Ultrafilter theorem refined

[William Elliot]

Let S be a set, F_A = { B subset S | A subset B }
the principal filter for S generated by {A}.
The principal ultrafilters are F_x = F_{x}, for x in S.

The usual ultrafilter theorem is for a filter F,
F = /\{ G ultrafilter | F subset G }

This leads to a corollary, if F is a free filter, then
F = /\{ G free ultrafilter | F subset G }.

Directly one can show that if F_A is a principal ultrafilter,
F_A = { F_x | x in A }.

There remains the case when F is not free and not principal.

From the ultratheorem come another corollary.
If A = /\F not empty, then
F = { F_x | x in A } /\ { G ultrafilter | F subset G }
. = F_A /\ G ultrafilter | F subset G }.

Can that be refined for a filter F for S, to:
F = F_/\F /\ { G free ultrafilter | F subset G }

with the understanding F_(empty set) = P(S)?

[William Elliot]

Yes, but it's too weak with too many ultrafilters.
Heare a generalization of all the case considered
with fewer ultrafilters, perhaps the lest number.

Conjecture. If F is a filter for S, then
F = F_/\F /\ /\ { G free ultrafilter | F subset G, S - /\F in G }

and the described collection of ultrafilters
is the smallest such colleciton possible.

True or False?

[Victor Porton]

Yes, it readily follows from the fact that every ultrafilter is either
principal of free.

> with the understanding F_(empty set) = P(S)?

--
Victor Porton - http://portonvictor.org

[Victor Porton]

It is true:

Using notation of my book:
http://www.mathematics21.org/algebraic-general-topology.html
(an yes, reverse order of filters):

You conjecture can be rewritten as:
F = Cor F \/ \/{G is a free ultrafilter | G<=F, Cor F notin G}
what is equivalent to:
F = Cor F \/ \/{G is a free ultrafilter | G<=F\Cor F}
F = Cor F \/ \/atoms Edg F
F = Cor F \/ Edg F
what is true by definition of Edg.

> and the described collection of ultrafilters
> is the smallest such colleciton possible.

It is an interesting conjecture whether it is the smallest. I will think
about this.

> True or False?

[Victor Porton]

Whether it is smallest appears to be a difficult problem because it amounts
to properties of ultrafilters (which we can't even construct).

In my opinion, it is *not* smallest, because among principal and free
ultrafilters there is many so to day "degrees of freedom" for free
ultrafilters.

[William Elliot]

My conjecture may be useful for your
conjecture that for each filter, there's a
strong copartition by ultrafilters.

What do you think?

It it a digression from your thesis of
generalizing continuity by reloids and funcoids?

Both because of the excessive length and
the strange notations (see below) I may
not be able to completely review your work and edit
it to a paper that will be noticed by sci.math
and elsewhere to finally perfect it to the point where
you could publish it.

Your notation is hopelessly unreadable.
Use instead, known definitions and know notations
and for some, if not most strange notation, use
several know symbols. Stephen Willard in "General Topology"
orders the collection of filters like I do, naturally
by F <= G when F subset G. Indeed, it's much easier to
prove a statement myself that to wade through your
cryptic proof. Your will continue to be a failure
until you use common mathematical language instead
of your own private language meaningful toe just you.

[Victor Porton]

William Elliot wrote:

> My conjecture may be useful for your
> conjecture that for each filter, there's a
> strong copartition by ultrafilters.
>
> What do you think?

Maybe. But I expect that your conjecture is false and thus not useful.

> It it a digression from your thesis of
> generalizing continuity by reloids and funcoids?

I research filters mainly as a basis for reloids and funcoids, but not only.
I also research filters for their own sake.

[William Elliot]

On Mon, 30 Dec 2013, Victor Porton wrote:
> William Elliot wrote:
>
> > My conjecture may be useful for your
> > conjecture that for each filter, there's a
> > strong copartition by ultrafilters.
> >
> > What do you think?
>
> Maybe. But I expect that your conjecture is false and thus not useful.

It is false as the Fletcher filter shows by a cardinality consideration.
To the point, if F is a free filter, is there a minimal collection
K, of ultrafilters for which F = /\K? I would think K would have at
most 2^|S| ultrafilters.

