Are algebraic irrationals countable?
Joona I Palaste
x in R\Q is algebraic iff there exists a countable set A so that:
oo
\bigsum (a_i * x^i) = 0,
i=0
where each a_i is rational and a member of A.
This means that x in R\Q is algebraic iff it's the zero of some one-
variable polynomial with rational coefficients.
Now the question is: are algebraic irrationals countable? Of course,
there are countably many one-variable polynomials with rational
coefficients, in fact there are countably many polynomials with rational
coefficients, regardless of how may variables are involved.
But does every such polynomial have countably many zeroes? If so, then
the set of all zeroes of all polynomials with rational coefficients is
the union of countably many countable sets and thus countable.
Otherwise it is not countable.
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"The day Microsoft makes something that doesn't suck is probably the day they
start making vacuum cleaners."
- Ernst Jan Plugge
The World Wide Wade
Joona I Palaste <pal...@cc.helsinki.fi> wrote:
> I assume algebraic irrationals are defined as:
>
> x in R\Q is algebraic iff there exists a countable set A so that:
> oo
> \bigsum (a_i * x^i) = 0,
> i=0
> where each a_i is rational and a member of A.
>
> This means that x in R\Q is algebraic iff it's the zero of some one-
> variable polynomial with rational coefficients.
>
> Now the question is: are algebraic irrationals countable? Of course,
> there are countably many one-variable polynomials with rational
> coefficients, in fact there are countably many polynomials with rational
> coefficients, regardless of how may variables are involved.
> But does every such polynomial have countably many zeroes? If so, then
> the set of all zeroes of all polynomials with rational coefficients is
> the union of countably many countable sets and thus countable.
> Otherwise it is not countable.
Algebraic irrationals are not defined as roots of power series with
rational coefficients, an infinite power series is not a polynomial, there
are uncountably many power series with rational coefficients, and a
polynomial has only finitely many zeros. So you have some rough sledding
here ...
--WWW.
Joona I Palaste
My representation was perhaps too ambiguous. I did NOT intend the power
series to be infinite. I intended it to be finite, but arbitrarily long.
That means when I wrote:
"All a_i (where i is from 0 to oo) are rational and members of A", I
also implied that there exists a number k in N so that all a_i are 0
when i>=k.
Sorry about that confusion!
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"A computer program does what you tell it to do, not what you want it to do."
- Anon
chris
Theorem (1): the polynomial a_n*x^n+...+a_1*x+a_0=0 has at most n roots
Now, consider the following sets..
S_m = all the roots of polynomials of the form a_m*x^m+...+a_1*x+a_0=0
Now obviously the algebraic numbers are included (and in fact are equal to)
the union of the S_m, m=1 to infinity.
I will use N to represent the set of natural numbers..
we can make a bijection N*N -> N (look up cantor's diagonal argument)
Therefore we can make a bijection N*N*...*N -> N
The sets of algebraic numbers that are in S_m can be considered as elements
of the set N*N*..*N*m , with m+1 copies of N.. that is the (m+1)
co-efficents, and then which of the m possible roots we have (if we miss any
roots, just put in 0).
So each S_m is mappable to N.
Now, can we map (all the S_m) -> N?
Yes we can, form a pair (x,y) to be the yth number in the mapping we have
all S_x->N.. this gives us a mapping from (all the S_m) -> N*N, which we can
then map -> N
so we can map (union S_m) -> N, and the union of the S_m contains all the
algeraic irrationals (in fact, it contains each of them an infinate number
of times, but we shall ignore that!)
"Joona I Palaste" <pal...@cc.helsinki.fi> wrote in message
news:adarcd$6ok$1...@oravannahka.helsinki.fi...
Joona I Palaste
> Yes they are. I'll state each theorem I use:
> Theorem (1): the polynomial a_n*x^n+...+a_1*x+a_0=0 has at most n roots
> Now, consider the following sets..
> S_m = all the roots of polynomials of the form a_m*x^m+...+a_1*x+a_0=0
> Now obviously the algebraic numbers are included (and in fact are equal to)
> the union of the S_m, m=1 to infinity.
> I will use N to represent the set of natural numbers..
> we can make a bijection N*N -> N (look up cantor's diagonal argument)
> Therefore we can make a bijection N*N*...*N -> N
> The sets of algebraic numbers that are in S_m can be considered as elements
> of the set N*N*..*N*m , with m+1 copies of N.. that is the (m+1)
> co-efficents, and then which of the m possible roots we have (if we miss any
> roots, just put in 0).
