2 on p. 168 of J. Mycielski,

"Algebraic independence and measure", Fund. Math. 61

(1967), 165-169]

If I'm correct in the inferences I gave above from the references

I cited (heck, if (a) and (b) are correct, let alone what

Neumann and Mycielski proved), then my guess that the

generalized Hausdorff dimension for a Hamel basis in R

can be arbitrarily small is immediate. [In fact, "Hausdorff"

could be improved to "upper Minkowski" ...]

Dave L. Renfro

Sep 26, 1999, 3:00:00 AM9/26/99

to

On Sun, 26 Sep 1999, Mike Oliver wrote:

> dlrenfro wrote:

> > I think what is true is that any Hamel basis for R^n

> > (R = reals, n > 1) over the rationals is not Borel

> > [this seems to be what Kormes proves in his dissertation

> > (cited in my previous post), but the archaic terminology

> > and notation makes it difficult to tell for sure],

> > while there exist perfect (necessarily nowhere dense)

> > Hamel bases for R over the rationals.

>

> Don't think so.

>

> Suppose B is a closed (or even analytic) Hamel basis for R over

> Q, and let b be its unique rational element. Then we can get

> an analytic set of representatives for the Vitali equivalence

> relation as follows: Any real r has a unique representation

> (up to order and zero coefficients) as a rational linear combination

> of elements of B. Let r be in the set of representatives S just

> in case the coefficient of b (the rational element of B) is zero.

>

> Now you can say r \in S just in case there exists a real coding

> a tuple of reals such that each real in the tuple is an element

> of B, r is a rational linear combination of the reals in the tuple,

> and none of the reals in the tuple equals b. Clearly this is a

> Sigma^1_1 boldface condition if B is, so S is analytic. But

> this is impossible by the standard argument.

>

> Anybody see any errors in the above?

What if the Hamel basis has no rational element?

Sep 27, 1999, 3:00:00 AM9/27/99

to

Nicolas Bray <br...@soda.CSUA.Berkeley.EDU> wrote in message

news:Pine.BSF.4.10.990924...@soda.CSUA.Berkeley.EDU...

Here's the ANTE to your dilemma: I program at a small company in the

ENGINEERING department which provides the FRAMEWORK for the software

developers and tax analysts. Karl M

Sep 27, 1999, 3:00:00 AM9/27/99

to

Nicolas Bray <br...@soda.CSUA.Berkeley.EDU> wrote in message

news:Pine.BSF.4.10.990924...@soda.CSUA.Berkeley.EDU...

(...)

> Well, as far as I can tell from their website all you have to do is send

> them some money and you get an acm.org email forwarding address. Not

> exactly a reliable sign of credibility.

Of course not. Membership is what counts. Yes, I am a member of the

Association, off and on since 1969 depending on my line of work, which is

DETERMINED by the BOURGEOISE. Karl M

Sep 27, 1999, 3:00:00 AM9/27/99

to

Oh, I suppose that is possible. It complicates the argument

a little but not much: Let the number 1 be expressed as

1 = q_0*d_0 + q_1*d_1 + q_2*d_2 + ... + q_n*d_n

where the q's are rational and the d's are taken from B.

Now for any real r, let it be expressed as

r = R*d_0 + r_0*b_0 + r_1*b_1 + ... + r_m*b_m

where R and all the r's are rational, and the b_s are taken from

B. Now let the representative of the Vitali equivalence class

of r be r-(R/q_0) (i.e. the unique element whose d_0 component is 0).

This should still give you an analytic set of reps if B is analytic.

Sep 28, 1999, 3:00:00 AM9/28/99

to

In article <37EFFA26...@math.ucla.edu>,

Or, shortcut: remove d_0 from the basis and replace it by 1.

A question. I'm pretty sure I've seen somewhere, maybe in an

old paper, the claim "a Hamel basis cannot have the property of Baire".

Does anybody know where?

