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Sep 17, 2008, 5:29:33 AM9/17/08

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This is a synopsis of a pair of papers by

Victor Porton <por...@narod.ru>

Victor Porton <por...@narod.ru>

regarding a (partial) order on binary relations f subset XxY

between two sets X and Y, that is an order on P(XxY) coarser than

the subset order.

Definition.

domain f = dom f = { x | some y with (x,y) in f } = p1(f)

image f = img f = { y | some x with (x,y) in f } = p2(f)

restriction f to A = f|A = { (x,y) in f | x in A } = f /\ AxS

(/\ \/, intersection union)

Propositions.

f subset dom f x rng f; dom f subset X; rng f subset Y

f|dom f = f. Proof: f|dom f = f /\ (dom f)xY = f

since f subset dom f x rng f subset dom f x Y

(f|A)|B = f|(A /\ B). Proof: (f|A)|B = f|A /\ BxY = f /\ AxY /\ BxY

= f /\ (A /\ B)xY = f|(A /\ B)

dom f|A = dom f /\ A. Proof

x in dom f|A iff some y with (x,y) in f|A

iff some y with (x,y) in f, x in A iff x in dom f /\ A

Definition

f <= g when f = g|dom f

Propositions

f <= g ==> f subset g, dom f subset dom g

<= is (partial) order (proof follows each property)

f <= f. f = f|dom f

f <= g, g <= f ==> f = g. f subset g; g subset f

f <= g, g <= h ==> f <= h. dom f subset dom g

f = g|dom f = (h|dom g)|dom f = h|(dom g /\ dom f) = h|dom f

Definition

The function induced by f is the function

f*:dom f -> P(Y), x -> { y | (x,y) in f }

Propositions

dom f = dom f*

f = { (x,y) | x in dom f*, y in f*(x) }. Proof.

(x,y) in f iff x in dom f, y in f*(x) iff x in dom f*, y in f*(x)

f = g iff f* = g*

(f|A)* = f* /\ AxP(Y) = f*|A

f <= g iff f* <= g* iff f* subset g*

Proof. f <= g iff f = g|dom f iff f* = (g|dom f)*

iff f* = g*|dom f iff f* subset g*

f <= g iff for all x in dom f, f*(x) = g*(x)

Let R = P(XxY), R* = { F:A -> P(Y) | A subset X }

F,G in R* ==> F /\ G in R

If F in R* then f = { (x,y) | x in dom f, y in F(x) } in R and F = f*

h:(R,<=) -> (R*,subset) is order isomorphism

(R,<=) is a down lattice or meet semi-lattice

f* /\ g* = inf f*, g*; h^-1(f* /\ g*) = inf f,g

-- conditions equivalent to f <= g

f <= g iff some A with f = g|A. Proof left to right.

If f = g|A, then dom f = dom g|A = dom g /\ A

g|dom f = g|(dom g /\ A) = (g|dom g)|A = g|A = f

I fail see the significance for the remaining, final portion of the

author's work.

Notation. f o g = fg = { (x,y) | some t with (x,t) in f, (t,y) in g }

Note. This composition for relations is

the reverse of composition for functions.

f^-1 g|dom f = f^-1 g. Proof: (x,y) in f^-1 g|dom f

iff some t with (t,x) in f, (t,y) in g|dom f

iff some t in dom f with (t,x) in f, (t,y) in g

iff some t with (t,x) in f, (t,y) in g

iff (x,y) in f^-1 g

Thus f <= g ==> f^-1 f = f^-1 g

f^-1 f = f^-1 g iff f^-1 f = g^-1 f. Proof left to right.

f^-1 f = (f^-1 f)^-1 = (f^-1 g)^-1 = g^-1 f

(1) f^-1 f = f^-1 g, f subset g

(2) f^-1 f = g^-1 f, f subset g

(3) f^-1 f = f^-1 g, dom f subset dom g

(4) f^-1 f = g^-1 f, dom f subset dom g

In conclusion

f <= g ==> (1) ==> (2) ==> (3) ==> (4)

The author claims (4) ==> f <= g.