----

[Niels Diepeveen]

Op 31-12-13 05:25, William Elliot schreef:

> To the point, if F is a free filter, is there a minimal collection
> K, of ultrafilters for which F = /\K? I would think K would have at
> most 2^|S| ultrafilters.
>

If F is the cofinite filter on an infinite set S and K a collection of
ultrafilters such that /\K = F, then for every G in K also
/\(K \ {G}) = F.

Proof: Recall that if G is an ultrafilter on S, then for every
A subset S either A in G or S\A in G.
It follows that /\K is the cofinite filter iff for every
infinite A subset S there is a G in K such that A in G.
Now suppose that there is an A for which there is exactly
one G in K for which A in G. A is infinite, so it can be partitioned
into infinite A_1 and A_2.
Since A_1 and A_2 are disjoint, they cannot both be in G, but since
they are subsets of A, neither can be in an other element of K. This
contradicts the assumption that /\K is the cofinite filter.

--
Niels Diepeveen

[William Elliot]

How so? All that proves is every infinite
subset is in more than one ultrafilter.

[Niels Diepeveen]

Op 02-01-14 07:43, William Elliot schreef:

Yes, and that implies that if /\K = F then /\(K \ {G}) = F,
hence no K is can be minimal in that respect.

[Victor Porton]

Niels Diepeveen, I will check your proof.

Assume it is correct.

It seems that from your statement it follows that there are no copartition
of the cofinite filter, and one my (relatively) long standing conjectures is
now solved.

I will check it after checking for a PHP or Perl programming job ads. (I've
lost a job recently and am not in searching.)

[Victor Porton]

Niels Diepeveen wrote:
> Op 31-12-13 05:25, William Elliot schreef:
>
>> To the point, if F is a free filter, is there a minimal collection
>> K, of ultrafilters for which F = /\K? I would think K would have at
>> most 2^|S| ultrafilters.
>>
>
> If F is the cofinite filter on an infinite set S and K a collection of
> ultrafilters such that /\K = F, then for every G in K also
> /\(K \ {G}) = F.
>
> Proof: Recall that if G is an ultrafilter on S, then for every
> A subset S either A in G or S\A in G.
> It follows that /\K is the cofinite filter iff for every
> infinite A subset S there is a G in K such that A in G.

Please elaborate why "/\K is the cofinite filter iff ...".

> Now suppose that there is an A for which there is exactly
> one G in K for which A in G. A is infinite, so it can be partitioned
> into infinite A_1 and A_2.
> Since A_1 and A_2 are disjoint, they cannot both be in G, but since
> they are subsets of A, neither can be in an other element of K. This
> contradicts the assumption that /\K is the cofinite filter.

--

[Victor Porton]

Karl Kronenfeld says the proof of Niels Diepeveen is definitely incorrect:

http://math.stackexchange.com/questions/625236/help-to-understand-a-proof-about-filters

[Victor Porton]

Karl Kronenfeld now have said that the proof is nevertheless correct.

I think, we will check the full proof tomorrow.

[Niels Diepeveen]

Op 02-01-14 20:04, Victor Porton schreef:

> Niels Diepeveen wrote:
>> Op 31-12-13 05:25, William Elliot schreef:
>>
>>> To the point, if F is a free filter, is there a minimal collection
>>> K, of ultrafilters for which F = /\K? I would think K would have at
>>> most 2^|S| ultrafilters.
>>>
>>
>> If F is the cofinite filter on an infinite set S and K a collection of
>> ultrafilters such that /\K = F, then for every G in K also
>> /\(K \ {G}) = F.
>>
>> Proof: Recall that if G is an ultrafilter on S, then for every
>> A subset S either A in G or S\A in G.
>> It follows that /\K is the cofinite filter iff for every
>> infinite A subset S there is a G in K such that A in G.
>
> Please elaborate why "/\K is the cofinite filter iff ...".

