From: dlrenfro
Subject: Re: PI an Integer?
Date: Sat, 25 Sep 1999 23:59:23 -0500
Newsgroups: sci.logic,sci.math,rec.arts.sf.science
Keywords: Hamel bases of R over Q cannot be 'nice'.
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
==============================================================================
From: Mike Oliver
Subject: Re: PI an Integer?
Date: Sun, 26 Sep 1999 18:12:33 -0700
Newsgroups: sci.logic,sci.math,rec.arts.sf.science
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?
==============================================================================
From: Fred Galvin
Subject: Re: PI an Integer?
Date: Sun, 26 Sep 1999 23:47:13 -0500
Newsgroups: sci.logic,sci.math,rec.arts.sf.science
On Sun, 26 Sep 1999, Mike Oliver wrote:
[previous article quoted --djr]
> Anybody see any errors in the above?
What if the Hamel basis has no rational element?
==============================================================================
From: Mike Oliver
Subject: Re: PI an Integer?
Date: Mon, 27 Sep 1999 16:13:42 -0700
Newsgroups: sci.logic,sci.math,rec.arts.sf.science
Fred Galvin wrote:
>
> On Sun, 26 Sep 1999, Mike Oliver wrote:
> > Anybody see any errors in the above?
>
> What if the Hamel basis has no rational element?
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.
==============================================================================
From: ikastan@uranus.uucp (Ilias Kastanas)
Subject: Hamel bases (was: Re: PI...)
Date: 28 Sep 1999 12:47:26 GMT
Newsgroups: sci.logic,sci.math,rec.arts.sf.science
In article <37EFFA26.F12D99A0@math.ucla.edu>,
Mike Oliver wrote:
>
>Fred Galvin wrote:
>>
>> On Sun, 26 Sep 1999, Mike Oliver wrote:
>> > Anybody see any errors in the above?
>>
>> What if the Hamel basis has no rational element?
>
>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.
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
==============================================================================
From: dlrenfro@gateway.net (Dave L. Renfro)
Subject: Re: Hamel bases (was: Re: PI...)
Date: 2 Oct 1999 10:16:20 -0400
Newsgroups: sci.math
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
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
==============================================================================
From: ikastan@uranus.uucp (Ilias Kastanas)
Subject: Re: Hamel bases (was: Re: PI...)
Date: 5 Oct 1999 11:39:54 GMT
Newsgroups: sci.math,sci.logic
In article ,
Dave L. Renfro 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