> So each S_m is mappable to N.
> Now, can we map (all the S_m) -> N?
> Yes we can, form a pair (x,y) to be the yth number in the mapping we have
> all S_x->N.. this gives us a mapping from (all the S_m) -> N*N, which we can
> then map -> N
> so we can map (union S_m) -> N, and the union of the S_m contains all the
> algeraic irrationals (in fact, it contains each of them an infinate number
> of times, but we shall ignore that!)
Thank you. Now since the set of all real numbers (i.e. R) is
uncountable, this must mean that transcendental irrationals exist,
because no set can be both countable and uncountable at the same time,
therefore the set of algebraic reals cannot equal R.
Now I just need to remember what the proof of the uncountability of R
was and I am all set...
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"You can pick your friends, you can pick your nose, but you can't pick your
relatives."
- MAD Magazine
David C. Ullrich
wrote:
>I assume algebraic irrationals are defined as:
>
>x in R\Q is algebraic iff there exists a countable set A so that:
> oo
>\bigsum (a_i * x^i) = 0,
> i=0
>where each a_i is rational and a member of A.
No, that's not the definition at all.
>This means that x in R\Q is algebraic iff it's the zero of some one-
>variable polynomial with rational coefficients.
_This_ is the definition. But this is not the same as what
you wrote above - that infinite sum is not a polynomial.
(Also the bit about "A is a countable set...each a_i is rational
and a member of A" is exactly the same as just saying "each a_i
is rational".)
>Now the question is: are algebraic irrationals countable?
Yes. (But they would not be countable by your first definition.
What made you think that that was the definition? People have
_given_ the definition several times, in replies to your posts.)
>Of course,
>there are countably many one-variable polynomials with rational
>coefficients, in fact there are countably many polynomials with rational
>coefficients, regardless of how may variables are involved.
>But does every such polynomial have countably many zeroes?
A polynomial in one variable has only finitely many zeroes.
>If so, then
>the set of all zeroes of all polynomials with rational coefficients is
>the union of countably many countable sets and thus countable.
>Otherwise it is not countable.
>
>--
>/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
>| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
>| http://www.helsinki.fi/~palaste W++ B OP+ |
>\----------------------------------------- Finland rules! ------------/
>"The day Microsoft makes something that doesn't suck is probably the day they
>start making vacuum cleaners."
> - Ernst Jan Plugge
David C. Ullrich
Joona I Palaste
>>I assume algebraic irrationals are defined as:
>>
>>x in R\Q is algebraic iff there exists a countable set A so that:
>> oo
>>\bigsum (a_i * x^i) = 0,
>> i=0
>>where each a_i is rational and a member of A.
> No, that's not the definition at all.
>>This means that x in R\Q is algebraic iff it's the zero of some one-
>>variable polynomial with rational coefficients.
> _This_ is the definition. But this is not the same as what
> you wrote above - that infinite sum is not a polynomial.
As I wrote in my reply to the WWW, I was really meaning an arbitrarily
long finite sum, and not an infinite sum.
> (Also the bit about "A is a countable set...each a_i is rational
> and a member of A" is exactly the same as just saying "each a_i
> is rational".)
Well, yes. I suppose the notation "a_i" implies that the set of all
a_i is countable.
>>Now the question is: are algebraic irrationals countable?
> Yes. (But they would not be countable by your first definition.
> What made you think that that was the definition? People have
> _given_ the definition several times, in replies to your posts.)
They are not countable by the definition I *wrote*. They are countable
by the definition I *meant to write*.
>>Of course,
>>there are countably many one-variable polynomials with rational
>>coefficients, in fact there are countably many polynomials with rational
>>coefficients, regardless of how may variables are involved.
>>But does every such polynomial have countably many zeroes?
> A polynomial in one variable has only finitely many zeroes.
Yes, of course. And a finite set is always countable. How silly of me
not to realise that.
>>If so, then
>>the set of all zeroes of all polynomials with rational coefficients is
>>the union of countably many countable sets and thus countable.
>>Otherwise it is not countable.
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"When a man talks dirty to a woman, that's sexual harassment. When a woman talks
dirty to a man, that's 14.99 per minute + local telephone charges!"