["pr. of Baire" = being in the sigma-algebra generated by the

open sets and the meager sets; "meager" = ctbl union of nowhere dense

sets. It turns out X has the pr. of Baire <=> X = open Delta meager...

hence X is either meager, or comeager on an open set]

The claim is stronger than Mike's statement (the sigma-algebra

properly contains the analytic sets.) The claim is also quite false!

What _is_ true is that a Hamel basis B cannot be comeager on

an open set... for the same reason it cannot have positive measure: either

property implies the difference set B-B contains an interval, and

thus some b_i - b_j = a non-0 rational.

But B can be meager just fine. Take a set K such that K+K =R,

and do the usual construction, but pick elements for B from K (let r

be outside the span of the part already built, write r = k1+k2, adjoin

one (if enough) or both of them as new elements, repeat). The resulting

B spans K, and thus R; and being a subset of K U {1} it will be

meager, or have measure 0, if K does.

E.g. say C is the Cantor set in [0,1]. C+C = [0,2] then...

so put a copy of C in each [j, j+1], j in Z, and their union works

as K. This K is both measure-0 and meager (in fact, nowhere dense),

so we get a nowhere-dense Hamel basis B that is also of measure 0.

For curiosity's sake, the other three possibilities are obtainable

as well, that is, B meager + non-measurable; no pr. of Baire + measure 0;

no pr. of Baire + non-measurable.

Ilias

Oct 2, 1999, 3:00:00 AM10/2/99

to

NOTE: I sent the following in to be posted over two days ago

but, as seems to be the case with several of my recent

posts, it doesn't seem to have appeared outside of my

own newserver. This time I am posting it via the

Swarthmore sci.math archieves at

<http://forum.swarthmore.edu/epigone/sci.math>

I apologize if this results in two posts of mine

being identical, but after the amount of time I spent

on this particular response I'm a bit iterated that

it might have been lost in cyberspace somewhere.

but, as seems to be the case with several of my recent

posts, it doesn't seem to have appeared outside of my

own newserver. This time I am posting it via the

Swarthmore sci.math archieves at

<http://forum.swarthmore.edu/epigone/sci.math>

I apologize if this results in two posts of mine

being identical, but after the amount of time I spent

on this particular response I'm a bit iterated that

it might have been lost in cyberspace somewhere.

*******************************************************

*******************************************************

Ilias Kastanas [sci.math 28 Sep 1999 12:47:26 GMT]

[snip]

> A question. I'm pretty sure I've seen somewhere,

>maybe in an old paper, the claim "a Hamel basis cannot have

>the property of Baire". Does anybody know where?

If so, the claim is false. Your example of a first category

Hamel basis below contradicts this.

[snip]

> What _is_ true is that a Hamel basis B cannot be

>comeager on an open set... for the same reason it cannot have

>positive measure: either property implies the difference set

>B-B contains an interval, and thus some b_i - b_j = a

>non-0 rational.

>

> But B can be meager just fine. Take a set K

>such that K+K =R, and do the usual construction, but pick

>elements for B from K (let r be outside the span of the

>part already built, write r = k1+k2, adjoin one (if enough)

>or both of them as new elements, repeat). The resulting B

>spans K, and thus R; and being a subset of K U {1} it

>will be meager, or have measure 0, if K does.

>

> E.g. say C is the Cantor set in [0,1]. C+C = [0,2]

>then... so put a copy of C in each [j, j+1], j in Z,

>and their union works as K. This K is both measure-0 and meager

>(in fact, nowhere dense), so we get a nowhere-dense Hamel basis

>B that is also of measure 0.

>

> For curiosity's sake, the other three possibilities are

>obtainable as well, that is, B meager + non-measurable;

>no pr. of Baire + measure 0; no pr. of Baire + non-measurable.

Some results involving Hamel bases that may be of

interest (i.e. How I spent this Tuesday afternoon.)

1. No Hamel basis is an analytic set.

F. B. Jones, "Measure and other properties of a Hamel

base", Bull. Amer. Math. Soc. 48 (1942), 472-481.

[Theorem 9 on page 476.]

2. The complement of any Hamel basis is everywhere

of the second category.