I should have mentioned that since /\K is a free filter, I am presuming
here that K is a collection of free ultrafilters.
Then /\K contains the cofinite sets, since they are in every free
filter and with that in mind.
With respect to the other subsets of S, the following are obviously
equivalent:
- /\K contains no set with an infinite complement
- for every A with infinite complement there is a G in K such that
A not in G
- for every infinite B subset S, there is a G in K such that S\B
not in G
- for every infinite B subset S, there is a G in K such that B in G
The equivalence of the last two steps depends on the fact that the
elements of K are ultrafilters.

[William Elliot]

On Fri, 3 Jan 2014, Victor Porton wrote:
> Niels Diepeveen wrote:
> > Op 31-12-13 05:25, William Elliot schreef:
> >
> >> To the point, if F is a free filter, is there a minimal collection
> >> K, of ultrafilters for which F = /\K? I would think K would have at
> >> most 2^|S| ultrafilters.
> >>
> >
> > If F is the cofinite filter on an infinite set S and K a collection of
> > ultrafilters such that /\K = F, then for every G in K also
> > /\(K \ {G}) = F.
> >
> > Proof: Recall that if G is an ultrafilter on S, then for every
> > A subset S either A in G or S\A in G.

> > It follows that /\K is the cofinite filter iff for every
> > infinite A subset S there is a G in K such that A in G.

As you removed the proposition that you wanted to prove,
I didn't answer when you asked but here where's the proposition
is stated. It's important that the problem be included in the
reply with the question. Otherwise, I'm not able to answer.

Assume F = /\K is the Frechet filter, A is infinite.
and that for all G in K, A not in G. Then for all G in K,
S\A in G. Thus S\A in F = /\K; S\A cofinite and A finite
which it isn't.

Now assume some infinite A with for all G in K, A not in G.
Thus for all G in K, S\A in G; S\A in F = /\K. As A is infinite,
S\A isn't cofinite and F isn't the Fletcher filter.

[William Elliot]

Yes, the proposition that if K subset Fu(S),
F = /\K is the Frechet filter and G in K, then
F = /\(K\{G}) shows there is no minimal collection
K of ultrafilter. That however is not my point.

My point is that your proof is incomplete as you
assume there's some infinite A for which there's
an unique G in K with A in B. That is what I'm
asking you to prove, that necessary fact.

[Niels Diepeveen]

Op 03-01-14 04:33, William Elliot schreef:

you mean A in G?

> asking you to prove, that necessary fact.
>

Sorry for being unclear. I am only supposing that there is such an A in
order to prove by contradiction that there isn't.

Then, having established that every infinite A occurs in at least two
elements of K, it is easy to conclude that every infinite A occurs in
at least one element of K \ {G}.

Does that clear it up?

[Victor Porton]

Niels Diepeveen wrote:
> Op 31-12-13 05:25, William Elliot schreef:
>
>> To the point, if F is a free filter, is there a minimal collection
>> K, of ultrafilters for which F = /\K? I would think K would have at
>> most 2^|S| ultrafilters.
>>
>
> If F is the cofinite filter on an infinite set S and K a collection of
> ultrafilters such that /\K = F, then for every G in K also
> /\(K \ {G}) = F.

Sorry, I don't understand your proof. Could you elaborate?

> Proof: Recall that if G is an ultrafilter on S, then for every
> A subset S either A in G or S\A in G.
> It follows that /\K is the cofinite filter iff for every
> infinite A subset S there is a G in K such that A in G.
> Now suppose that there is an A for which there is exactly
> one G in K for which A in G. A is infinite, so it can be partitioned

Why we suppose it? What is proved by contradiction with this supposition?

> into infinite A_1 and A_2.
> Since A_1 and A_2 are disjoint, they cannot both be in G, but since
> they are subsets of A, neither can be in an other element of K. This
> contradicts the assumption that /\K is the cofinite filter.

See also
http://math.stackexchange.com/questions/625236/help-to-understand-a-proof-about-filters

[Niels Diepeveen]

Op 03-01-14 18:13, Victor Porton schreef:

> Niels Diepeveen wrote:
>> Op 31-12-13 05:25, William Elliot schreef:
>>
>>> To the point, if F is a free filter, is there a minimal collection
>>> K, of ultrafilters for which F = /\K? I would think K would have at
>>> most 2^|S| ultrafilters.
>>>
>>
>> If F is the cofinite filter on an infinite set S and K a collection of
>> ultrafilters such that /\K = F, then for every G in K also
>> /\(K \ {G}) = F.
>
> Sorry, I don't understand your proof. Could you elaborate?

