PI an Integer?

23 views

Russell Easterly

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.

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

Dr. Michael Albert

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.

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

Doug Norris

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

----------------------------------------------------------------------------
----------------------------------------------------------------------------
"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

bo...@my-deja.com

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.
>
> 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"?
>
The following is meant for entertainment purposes only. Don't take
it too seriously.

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"

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

William L. Bahn

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.

Clive Tooth

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"

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

Anatoly Vorobey

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?

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

Daniel Longley

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?

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

karl malbrain

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

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

> > 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

Remco Gerlich

Sep 23, 1999, 3:00:00 AM9/23/99
to
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, scar...@pino.selwerd.cx
"This gubblick contains many nonsklarkish English flutzpahs, but the
overall pluggandisp can be glorked from context" (David Moser)

Jake Wildstrom

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...

<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 |
+-------------------------------------------------------------+

karl malbrain

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

Jake Wildstrom <wil...@mit.edu> wrote in message
news:7se5e4$g...@senator-bedfellow.MIT.EDU... (...) > <shrug> I've read karl's other posts, and he always does this. Combined with > the often illogical statements the capitalized words are in, I get a distinct > "Zippy the Math Geek" feel from them--very surreal. Perhaps you missed the post where I declared the label NOMINATIVE. And, what exactly do YOU do for a living??? Karl M Riboflavin unread, Sep 23, 1999, 3:00:00 AM9/23/99 to bo...@my-deja.com wrote in message <7sdh0d$aa9$1...@nnrp1.deja.com>... [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 Thomas unread, Sep 24, 1999, 3:00:00 AM9/24/99 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?

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

Nicolas Bray

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?

Nicolas Bray

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.

Ulrich Sondermann

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

Rory Molinari

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

dlrenfro

Sep 25, 1999, 3:00:00 AM9/25/99
to
dlrenfro [that's me, folks!]

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

Fred Galvin

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?

karl malbrain

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

karl malbrain

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

Mike Oliver

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.

Ilias Kastanas

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

Dave L. Renfro

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.

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

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

[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
>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

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
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
Krakow-Katowice, 1985.

I don't have a copy of this book, but supposedly
chaper 11 deals with Hamel bases.

Dave L. Renfro

Ilias Kastanas

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
@
@[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
@>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

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
@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
@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