W. Sierpinski, "La base de M. Hamel et la propriete de

Baire", Publications de l'Institut Mathematique

(Beograd) 4 (1935), 220-224.

A corollary of this is that any Hamel basis with

the Baire property is a first category set. (Such

sets do exist.) The measure analoge of this result

(given in my Sept. 25 post) is that any measurable

Hamel basis has measure zero.

3. There exists a Hamel basis that has nonempty

intersection with each perfect set.

C. Burstin, "Die Spaltung des Kontinuum in c in L

Sinne nichtmessbare Mengen", Sitzungsberichte der

Akademie der Wissenschaften, Vienna, Mathematisch-

naturwissenschaftliche Klasse, Abt. IIa, 125

(part 3) (1916).

[Reference in footnote 15, page 476 of the Jones

paper and in footnote on page 9 of Kormes' dissertation

cited in my Sept. 25 post. (I don't have a copy of

Burstin's paper.)]

Jones observes that this particular Hamel basis does

not contain an uncountable analytic set, and then procedes

to construct a Hamel basis that does contain a perfect set

(i.e. contains an uncountable closed set).

4. Although there exist Hamel bases having measure zero

(Sierpinski, 1920; see my Sept. 25 post), the set of

all *integer* linear combinations of elements in any

Hamel basis is nonmeasurable. Indeed, it has zero

inner measure and maximal outer measure in every

interval. (Note, of course, that the set of

*rational* linear combinations of elements in any

Hamel basis gives all the reals.)

P. Erdos, "On some properties of Hamel bases", Colloq.

Math. 10 (1963), 267-269. [Theorem 1]

This implies that every Hamel basis has inner measure

zero, which was first proved in Sierpinski's 1920 paper.

(The Baire category analog of this result is given

in #2 above.)

The Baire category version of this result [There

exists a Hamel basis such that neither it nor its

complement contains a second category set.],

is given in

Marek Kuczma, "On some properties of Erdos sets",

Colloq. Math. 48 (1984), 127-133. [Theorem 6, page 133]

5. CH implies there exists a Hamel basis for which the

set of all its *nonnegative* rational linear combinations

has measure zero.

Theorem 2 in Erdos paper above. In fact, Erdos proves this

set is a Lusin set (has countable intersection with every

nowhere dense set). [Recall that the existence of a Lusin

set is independent of ZFC. I don't know if the weaker

"measure zero" statement is independent of ZFC, but I

would guess it is.] Thus, this set, and hence the

associated Hamel basis, has strong measure zero. [This

implies it has measure zero relative to any countably

additive nonatomic measure (i.e. has universal measure

zero).] Since any uncountable subset of a Lusin set is

a Lusin set and any Lusin set is a second category set

(present usage: second category = not first category;

some people use this term when they really mean

"complement of a first category set"), the Hamel basis

associated with this result is a second category

subset of the reals.

Apparently unaware of Erdos' paper, Darst proved that

CH implies there exists a Hamel basis that is a

Lusin set in

R. B. Darst, "On measure and other properties of a

Hamel basis", Proc. Amer. Math. Soc. 16 (1965), 645-646.

6. CH implies there exists a Hamel basis for which the

set of all its *nonnegative* rational linear combinations

is first category in the reals.

Harry I. Miller, "On a property of Hamel bases",

Bollettino U.M.I. (7) 3-A (1989), 39-43.

Miller actually makes use of a weaker hypothesis than

CH, namely that any union of less than continuum many

sets of measure zero has measure zero.

A sketch of a proof of this result can also be

found in the exercises on page 275 of

M. Kuczma, AN INTRODUCTION TO THE THEORY OF

FUNCTIONAL EQUATIONS AND INEQUALITIES, Panstwowe

Wydawnictwo Naukowe, Uniwersytet Slaski, Warszawa-

Krakow-Katowice, 1985.

I don't have a copy of this book, but supposedly

chaper 11 deals with Hamel bases.