- Ruben Stiller
David C. Ullrich
wrote:
>David C. Ullrich <ull...@math.okstate.edu> scribbled the following:
>> On 1 Jun 2002 16:05:33 GMT, Joona I Palaste <pal...@cc.helsinki.fi>
>> wrote:
>
>>>I assume algebraic irrationals are defined as:
>>>
>>>x in R\Q is algebraic iff there exists a countable set A so that:
>>> oo
>>>\bigsum (a_i * x^i) = 0,
>>> i=0
>>>where each a_i is rational and a member of A.
>
>> No, that's not the definition at all.
>
>>>This means that x in R\Q is algebraic iff it's the zero of some one-
>>>variable polynomial with rational coefficients.
>
>> _This_ is the definition. But this is not the same as what
>> you wrote above - that infinite sum is not a polynomial.
>
>As I wrote in my reply to the WWW, I was really meaning an arbitrarily
>long finite sum, and not an infinite sum.
>
>> (Also the bit about "A is a countable set...each a_i is rational
>> and a member of A" is exactly the same as just saying "each a_i
>> is rational".)
>
>Well, yes. I suppose the notation "a_i" implies that the set of all
>a_i is countable.
That was not my point. The point is that Q _is_ countable, so
saying the a_i lie in some countable subset of Q is the same
as saying that they lie in Q.
(And when you say A is countable, and the a_i are rationals in
A, all you're saying is "the a_i lie in some countable subset
of Q", that set being the intersection of Q and A.)
In some situations the notation "a_i" would be taken to
imply that there were only countably many, but in some
situations it would _not_ imply this. But saying the
a_i are rational certainly implies that the set of
all the a_1 is countable. (For that matter, a _polynomial_
has only _finitely_ many coefficients...)
David C. Ullrich
Joona I Palaste
Thanks. NOW I understand your point. Because I said "all a_i are
rational", that means that the set of all a_i is a subset of Q, and
every subset of Q is countable.
So it would have sufficed to say: There exist some numbers a_i, where
i is a natural number and all a_i are rational...
> In some situations the notation "a_i" would be taken to
> imply that there were only countably many, but in some
> situations it would _not_ imply this. But saying the
> a_i are rational certainly implies that the set of
> all the a_1 is countable. (For that matter, a _polynomial_
> has only _finitely_ many coefficients...)
Yes, of course. If we have an infinite series of values in some set S,
doesn't that also mean that we have a function from N to S? I.e. if
for all n in N there exists an s_n in S, that also means that there
exists a function f: N->S so that for all n in N, f(n)=s_n? Then the
notations f(n) and s_n would be equivalent.
But if we instead have a function f from R to some set S, then by the
above claim we can say that s_r means the same as f(r), but since R is
not countable, then the set of all s_r doesn't have to be countable
either... (It *can* be countable, though. Trivial example: Assume S=R
and let f be f(r)=0 for all r in R.)
PS. Please do not quote parts of my post that you do not reply to, in
particular do not quote my signature. Thanks.
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"The obvious mathematical breakthrough would be development of an easy way to
factor large prime numbers."
- Bill Gates
David C. Ullrich
wrote:
[...]
>
>PS. Please do not quote parts of my post that you do not reply to, in
>particular do not quote my signature. Thanks.
What a curious request.
Hint: This is usenet - hang out here long enough and you'll have
people quoting you as saying things that you simply never said.
When people are quoting things you actually did write you don't
have any cause to complain. (And if you think people are going to
make a point of scrolling down and snipping your sig you're
wrong...)
Never noticed your sig until just now. The 'Kingpriest of "The
Flying Lemon Tree"' does look a little curious, but if you don't
want people to know your sig reads that way there's a simple
solution.
>--
>/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
>| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
>| http://www.helsinki.fi/~palaste W++ B OP+ |
>\----------------------------------------- Finland rules! ------------/
>"The obvious mathematical breakthrough would be development of an easy way to
>factor large prime numbers."
> - Bill Gates
David C. Ullrich
Ben Webster
> Thank you. Now since the set of all real numbers (i.e. R) is
> uncountable, this must mean that transcendental irrationals exist,
> because no set can be both countable and uncountable at the same time,
> therefore the set of algebraic reals cannot equal R.
> Now I just need to remember what the proof of the uncountability of R
> was and I am all
set...
Heck, we can just use the interval If the real numbers between 0 and 1
are countable, we can make an infinite list which contains every real
number in this interval somewhere on it. Write out this list with the
numbers written as decimal numbers. now make a new real number, where
the nth digit after the decimal point of our new number is simply the
nth digit of the nth number in our list, plus 1 (take 9 to 0). Now if
we stick a 0 in front of the decimal point, this is clearly a real
number between 0 and 1. but it cannot be list because it differents
from every element in our list in at least one place.