Here is a breakdown of the line of reasoning. Assume that K is
a collection of free ultrafilters on S. Then

/\K is the cofinite filter

==>
every infinite subset of S occurs in at least one element of K
==>
every infinite subset of S occurs in at least two elements of K
==>
every infinite subset of S occurs in at least one element of K \ {G}
(regardless of what G is)
==>
/\(K \ {G}) is the cofinite filter

I have already explained the first and the last step in a previous post.
I should hope that I do not have to explain the third step:)

>
>> Proof: Recall that if G is an ultrafilter on S, then for every
>> A subset S either A in G or S\A in G.
>> It follows that /\K is the cofinite filter iff for every
>> infinite A subset S there is a G in K such that A in G.
>> Now suppose that there is an A for which there is exactly
>> one G in K for which A in G. A is infinite, so it can be partitioned
>
> Why we suppose it? What is proved by contradiction with this supposition?

After showing every infinite set occurs at least once, we show that it
it occurs at least twice by demonstrating that it cannot occur
exactly once.

>
>> into infinite A_1 and A_2.
>> Since A_1 and A_2 are disjoint, they cannot both be in G, but since
>> they are subsets of A, neither can be in an other element of K. This
>> contradicts the assumption that /\K is the cofinite filter.
>
> See also
> http://math.stackexchange.com/questions/625236/help-to-understand-a-proof-about-filters
>

I have seen it an commented on it, but I don't think it is helpful to
fragment the discussion in this way.

[William Elliot]

On Thu, 2 Jan 2014, Victor Porton wrote:
> It seems that from your statement it follows that there are no copartition
> of the cofinite filter, and one my (relatively) long standing conjectures is
> now solved.

He has show my conjecture that there's a minimal collection
of ultrafilters whose intersection is the Fechet filter, is
false. Yes, that indeed, directly shows there is no weak
copartition of the Frechet filter.

However there's a simple copartition of any filter by
ultrafilters and a strong copartition of principal filters
by principal ultrafilters. Does that adequately summarize
the subtopic of copartitiioning filters?

> I will check it after checking for a PHP or Perl programming job ads. (I've
> lost a job recently and am not in searching.)

For what sort of wark are you looking?

Upon closer reading of your chapter on flitrators
it appears a mess which to understand the definitions
requires casting them into known terms so that they
make sense and make it much easier to follow and read
the chapter.

The concepts involed are very tenuous, more so than
the not very useful concepts of semiopen, preopen
alpha open, and semi-preopen or beta open.

I'd call them partial infinum preserved embeddings
and the like. The distinctions are real but are they
of any use?

[Victor Porton]

William Elliot wrote:

> On Thu, 2 Jan 2014, Victor Porton wrote:
>> It seems that from your statement it follows that there are no
>> copartition of the cofinite filter, and one my (relatively) long standing
>> conjectures is now solved.
>
> He has show my conjecture that there's a minimal collection
> of ultrafilters whose intersection is the Fechet filter, is
> false. Yes, that indeed, directly shows there is no weak
> copartition of the Frechet filter.
>
> However there's a simple copartition of any filter by
> ultrafilters and a strong copartition of principal filters
> by principal ultrafilters. Does that adequately summarize
> the subtopic of copartitiioning filters?

What? You've above confirmed that it is false for Frechet filter and now say
that it is true for "any" filter. Don't you contradict to yourself?

>> I will check it after checking for a PHP or Perl programming job ads.
>> (I've lost a job recently and am not in searching.)
>
> For what sort of wark are you looking?

PHP or Perl programming job.

[Victor Porton]

Niels Diepeveen wrote:

> Op 03-01-14 18:13, Victor Porton schreef:
>> Niels Diepeveen wrote:
>>> Op 31-12-13 05:25, William Elliot schreef:
>>>
>>>> To the point, if F is a free filter, is there a minimal collection
>>>> K, of ultrafilters for which F = /\K? I would think K would have at
>>>> most 2^|S| ultrafilters.
>>>>
>>>
>>> If F is the cofinite filter on an infinite set S and K a collection of
>>> ultrafilters such that /\K = F, then for every G in K also
>>> /\(K \ {G}) = F.
>>
>> Sorry, I don't understand your proof. Could you elaborate?
>
> Here is a breakdown of the line of reasoning. Assume that K is
> a collection of free ultrafilters on S. Then
>
> /\K is the cofinite filter
> ==>
> every infinite subset of S occurs in at least one element of K
> ==>

I don't understand this implication.

