Borel set
Andersen
What are the Borel sets (or Borel-algebra). As I understand it, it is
the smallest sigma-algebra which contains all the open intervals.
But being a sigma-algebra, it must also contain all the complements,
which are the closed sets. Hence, it will contain all the open and
closed sets. Since it must contain and unions/intersections, it will
contain all sets that are neither open nor closed, e.g. [1,1]U(1,2)=[1,2).
So it seems to me that this generates all the reals. How is this set
smaller than the power set of the reals?
Andersen
Glen Wheeler
news:465175c2$0$97272$892e...@authen.yellow.readfreenews.net...
Ha! It is much smaller. Easiest is to do a cardinality argument. Second
easiest is to find an example. Royden has an example as an exercise, you
might do well to look that up.
Borel sets are fun.
--
Glen
se...@btinternet.com
A sigma-algebra only requires closure under *countable* unions (and,
as you say, complements).
So for example a Vitali set will not be a Borel set, i.e. not a member
of a Borel algebra, even though it is a subset of the reals. So "it
will contain *all* sets that are neither open nor closed" (my
emphasis) is wrong.
Andersen
> Ha! It is much smaller. Easiest is to do a cardinality argument. Second
> easiest is to find an example. Royden has an example as an exercise, you
> might do well to look that up.
Could you please do any one of them? I do not have Royden, and I can't
really think of how I could do a cardinality argument.
Andersen
Andersen
> A sigma-algebra only requires closure under *countable* unions (and,
> as you say, complements).
> So for example a Vitali set will not be a Borel set, i.e. not a member
> of a Borel algebra, even though it is a subset of the reals. So "it
> will contain *all* sets that are neither open nor closed" (my
> emphasis) is wrong.
>
Could you explain this a bit more. I have looked up the definition of a
Vitaly set from Wiki, which is as follows:
"If x and y are real numbers and x − y is a rational number, then we
write x ~ y and we say that x and y are equivalent; ~ is an equivalence
relation. For each x, there is a subset [x] = {y in R : x ~ y} called
the equivalence class of x. The set of these equivalence classes
partitions R. By the axiom of choice, we are able to choose a set V ⊂
[0, 1] containing exactly one representative out of each equivalence
class (for any equivalence class [x], the set V ∩ [x] is a singleton).
We say that V is a Vitali set."
I think I understand the above. Could you please explain why such a V
cannot be a Borel set?
Is it that there will be uncountably infinite number of members in V,
and you cannot therefore construct them by taking countable unions?
Please explain more.
Andersen
José Carlos Santos
I don't know whether or not this is the example from Royden's textbook,
but one possibility is: the set of those numbers _x_ which can be
written as 1/(m_1 + 1/(m_2 + 1/(m_3 + ...))), where each m_k is a
natural number and, for each _k_, m_k divides m_{k + 1}.
Best regards,
Jose Carlos Santos
José Carlos Santos
>> A sigma-algebra only requires closure under *countable* unions (and,
>> as you say, complements).
>> So for example a Vitali set will not be a Borel set, i.e. not a member
>> of a Borel algebra, even though it is a subset of the reals. So "it
>> will contain *all* sets that are neither open nor closed" (my
>> emphasis) is wrong.
>>
>
> Could you explain this a bit more. I have looked up the definition of a
> Vitaly set from Wiki, which is as follows:
>
> "If x and y are real numbers and x − y is a rational number, then we
> write x ~ y and we say that x and y are equivalent; ~ is an equivalence
> relation. For each x, there is a subset [x] = {y in R : x ~ y} called
> the equivalence class of x. The set of these equivalence classes
> partitions R. By the axiom of choice, we are able to choose a set V ⊂
> [0, 1] containing exactly one representative out of each equivalence
> class (for any equivalence class [x], the set V ∩ [x] is a singleton).
> We say that V is a Vitali set."
>
> I think I understand the above. Could you please explain why such a V
> cannot be a Borel set?
Because every Borel set is measurable and the Vitali set isn't.
se...@btinternet.com
>
> - Show quoted text -
That is a proof - though as for a reason, a Vitali set is an
uncountable union of arbitrary points (so arbitrary that the Axiom of
Choice has to be used and so Vitali sets are unconstructable) and so
need not appear in a sigma-algebra, and will not be a countable union
of (open, closed or half open/half closed] intervals and points.
Glen Wheeler
> se...@btinternet.com wrote:
>
>> A sigma-algebra only requires closure under *countable* unions (and,
>> as you say, complements).