Dave L. Renfro

Oct 5, 1999, 3:00:00 AM10/5/99

to

In article <zun9nh...@forum.swarthmore.edu>,

Dave L. Renfro <dlre...@gateway.net> wrote:

@

@Ilias Kastanas [sci.math 28 Sep 1999 12:47:26 GMT]

@

@[snip]

@

@> A question. I'm pretty sure I've seen somewhere,

@>maybe in an old paper, the claim "a Hamel basis cannot have

@>the property of Baire". Does anybody know where?

@

@If so, the claim is false. Your example of a first category

@Hamel basis below contradicts this.

@

@[snip]

@

@> What _is_ true is that a Hamel basis B cannot be

@>comeager on an open set... for the same reason it cannot have

@>positive measure: either property implies the difference set

@>B-B contains an interval, and thus some b_i - b_j = a

@>non-0 rational.

@>

@> But B can be meager just fine. Take a set K

@>such that K+K =R, and do the usual construction, but pick

@>elements for B from K (let r be outside the span of the

@>part already built, write r = k1+k2, adjoin one (if enough)

@>or both of them as new elements, repeat). The resulting B

@>spans K, and thus R; and being a subset of K U {1} it

@>will be meager, or have measure 0, if K does.

@>

@> E.g. say C is the Cantor set in [0,1]. C+C = [0,2]

@>then... so put a copy of C in each [j, j+1], j in Z,

@>and their union works as K. This K is both measure-0 and meager

@>(in fact, nowhere dense), so we get a nowhere-dense Hamel basis

@>B that is also of measure 0.

@>

@> For curiosity's sake, the other three possibilities are

@>obtainable as well, that is, B meager + non-measurable;

@>no pr. of Baire + measure 0; no pr. of Baire + non-measurable.

@

@

@Some results involving Hamel bases that may be of

@interest (i.e. How I spent this Tuesday afternoon.)

@

@1. No Hamel basis is an analytic set.

@

@F. B. Jones, "Measure and other properties of a Hamel

@base", Bull. Amer. Math. Soc. 48 (1942), 472-481.

@[Theorem 9 on page 476.]

@

@

@2. The complement of any Hamel basis is everywhere

@ of the second category.

@

@W. Sierpinski, "La base de M. Hamel et la propriete de

@Baire", Publications de l'Institut Mathematique

@(Beograd) 4 (1935), 220-224.

Dave L. Renfro <dlre...@gateway.net> wrote:

@

@Ilias Kastanas [sci.math 28 Sep 1999 12:47:26 GMT]

@

@[snip]

@

@> A question. I'm pretty sure I've seen somewhere,

@>maybe in an old paper, the claim "a Hamel basis cannot have

@>the property of Baire". Does anybody know where?

@

@If so, the claim is false. Your example of a first category

@Hamel basis below contradicts this.

@

@[snip]

@

@> What _is_ true is that a Hamel basis B cannot be

@>comeager on an open set... for the same reason it cannot have

@>positive measure: either property implies the difference set

@>B-B contains an interval, and thus some b_i - b_j = a

@>non-0 rational.

@>

@> But B can be meager just fine. Take a set K

@>such that K+K =R, and do the usual construction, but pick

@>elements for B from K (let r be outside the span of the

@>part already built, write r = k1+k2, adjoin one (if enough)

@>or both of them as new elements, repeat). The resulting B

@>spans K, and thus R; and being a subset of K U {1} it

@>will be meager, or have measure 0, if K does.

@>

@> E.g. say C is the Cantor set in [0,1]. C+C = [0,2]

@>then... so put a copy of C in each [j, j+1], j in Z,

@>and their union works as K. This K is both measure-0 and meager

@>(in fact, nowhere dense), so we get a nowhere-dense Hamel basis

@>B that is also of measure 0.

@>

@> For curiosity's sake, the other three possibilities are

@>obtainable as well, that is, B meager + non-measurable;

@>no pr. of Baire + measure 0; no pr. of Baire + non-measurable.

@

@

@Some results involving Hamel bases that may be of

@interest (i.e. How I spent this Tuesday afternoon.)

@

@1. No Hamel basis is an analytic set.