Joona I Palaste
> set...
Thank you. Now we have proven that algebraic reals are countable, even
if they are irrational, and all reals are uncountable.
Because an uncountable set can have a countable subset, but a countable
set can not have an uncountable subset, this means that the set of
algebraic reals is a proper subset of the set of all reals. Because it
is a proper subset its complement with respect to R must be non-empty,
which proves that transcendental numbers exist.
Actually *finding* transcendental numbers is a whole different kettle
of fish, though...
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"We sorcerers don't like to eat our words, so to say."
- Sparrowhawk
Dik T. Winter
> Actually *finding* transcendental numbers is a whole different kettle
> of fish, though...
Not really. There are quite a few known, and given a transcendental and
an algebraic number, their sum is transcendental. (There product is
also transcendental if the algebraic number is nonzero.)
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
The World Wide Wade
ull...@math.okstate.edu (David C. Ullrich) wrote:
> Never noticed your sig until just now. The 'Kingpriest of "The
> Flying Lemon Tree"' does look a little curious, but if you don't
> want people to know your sig reads that way there's a simple
> solution.
If I may make a suggestion: There is a point of Usenet etiquette here.
Leaving the tail end of someone's post and their sig in your reply as you
do - unresponded to and then followed by your own auto sig (that's how it
appears) - is a waste of cyberspace and slightly confusing and/or annoying
to readers.
--WWW.
The World Wide Wade
1. Which real numbers x are roots of a power series with rational
coefficients, ie, for which x in R do there exist rational a_0, a_1, ...
such that
a_0 + a_1*x + a_2*x^2 + ... = 0?
2. Same question, except restrict the a_n's to be integers.
I have solutions, but I won't spoil your Saturday fun.
--WWW.
Arturo Magidin
>
>x in R\Q is algebraic iff there exists a countable set A so that:
> oo
>\bigsum (a_i * x^i) = 0,
> i=0
>where each a_i is rational and a member of A.
No, they are not. That would be power series.
An algebraic number is an element of C (the complex numbers) which is
a root of a polynomial with rational (or integer, amounts to the same
thing) coefficients.
That is:
DEF: A complex number c is algebraic iff there exists a positive
integer n, and rational a_0,...,a_n, a_n not zero, such that
a_n*c^n + a_{n-1}*c^{n-1} + ... + a_1*c + a_0 = 0
Then, an "irrational algebraic" would be a real number which is: (a)
algebraic; and (b) not rational.
Since an algebraic number is rational if and only if it is a root of
polynomial of degree 1 with rational coefficients, then an irrational
algebraic is a real number which is a root of an irreducible
polynomial of degree n>1 with rational coefficients.
>This means that x in R\Q is algebraic iff it's the zero of some one-
>variable polynomial with rational coefficients.
This does not agree with your definition above. Polynomials have
finite degree, you gave a formal power series.
As to your question: yes, algebraic irrationals are countable. There
are clearly at least countably many of them, since the square root of
any square free positive integer is an algebraic irrational. And since
there are at most countably many algebraic numbers altogether (there
are only countably many polynomials, since they must have finite
degree), there must be a countable number of them.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
Joona I Palaste <pal...@cc.helsinki.fi> wrote:
>My representation was perhaps too ambiguous. I did NOT intend the power
>series to be infinite. I intended it to be finite, but arbitrarily long.
>That means when I wrote:
>"All a_i (where i is from 0 to oo) are rational and members of A", I
>also implied that there exists a number k in N so that all a_i are 0
>when i>=k.
>Sorry about that confusion!
The easiest way to do that if you want to stick to the infinite sum is
to say "where the a_i are all rationals, and almost all are zero", or
"and all but a finite number of the a_i are equal to zero."
Dave L. Renfro
[sci.math 1 Jun 2002 16:05:33 GMT]
http://mathforum.org/epigone/sci.math/twalpougo
wrote
> I assume algebraic irrationals are defined as:
>
> x in R\Q is algebraic iff there exists a countable set A so that:
> oo\bigsum (a_i * x^i) = 0,
> i=0
> where each a_i is rational and a member of A.
You want "finite", not "countable". This aside, what you've
written is overly formal. There is no need to define
"algebraic irrational". Just define "algebraic" and then say
"algebraic irrational". There is no need to introduce the set
'A', no need for something like \bigsum (a_i * x^i) = 0 (that's
what the word "polynomial" is for [Now I'm starting to sound like
David Ullirch . . .] ), etc.