>> So for example a Vitali set will not be a Borel set, i.e. not a member
>> of a Borel algebra, even though it is a subset of the reals. So "it
>> will contain *all* sets that are neither open nor closed" (my
>> emphasis) is wrong.
>>
>
> Could you explain this a bit more. I have looked up the definition of a
> Vitaly set from Wiki, which is as follows:
>
> relation. For each x, there is a subset [x] = {y in R : x ~ y} called the
> equivalence class of x. The set of these equivalence classes partitions R.
>
> I think I understand the above. Could you please explain why such a V
> cannot be a Borel set?
>
> Is it that there will be uncountably infinite number of members in V, and
> you cannot therefore construct them by taking countable unions? Please
> explain more.
>
There is a proof on Wikipedia (unfortunately). This is the standard
example. Drawing a picture might help ;).
--
Glen
Glen Wheeler
news:5bdht5F...@mid.individual.net...
But, can you think of a set which is Lebesgue measurable, and not Borel?
This is fun :).
--
Glen
µ
> So it seems to me that this generates all the reals. How is this set
> smaller than the power set of the reals?
It can be shown that the set of all the Borel sets of real numbers has
the cardinality of the set of reals numbers, hence is much more smaller
than the set of all subsets of the reals.
The proof I read a long time ago uses Suslin sets: it seems to be quite
easy to show that there are no more Suslin sets than reals, then to see
that every Borel set is a Suslin set.
--
Mū
José Carlos Santos
>>>> Ha! It is much smaller. Easiest is to do a cardinality argument.
>>>> Second easiest is to find an example. Royden has an example as an
>>>> exercise, you might do well to look that up.
>>> Could you please do any one of them? I do not have Royden, and I can't
>>> really think of how I could do a cardinality argument.
>> I don't know whether or not this is the example from Royden's textbook,
>> but one possibility is: the set of those numbers _x_ which can be
>> written as 1/(m_1 + 1/(m_2 + 1/(m_3 + ...))), where each m_k is a
>> natural number and, for each _k_, m_k divides m_{k + 1}.
>>
>
> But, can you think of a set which is Lebesgue measurable, and not Borel?
The example that I described has that property. It must have, because,
since the existence of non-Lebesgue measurable sets depends upon the
axiom of choice, no explicit example of a set of real numbers can be
non-measurable.
fjb...@yahoo.com
One way to see it is via Lebesgue measure. If V were a Borel set, it
would be Lebesgue measurable. You can show that [0,1] can be written
as a countable disjoint union of V translated by rationals (mod 1);
then any value you try for m(V) will contradict countable additivity
of m.
Stephen J. Herschkorn
In the standard cardinality argument, one builds up the Borel sets
through the ordinals. Let B0 be the set of all open intervals of real
numbers. Let B_(a+1) be the set of all countable unions of sets which
are in B_a or whose complements are in B_a. If a is a limit ordinal, let
B_a = union(B_b: b < a}. Show that the least uncountable ordinal w1 is
the least ordinal a such that B_a is a sigma-field, and that the
cardinality of B_w1 is c = 2^w, the cardinality of the reals (where w =
omega, the cardinality of the natural numbers). This is strictly less
than 2^c, the cardinality of the power set of the reals.
> I have looked up the definition of a Vitaly set from Wiki, which is as
> follows:
>
> "If x and y are real numbers and x − y is a rational number, then we
> write x ~ y and we say that x and y are equivalent; ~ is an
> equivalence relation. For each x, there is a subset [x] = {y in R : x
> ~ y} called the equivalence class of x. The set of these equivalence
> classes partitions R. By the axiom of choice, we are able to choose a
> set V ⊂ [0, 1] containing exactly one representative out of each
> equivalence class (for any equivalence class [x], the set V ∩ [x] is a
> singleton). We say that V is a Vitali set."
>
> I think I understand the above. Could you please explain why such a V
> cannot be a Borel set?
For each rational q in [0,1), let V_q = {v + q mod 1: v in V}. Show that
[0,1] = union(q, V_q), a countable union. Conclude that the Lebesgue
mesaure of V is neither 0 nor positive, i.e., V is not Lebesgue
measurable, hence not Borel.
(Every Borel set is a Lebesgue set. In fact, the Lebesgue sigma-field is
the completion of the Borel sigma-field under Lebesgue meausure. That
is, every Lebesgue set is the union of a Borel set and a subset of a
null Borel set.)
