Is my open problem about equality of filters difficult?

[Victor Porton]

How do you think, is this open problem difficult?

http://www.openproblemgarden.org/op/pseudodifference_of_filter_objects

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

[William Elliot]

> How do you think, is this open problem difficult?
> http://www.openproblemgarden.org/op/pseudodifference_of_filter_objects
>

Yes, just reading it is difficult.

Just what, in plain text is the difference between a filter and a filter
object? A filter object is a filter that's a member of an (partially)
ordered collection of filters?

What does bigcap^{mathfrak{F}}, bigcup^{mathfrak{F}},
cap^{mathfrak{F}} and cup^{mathfrak{F}} mean?

> Pseudodifference of filter objects

> Let U is a set. A filter mathcal{F} (on U ) is a
> non-empty set of subsets of U such that A, B in mathcal{F}
> Leftrightarrow A cap B in mathcal{F} . Note that unlike some
> other authors I do not require emptyset notin mathcal{F} .

> I will call the set of filter objects the set of filters ordered
> reverse to set theoretic inclusion of filters, with principal filters
> equated to the corresponding sets. I will denote (operatorname{up} a)
> the filter corresponding to a filter object a . I will denote the
> set of filter objects (on U ) as mathfrak{F} .

> I will denote (operatorname{atoms} a) the set of atomic lattice
> elements under a given lattice element a . If a is a filter
> object, then (operatorname{atoms} a) is essentially the set of
> ultrafilters over a .

> Problem> Which of the following expressions are pairwise equal for
> all a, b in mathfrak{F} for each set U ? (If some are not
> equal, provide counter-examples.)

> 1. bigcap^{mathfrak{F}} { z in mathfrak{F} | a subseteq
> . . b cup^{mathfrak{F}} z } ;

> 2. bigcup^{mathfrak{F}} { z in mathfrak{F} | z subseteq
> . . a wedge z cap^{mathfrak{F}} b = emptyset } ;

> 3. bigcup^{mathfrak{F}} (operatorname{atoms} a setminus
> . . operatorname{atoms} b) ;

> 4. bigcup^{mathfrak{F}} { a cap^{mathfrak{F}} (Usetminus
> . . B) | B in operatorname{up} b }.

[Victor Porton]

William Elliot wrote:

>> How do you think, is this open problem difficult?
>> http://www.openproblemgarden.org/op/pseudodifference_of_filter_objects
>>
>
> Yes, just reading it is difficult.
>
> Just what, in plain text is the difference between a filter and a filter
> object? A filter object is a filter that's a member of an (partially)
> ordered collection of filters?
>
> What does bigcap^{mathfrak{F}}, bigcup^{mathfrak{F}},
> cap^{mathfrak{F}} and cup^{mathfrak{F}} mean?

The order \subseteq is the reverse for filter objects, rather than for
filter:
$a\subseteq b \Leftrightarrow
\operatorname{up}a\supseteq\operatorname{up}b$.

bigcap^{mathfrak{F}} is the infimum on the set of filter objects (that is
supremum on the set of filters).

This is an old notation. I changed to write without the concept of filter
objects in my book:

http://www.mathematics21.org/algebraic-general-topology.html

[William Elliot]

What does wedge mean in

[Victor Porton]

\wedge is logical and.

[William Elliot]

On Tue, 21 May 2013, Victor Porton wrote:
> William Elliot wrote:
>
> > Just what, in plain text is the difference between a filter and a filter
> > object? A filter object is a filter that's a member of an (partially)
> > ordered collection of filters?
> >
> > What does bigcap^{mathfrak{F}}, bigcup^{mathfrak{F}},
> > cap^{mathfrak{F}} and cup^{mathfrak{F}} mean?
>
> The order \subseteq is the reverse for filter objects, rather than for
> filter: $a\subseteq b \Leftrightarrow
> \operatorname{up}a\supseteq\operatorname{up}b$.
>
> bigcap^{mathfrak{F}} is the infimum on the set of filter objects (that is
> supremum on the set of filters).
>
> This is an old notation. I changed to write without the concept of filter
> objects in my book:

I should think so as it seem superfluous.

> >> Problem> Which of the following expressions are pairwise equal for
> >> all a, b in mathfrak{F} for each set U ?

Let F(S) be the set of filters over S and the empty set.
For all A,B in F(S) does

/\{ F in F(S) : A subset B \/ F }
=
\/{ F in F(S) | F subset A, F /\ B = emptyset }?

I pressume you don't allow empty filters.
Since you allow P(S) in F(S), let B = P(S) and the question now is does

/\{ F in F(S) : A subset P(S) }
=
\/{ F in F(S) | F subset A, F /\ P(S) = emptyset }?

Does /\F(S) = /\{ F | F in F(S) } = \/emptyset = emptyset?

Is /\F(S) empty? No. Since for all F in F(S), S in F
we have S in /\F(S). So 1 /= 2.

[Victor Porton]

No, \/emptyset = P(S).

This is an error and I don't read further.

> Is /\F(S) empty? No. Since for all F in F(S), S in F
> we have S in /\F(S). So 1 /= 2.
>
>> >> 1. bigcap^{mathfrak{F}} { z in mathfrak{F} | a subseteq
>> >> . . b cup^{mathfrak{F}} z } ;
>> >
>> >> 2. bigcup^{mathfrak{F}} { z in mathfrak{F} | z subseteq
>> >> . . a wedge z cap^{mathfrak{F}} b = emptyset } ;

[William Elliot]

No, /\emptyset = P(S)

\/emptyset = emptyset. Proof.

If x in \/emptyset, then there's some set A in emptyset with x in A.
But since no set can be in emptyset, that there's some A is emptyset,
is a contradictionn. Since assuming x in \/emptyset leads to a
contradiction, it must be that \/emptyset is empty.

> This is an error and I don't read further.

It is not. So read on, I've shown 1 and 2 aren't always equal.

[Victor Porton]

You have confused \/ on the set of filter objects (\/^{F}) with \/ on the
set of subsets of S.

In this content \/emptyset is the supremum of an empty set of filter objects
and it is the minimal filter object (corresponding to the filter P(S)).