@

@F. B. Jones, "Measure and other properties of a Hamel

@base", Bull. Amer. Math. Soc. 48 (1942), 472-481.

@[Theorem 9 on page 476.]

@

@

@2. The complement of any Hamel basis is everywhere

@ of the second category.

@

@W. Sierpinski, "La base de M. Hamel et la propriete de

@Baire", Publications de l'Institut Mathematique

@(Beograd) 4 (1935), 220-224.

Yes; as mentioned, B cannot contain a comeager

subset of any open set.

@A corollary of this is that any Hamel basis with

@the Baire property is a first category set. (Such

@sets do exist.) The measure analoge of this result

@(given in my Sept. 25 post) is that any measurable

@Hamel basis has measure zero.

@

@

@3. There exists a Hamel basis that has nonempty

@ intersection with each perfect set.

Straightforward; for each perfect set, we put a point of

it in the basis, and keep a point of it out -- by putting in a

rational multiple. So the basis is totally imperfect, and thus

non-measurable and without the property of Baire.

@C. Burstin, "Die Spaltung des Kontinuum in c in L

@Sinne nichtmessbare Mengen", Sitzungsberichte der

@Akademie der Wissenschaften, Vienna, Mathematisch-

@naturwissenschaftliche Klasse, Abt. IIa, 125

@(part 3) (1916).

@[Reference in footnote 15, page 476 of the Jones

@paper and in footnote on page 9 of Kormes' dissertation

@cited in my Sept. 25 post. (I don't have a copy of

@Burstin's paper.)]

@

@Jones observes that this particular Hamel basis does

@not contain an uncountable analytic set, and then procedes

Of course; since any uncountable analytic set has

a perfect subset.

@to construct a Hamel basis that does contain a perfect set

@(i.e. contains an uncountable closed set).

@

E.g. you can construct a perfect set linearly independent

over Q, and extend it to a basis.

@4. Although there exist Hamel bases having measure zero

@ (Sierpinski, 1920; see my Sept. 25 post), the set of

@ all *integer* linear combinations of elements in any

@ Hamel basis is nonmeasurable. Indeed, it has zero

@ inner measure and maximal outer measure in every

@ interval. (Note, of course, that the set of

@ *rational* linear combinations of elements in any

@ Hamel basis gives all the reals.)

@

@P. Erdos, "On some properties of Hamel bases", Colloq.

@Math. 10 (1963), 267-269. [Theorem 1]

That's nice. Call the set D; its part in [0,1], D_0, gives

by translation D in any [k,k+1]. D cannot have positive inner measure,

for then D-D (which is D) would contain an interval, violating lin.

independence of B. So D, and D_0, have inner measure 0. If D_0 had

outer measure 0 it would be measurable, and {1/n D}, n=1,2,... would

be countably many measure 0 sets covering R. Thus m*(D_0) = c, 0 < c <= 1.

Now [0,1/2], [1/2,1] split D_0 into two sets, each of outer measure

c/2 (one set is 1 - the other). Likewise, 1/2 D_0, of outer measure c/2

in [0,1/2], splits into two sets having m* = c/4; and D_0, being a subset

of 1/2 D_0, splits likewise. And so on; for any k and any one of the

2^k binary intervals I, m*(D_0 intersect I) / m(I) = c. But if c were

< 1 there would exist I's making this density ratio smaller than any po-

sitive number desired. Hence m*(D_0) = 1, and D has full outer measure.

(I don't know whether this was Erdos's argument; I haven't seen

his paper).

@This implies that every Hamel basis has inner measure

@zero, which was first proved in Sierpinski's 1920 paper.

@(The Baire category analog of this result is given

@in #2 above.)

@

@The Baire category version of this result [There

@exists a Hamel basis such that neither it nor its

@complement contains a second category set.],

@is given in

@

@Marek Kuczma, "On some properties of Erdos sets",

@Colloq. Math. 48 (1984), 127-133. [Theorem 6, page 133]

@

@

@5. CH implies there exists a Hamel basis for which the

@ set of all its *nonnegative* rational linear combinations

@ has measure zero.