> This means that x in R\Q is algebraic iff it's the zero of some
> one-variable polynomial with rational coefficients.
Again, I don't see why you want "x in R\Q". What you wrote is true,
but you can give a statement that is both logically stronger
and logically simpler by writing "x in R is algebraic iff ...".
Even if you're only interested in "algebraic irrational", you risk
confusing an attentive reader with your unnecessary emphasis on
"irrational" at this point.
Here's a quote from the beginning of Section 10 of Halmos' well-known
essay "How to write mathematics" that addresses the point I'm making:
"Sometimes a proposition can be so obvious that it
needn't even be called obvious and still the sentence
that announces it is bad exposition, bad because it
makes for confusion, misdirection, delay. I mean
something like this: "If R is a commutative semisimple
ring with unit and if x and y are in R, then
x^2 - y^2 = (x-y)(x+y)." The alert reader will ask
himself what semisimplicity and a unit have to do
with what he had always thought was obvious. Irrelevant
assumptions wantonly dragged in, incorrect emphasis, or
even just the absence of correct emphasis can wreak havoc."
> Now the question is: are algebraic irrationals countable? Of course,
> there are countably many one-variable polynomials with rational
> coefficients, in fact there are countably many polynomials with
> rational coefficients, regardless of how may variables are involved.
> But does every such polynomial have countably many zeroes? If so,
> then the set of all zeroes of all polynomials with rational
> coefficients is the union of countably many countable sets and
> thus countable. Otherwise it is not countable.
You've got the hard part of the proof correct, at least in outline
form. [Whether what you wrote is enough will depend on your audience.]
Do you remember the result from high school algebra that a nonzero
polynomial of degree n has at most n roots? Usually this is stated
in high school, not proved, but in case someone posts anything about
this being a consequence of the fundamental theorem of algebra, let
me point out that you DON'T need the fundamental theorem of algebra
to prove this. [Recall that the fundamental theorem of algebra says
every polynomial has at least one (possibly complex) root.]
Let P(x) be a polynomial of degree n. If P(x) has no roots, then
we're done (i.e. P(x) has at most n roots). If not, let r be a
root of P(x). Then we can write P(x) = Q(x) * (x-r) for some
polynomial Q(x). [Why? Suppose we divide P(x) by (x-r). Then we'll
get P(x) = Q(x) * (x-r) + R, where R is a constant. Evaluate both
sides of P(x) = Q(x) * (x-r) + R for x=r and you'll get R=0.] Now
repeat this process for Q if Q isn't a constant. Since the degree
of the quotient (when it's a polynomial of positive degree) drops
by 1 each time, we'll get at most n roots before the process stops
(i.e. when we get a quotient having no roots).
[[ Actually, all we need is that a nonzero polynomial has
countably many roots, but I don't know how to prove this
weaker statement in a way that's easier than how I proved
the stronger statement above. (Observing that a nonzero
analytic function has countably many zeros doesn't
qualify as easier, in my opinion.) ]]
Incidentally, the paper that Cantor first proved this result is
Cantor, "Über eine Eigenschaft des Inbegriffs aller reellen
algebraischen Zahlen" [On a property of the set of real algebraic
numbers], Journal f. reine und angew. Math. 77 (1874), 258-262.
There is an English translation on pp. 839-843 of volume 2 of
William Ewald's book "From Kant to Hilbert: A Source Book in the
Foundations of Mathematics", Oxford: Clarendon Press, 1999.
Cantor's proof had two main parts -- the algebraic numbers
are countable and the real numbers are uncountable. The latter
part wasn't proved by diagonalization in this paper. [Cantor's
diagonalization proof didn't appear until 1891.] Instead, Cantor
proved this using what is sometimes called the nested interval
property. For the details of Cantor's earlier proof of the
uncountability of the reals (using modern notions), and a way of
using a similar method to prove that the reals cannot be expressed
as a countable union of sets each of which has cardinality less than
the reals, see my posts at
http://mathforum.org/epigone/sci.math/skeldtwaxfrul/5i9kx3le0ndl@legacy
http://mathforum.org/epigone/sci.math/clanfrendmel/oup56h...@forum.mathforum.com
Incidentally, at the end of this paper Cantor observes that his
method can be easily adapted to show the existence of other types
of numbers as well. Thus, Cantor was aware of the power behind
his method even at this time. For instance, Cantor observes (I think)
that every interval contains infinitely many numbers that cannot be
expressed as a quotient of two finite Q-linear combinations of
numbers from a given countable set (such that no finite subset is
linearly dependent over Q). [Ewald's translation has the phrase
"rational functions", but I don't think this use of "rational
function" is what we now mean by rational function.] Cantor mentions
that a weaker version of this was proved in 1871 by B. Minnigerode.