Glen Wheeler wrote:
>Can you think of a set which is Lebesgue measurable, and not Borel?
>
The Cantor set is an uncountable Borel set of mesasure 0. Since the
Lebesgue sets are complete, every subset of the Cantor set is Lebesgue
measurable. This provides 2^c Lebesgue sets, which hence cannot all be
Borel.
--
Stephen J. Herschkorn sjher...@netscape.net
Math Tutor on the Internet and in Central New Jersey and Manhattan
Dave L. Renfro
> It can be shown that the set of all the Borel sets of
> real numbers has the cardinality of the set of reals
> numbers, hence is much more smaller than the set of
> all subsets of the reals. The proof I read a long
> time ago uses Suslin sets: it seems to be quite easy
> to show that there are no more Suslin sets than reals,
> then to see that every Borel set is a Suslin set.
Offhand, I can think of two proofs. One proof makes use
of the transfinite construction of the Borel sets. Because
this hierarchy has length aleph_1, and it begins with
a collection having cardinality c, and it is easy to
see that both the successor and limit ordinal stages
don't change the cardinality for collections with
c many sets, the Borel sets have cardinality no more
than that of the union of aleph_1 many sets each
having cardinality c, which is c (or, if you want to
avoid brining transfinite sums into the picture,
just use a straightforward transfinite induction).
The other proof I know of makes use of the fact
(which can be proved by what is sometimes called
the "good sets principle") that if C is a generator
for a sigma-algebra B and E is an element of B,
then there exists a countable subset C' of C
such that E belongs to the sigma-algebra generated
by C'. Now simply observe that the collection
of Borel sets has a countable generator B (open
intervals with rational number endpoints, for
instance) and every countable set has c many
subsets.
Dave L. Renfro
Glen Wheeler
That's an indirect argument at best :). An explicit proof would be nice,
but I suspect it is difficult.
--
Glen
Glen Wheeler
news:4651E95C...@netscape.net...
> Glen Wheeler wrote:
>
>>Can you think of a set which is Lebesgue measurable, and not Borel?
>>
>
> The Cantor set is an uncountable Borel set of mesasure 0. Since the
> Lebesgue sets are complete, every subset of the Cantor set is Lebesgue
> measurable. This provides 2^c Lebesgue sets, which hence cannot all be
> Borel.
>
But this is just the cardinality argument again. Although one of the
examples I am thinking of does involve the Cantor function. This is from
Royden, again. IIRC it is to do with the inverse images of the cantor
function.
--
Glen
Dave L. Renfro
>>> But, can you think of a set which is Lebesgue measurable,
>>> and not Borel?
Stephen J. Herschkorn wrote:
>> The example that I described has that property.
>> It must have, because, since the existence of
>> non-Lebesgue measurable sets depends upon the
>> axiom of choice, no explicit example of a set
>> of real numbers can be non-measurable.
Glen Wheeler wrote:
> That's an indirect argument at best :). An explicit
> proof would be nice, but I suspect it is difficult.
If we're going to toss out the axiom of choice, you
may as well toss out most everything in real analysis,
including the fact that a countable union of measurable
sets is measurable, a function is epsilon-delta continuous
at a point if and only if it is sequentially continuous
at that point, a countable union of sets is countable
(this can fail even for sets of real numbers), etc.
As for proving there are only c many Borel sets,
I indicated two proofs in my previous post in this
thread.
However, there are fairly explicit examples of analytic
sets that are not Borel in the reals, if that's what
you want -- no need to go all the way out into the
non-measurable wastelands. If you want to know why
the explicitly defined set posted earlier is not Borel,
again, there's no need to go all the way into the
non-measurable wastelands. I believe this is Lusin's
example, given around 1927, of an analytic set
that isn't Borel. One can give direct proofs that
it isn't Borel (but I don't know the subject well
enough to give one now), and I'm sure you can find
such a proof in any of the standard texts on the subject:
Kuratowski's "Topology", volume 1; Lusin's "Lecons Sur
les Ensembles. Analytiques et Leurs Applications";
Kechris' "Classical Descriptive Set Theory"; Srivastava's
"A Course on Borel Sets"; Sierpinski's "General Topology";
Van Rooij/Schikhof's "A Second Course on Real Functions".
Dave L. Renfro
Stephen J. Herschkorn
>Glen Wheeler wrote:
>
>
>
>>>>But, can you think of a set which is Lebesgue measurable,
>>>>and not Borel?