@

@Theorem 2 in Erdos paper above. In fact, Erdos proves this

@set is a Lusin set (has countable intersection with every

@nowhere dense set). [Recall that the existence of a Lusin

@set is independent of ZFC. I don't know if the weaker

@"measure zero" statement is independent of ZFC, but I

@would guess it is.] Thus, this set, and hence the

@associated Hamel basis, has strong measure zero. [This

@implies it has measure zero relative to any countably

@additive nonatomic measure (i.e. has universal measure

@zero).] Since any uncountable subset of a Lusin set is

@a Lusin set and any Lusin set is a second category set

@(present usage: second category = not first category;

@some people use this term when they really mean

@"complement of a first category set"), the Hamel basis

@associated with this result is a second category

@subset of the reals.

@

@Apparently unaware of Erdos' paper, Darst proved that

@CH implies there exists a Hamel basis that is a

@Lusin set in

@

@R. B. Darst, "On measure and other properties of a

@Hamel basis", Proc. Amer. Math. Soc. 16 (1965), 645-646.

@

@

@6. CH implies there exists a Hamel basis for which the

@ set of all its *nonnegative* rational linear combinations

@ is first category in the reals.

@

@Harry I. Miller, "On a property of Hamel bases",

@Bollettino U.M.I. (7) 3-A (1989), 39-43.

@

@Miller actually makes use of a weaker hypothesis than

@CH, namely that any union of less than continuum many

@sets of measure zero has measure zero.

@

@A sketch of a proof of this result can also be

@found in the exercises on page 275 of

@

@M. Kuczma, AN INTRODUCTION TO THE THEORY OF

@FUNCTIONAL EQUATIONS AND INEQUALITIES, Panstwowe

@Wydawnictwo Naukowe, Uniwersytet Slaski, Warszawa-

@Krakow-Katowice, 1985.

@

@I don't have a copy of this book, but supposedly

@chaper 11 deals with Hamel bases.

@

@

@Dave L. Renfro

@

An amusing variant is: construct a function f s.t. its graph

G meets every closed subset K of the plane whose projection on the x-

axis has uncountably- (and hence continuum-) many points. (As icing on

the cake, we can also arrange that f(x_1) + f(x_2) = f(x_1 + x_2);

define f on B, and extend by linearity). Namely, enumerate the K's

in type c. Find a b in B s.t. q*b is in the projection of K_0,

q some rational, and define f(q*b) (= q*f(b)) so that (q*b, f(q*b))

lies in K_0. At stage a, |span of b's already used| is < c, so there

is a point in the projection of K_a not in the span; use that point.

(The construction involves c -many b's; if they don't exhaust B, let

f be 0 on the rest of B).

The resulting G is dense on the plane... and very much so: on

any interval, f assumes as value every single number in R.

G is connected. (If not, there would be open sets M, N s.t.

M intersect G, N intersect G are disjoint and cover G. But then M, N

would themselves be disjoint [because G is dense; if they intersected,

G would get in!]. Thus M, N would disconnect M U N. However, M U N

is connected. Hint: the complement of M U N is a closed set C with

countable projection, i.e. a subset of a countable family of vertical

lines. So M U N contains co-countably many vertical lines; use them

to arcwise connect open disks in M U N. You get connected vertical

"slices"... and M U N is union of an increasing family of such slices).

Of course G - {any single point of G} is disconnected (by the

vertical line through that point). And the part of G inside any bounded

set is totally disconnected (connected components are just the points).

The complement of G, ~G, is also connected (and dense).

Any horizontal line cuts G in a (linear) totally imperfect set.

Taking them all, we get a decomposition of R into continuum-many disjoint

totally imperfect sets, each having continuum-many points.

If we carry G, ~G over to the open unit square (homeomorph of

the plane), we get an answer to a generalization of a recent question

about connectedness. Let S, T be any two disjoint subsets of the

boundary of the unit square; then there is a connected set containing

S, and a disjoint connected set containing T (G U S, ~G U T). We can

link up a lot more than just opposite vertices!

Ilias

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