In Minnigerode's weaker version the given countable set is a
given finite Q-linearly independent set. Minnigerode's paper
(but not Cantor's 1873 paper, at least not that I could find) is
actually on the internet, at
http://134.76.163.65:80/agora_docs/26030TABLE_OF_CONTENTS.html
One of my favorite topics is this way of proving the existence of an
object having some specified property P by showing in some sense that
almost all objects under consideration have property P --->>>
** The existence of absolutely normal numbers using Lebesgue measure.
** The existence of nowhere differentiable continuous functions using
Baire category.
** The existence of non-Borel sets using cardinality.
** The existence of an embedding of a given n-dimensional separable
metric space X into the (2n+1)-dimensional cube [0,1]^(2n+1)
(the Menger-Nöbeling theorem) using Baire category. [Hurewicz
proved in 1933 that "most" continuous maps from X into I are
embeddings.]
** Most lines in the plane containing the origin do not pass through
any point both of whose coordinates are nonzero rational numbers.
** The nonexistence of nowhere differentiable monotone functions.
[Every monotone function is differentiable a.e.]
** Every pointwise limit of continuous functions has a point of
continuity. [Every such limit is continuous on all but a first
category set of points.]
** Every uncountable closed set has a condensation point. [Indeed,
all but countably many points of such a set will be a condensation
point of the set.]
** The set of points a given distance away from a given closed
set in R^n has Lebesgue measure zero. [The set in question has
no points of Lebesgue density. (In fact, it satisfies at each
of its points what in potential theory is called the "external
sphere condition".) Now use the Lebesgue density theorem.]
O-K, I guess I've gone on long enough. Time to get back to work.
Dave L. Renfro
Robert Israel
The World Wide Wade <wrame...@attbi.remove13.com> wrote:
>My original interpretation of this problem leads to two questions.
>1. Which real numbers x are roots of a power series with rational
>coefficients, ie, for which x in R do there exist rational a_0, a_1, ...
>such that
> a_0 + a_1*x + a_2*x^2 + ... = 0?
Assuming you're assuming at least one a_i <> 0 to make the question
nontrivial:
Every real x will do. We can assume x <> 0, and take a_0 = 1. Choose
a_n inductively so that a_n is rational and
|sum_{j=0}^n a_j x^j| is as small as you like; if it goes to 0
we have sum_{j=0}^infinity a_j x^j = 0.
>2. Same question, except restrict the a_n's to be integers.
Well, the series will diverge if |x| >= 1 unless the a_n are eventually 0,
which brings us back to polynomials and algebraic numbers.
If |x| < 1, we can take a_0 = 1 and choose integers a_n inductively
to make |sum_{j=0}^n a_j x^j| <= |x^n|/2. Since that goes to 0 as n ->
infinity, we get sum_{j=0}^infinity a_j x^j = 0.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Tim Mellor
polynomials over Q.
A quick proof that they are countable:
Let T be the theory of R in the first order language of fields, then T
has a countable model M by downward Lowenheim-Skolem. Any member of
the algebraic numbers is dy definition 0 definable, and so included in
every model, so Q is a subset of M (and thus countable).
(The same argument on the theory of C shows that the algebraic closure
of Q is countable)
David C. Ullrich
wrote:
>The algebraic numbers are more usually defined to be the real roots of
>polynomials over Q.
>A quick proof that they are countable:
>
>Let T be the theory of R in the first order language of fields, then T
>has a countable model M by downward Lowenheim-Skolem. Any member of
>the algebraic numbers is dy definition 0 definable, and so included in
>every model, so Q is a subset of M (and thus countable).
That's probably the simplest proof, no doubt about it.
(What does "0 definable" mean?)
>(The same argument on the theory of C shows that the algebraic closure
>of Q is countable)
David C. Ullrich
Tim Mellor
There is some formula psi(x) (with no parameters) so that T prooves
there exists a unique x (psi(x))
The 0-definable, strictly emptyset definable, (the zero is lazy
notation but established) is important here because every 0-definable
element is in every model of the theory.