>>>>
>>>>
>
>Stephen J. Herschkorn wrote:
>
>
>
>>>The example that I described has that property.
>>>It must have, because, since the existence of
>>>non-Lebesgue measurable sets depends upon the
>>>axiom of choice, no explicit example of a set
>>>of real numbers can be non-measurable.
>>>
[...]
Correction: I did not write the last quote. That is from a post of
José Carlos Santos.
José Carlos Santos
Indeed there are! And the example that I posted earlier at this thread
is precisely an example of an analytic set which is not a Borel set.
Glen Wheeler
news:1179842915.8...@x18g2000prd.googlegroups.com...
> [lots of good stuff]
Thanks for the recommendations. I just want to clarify that I am not
looking for any answers, I've studied these areas thoroughly in my own work,
and feel capable. I'm definitely not throwing out Choice, or suggesting it.
Just asking questions to the group.
--
Glen
Martin Vaeth
>
> If we're going to toss out the axiom of choice, you
> may as well toss out most everything in real analysis,
That's an unfair argument in this connection, since
countable choice (and even dependent choice) should be
considered independently:
I think no one working in analysis would seriously reject DC.
The examples which you give are no problem with DC.
However, AC in general (or at least some weaker form, leading
to non-Lebesgue measurable sets) is a different story.
Dave L. Renfro
>> If we're going to toss out the axiom of choice, you
>> may as well toss out most everything in real analysis,
Martin Vaeth wrote:
> That's an unfair argument in this connection, since
> countable choice (and even dependent choice) should be
> considered independently:
> I think no one working in analysis would seriously reject DC.
> The examples which you give are no problem with DC.
> However, AC in general (or at least some weaker form, leading
> to non-Lebesgue measurable sets) is a different story.
This is a point I've been making for many years in sci.math.
If it's uncountable choice that someone is concerned about,
then they should say so. Speaking of the axiom of choice,
I think Jacques Hadamard argued that, from a plausibility
standpoint, arbitrary choice is more acceptable for existence
than explicitly defined choice, since there is more freedom
for the object to exist (i.e. there are fewer requirements
that we have to satisfy to know that the object exists),
or something like this.
Dave L. Renfro
Aatu Koskensilta
> This is a point I've been making for many years in sci.math.
> If it's uncountable choice that someone is concerned about,
> then they should say so. Speaking of the axiom of choice,
> I think Jacques Hadamard argued that, from a plausibility
> standpoint, arbitrary choice is more acceptable for existence
> than explicitly defined choice, since there is more freedom
> for the object to exist (i.e. there are fewer requirements
> that we have to satisfy to know that the object exists),
> or something like this.
The justification for the axiom of choice -- for me, in any case -- is that
it is a mathematical expression of the idea that sets are arbitrary
extensional collections. On a conception of sets based on definability it is
indeed not at all obvious that choice is justified, though I don't see how
it would justify the *failure* of choice either. (Of course, we might be
able to work out whether or not choice holds for some mathematically defined
notion of definability, but that's another matter).
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Martin Vaeth
> then they should say so. Speaking of the axiom of choice,
> I think Jacques Hadamard argued that, from a plausibility
> standpoint, arbitrary choice is more acceptable for existence
> than explicitly defined choice, since there is more freedom
> for the object to exist
This is true from a purely mathematical standpoint.
However, as soon as one uses (parts of) mathematics as a "model"
for some physical phenomenon, this "model" is already rather idealized,
even if just such a "familiar" thing as real numbers are involved.
Now if one uses even uncountable choice, this model is IMHO
"too idealized" to lead to the correct predictions in all cases.
Of course, this is currently only an intuition - a convincing way to
test whether this intuition is true or not is still unknown AFAIK.
Shmuel (Seymour J.) Metz
05/21/2007
at 12:34 PM, Andersen <anders...@hotmail.com> said:
>But being a sigma-algebra, it must also contain all the complements,
>which are the closed sets. Hence, it will contain all the open and
>closed sets. Since it must contain and unions/intersections, it will
>contain all sets that are neither open nor closed,
No; it contains all the semi-open intervals, but not all sets that are
neither open nor closed.
>So it seems to me that this generates all the reals.
It generates all of the one-point subsets of the reals. It does not
generate all of the subsets of the reals. Consider the difference
between "all unions" and "all countable unions".
> How is this set smaller than the power set of the reals?
Because not every set of reals can be generated that way.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>
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