> every infinite subset of S occurs in at least two elements of K
> ==>
> every infinite subset of S occurs in at least one element of K \ {G}
> (regardless of what G is)
> ==>
> /\(K \ {G}) is the cofinite filter
>
> I have already explained the first and the last step in a previous post.
> I should hope that I do not have to explain the third step:)

I also don't understand why /\K being not cofinite implies existence G in K
such as A in G for every infinite set A.

>>> Proof: Recall that if G is an ultrafilter on S, then for every
>>> A subset S either A in G or S\A in G.
>>> It follows that /\K is the cofinite filter iff for every
>>> infinite A subset S there is a G in K such that A in G.
>>> Now suppose that there is an A for which there is exactly
>>> one G in K for which A in G. A is infinite, so it can be partitioned
>>
>> Why we suppose it? What is proved by contradiction with this supposition?
>
> After showing every infinite set occurs at least once, we show that it
> it occurs at least twice by demonstrating that it cannot occur
> exactly once.
>
>>
>>> into infinite A_1 and A_2.
>>> Since A_1 and A_2 are disjoint, they cannot both be in G, but since
>>> they are subsets of A, neither can be in an other element of K. This
>>> contradicts the assumption that /\K is the cofinite filter.
>>
>> See also
>> http://math.stackexchange.com/questions/625236/help-to-understand-a-proof-about-filters
>>
>
> I have seen it an commented on it, but I don't think it is helpful to
> fragment the discussion in this way.

[Victor Porton]

Oh, I see a proof for this implication (about two infinite sets) here. I yet
don't understand why "neither can be in an other element of K".

>> See also
>> http://math.stackexchange.com/questions/625236/help-to-understand-a-proof-about-filters
>>
>
> I have seen it an commented on it, but I don't think it is helpful to
> fragment the discussion in this way.

[Victor Porton]

I also don't understand why This contradicts the assumption that /\K is the

[Niels Diepeveen]

Op 04-01-14 14:30, Victor Porton schreef:

In retrospect there really is no need to do this by contradiction. Here
is a direct proof of the second step.

Proposition: Let K be a collection of filters on S. If every infinite
subset of S occurs in at least one element of K, then every infinite
subset of S occurs in at least two elements of K.

Proof: Let A be any infinite subset of S. A can be split into two
disjoint infinite subsets A_1 and A_2. By hypothesis there are G_1
and G_2 such that A_1 in G_1 in K and A_2 in G_2 in K. Since
A_1 and A_2 are disjoint, G_1 and G_2 are distinct.
Because A_1 subset A subset S and A_1 in G_1 we also have
A in G_1, and similarly A in G_2.
Since A was arbitrary, this proves the proposition.

[William Elliot]

On Sat, 4 Jan 2014, Victor Porton wrote:
> William Elliot wrote:
> > On Thu, 2 Jan 2014, Victor Porton wrote:

> >> It seems that from your statement it follows that there are no
> >> copartition of the cofinite filter, and one my (relatively) long standing
> >> conjectures is now solved.
> >
> > He has show my conjecture that there's a minimal collection
> > of ultrafilters whose intersection is the Fechet filter, is
> > false. Yes, that indeed, directly shows there is no weak
> > copartition of the Frechet filter.
> >
> > However there's a simple copartition of any filter by
> > ultrafilters and a strong copartition of principal filters
> > by principal ultrafilters. Does that adequately summarize
> > the subtopic of copartitiioning filters?

> What? You've above confirmed that it is false for Frechet filter and now say
> that it is true for "any" filter. Don't you contradict to yourself?

No, simple partitions are the usual partitions of pairwise disjoint elements I
forget the stange term you used for them. The terms weak and strong
partitions are easily remembered.