-- This short section notes to be skipped

f^-1 f = f^-1 g ==> img f subset img g. Proof:

if y in img f: some x with (x,y) in f; (y,y) in f^-1 f

some t with (t,x) in f, (t,y) in g; t in img g

function f ==> f^-1 f = { (y,y) | y in img f }

(x,y) in f^-1 f iff some t with (t,x), (t,y) in f

iff some t with (t,x) in f, x = y iff x = y in img f

-- To continue,

the author claims (4) ==> f <= g.

He uses a definition for composition of relations that is the reverse of

the definition I've chosen. He also has reversed f and g. To continue

with his notation, he claims

gg^-1 = gf^-1, dom g subset dom f implies g <= f

He claims when dom g subset dom f, that

gg^-1 = gf^-1

is equivalent to

g* (g*)^-1 = g* (f*)^-1

Actually only implication is needed. How is that substantiated?

He states

g* g*^-1 = l(img g)

where l is the identity function. I can make no sense of that.

What I can show is

g* g*^-1 = { (y,y) | y in img g }

Thus if x in dom g*, f*(x) = g* g*^-1 f*(x)

(since img g* subset img f* as hinted at in short section)

= g* (f*)^-1 f*(x) = g*(x). This shows g <= f as show above.

----

Theorem about limiting binary relations

Theorem. Let f and g are binary relations. Then the following

statements are equivalent:

1. g = f|[dom g].

2. Exists a set K such that g = f|[K].

3. gg-1 = fg-1 and g is a subset of f.

4. gg-1 = gf^-1 and g is a subset of f.

5. gg-1 = fg-1 and dom g is a subset of dom f.

6. gg-1 = gf^-1 and dom g is a subset of dom f.

An other, a little less convenient for usage but much easier to

remember (mnemonic), equivalent form of this theorem is:

Theorem (mnemonic form). Given that f and g are binary relations and

dom g is a subset of dom f, right "multiplying" (composing) both sides

of the equation g = f|[dom g] with g-1 produces a formula equivalent

to this equation.

I have [9]stated this theorem as a conjecture in this weblog post. Now

I have proved it. (The proof is below.)

Note that qp denotes the composition of binary relations p and q.

Lemma

First we will prove the following lemma:

Lemma. If gg-1 = fg-1 and dom g is a subset of dom f then f|[dom g] =

g for any binary relations f and g.

Proof. Let gg-1 = fg-1.

For any relation p I will denote p* the corresponding function mapping

an X to a set of Y. That is p*(x) = {y : (x, y) in p}. Note that dom

p* = dom p.

I will denote L the multivalued function (relation) which maps a set

to its elements (that is the reverse of set membership relation).

Obviously p = Lp* and p = q <=> p* = q* for any relations p and q.

Our equation can be equivalently rewritten as

Lg* (Lg*)-1 = Lf* (Lg*)-1; Lg* g*-1 L-1 = Lf* g*-1 L-1;

what accordingly noted above is equivalent to

g*g*-1 = f*g*-1.

By well known properties of monovalued functions, we have

g* g*-1 = I[im g*]

g* g*^-1 = { (y,y) | y in img g* }

(I[im g*] denotes the identity relation on the set im g*); so

f*g*-1 = I[im g*],

that is for any y in im g* we have f*g*-1y = y.

So for any x in dom g* we have f*g*-1g*x = g*x. Because g*-1g*x

contains x and x in dom f* (dom f* is a superset of dom g*), we have

f*x = g*x.

So f*|[dom g*] = g* what is the same as f|[dom g] = g. End of proof.

Proof of the main theorem

Now we can easily prove the main theorem. First we will prove its

mnemonic form (repeated below):

Theorem (mnemonic form). Given that f and g are binary relations and

dom g is a subset of dom f, right multiplying both sides of the

equation g = f|[dom g] with g-1 produces a formula equivalent to this

equation.

Proof of the main theorem (mnemonic form).

Multiplying both sides of the equation g = f|[dom g] with g-1 produces

the formula gg-1 = fg-1. We need to prove that this formula is

equivalent to g = f|[dom g] (under the condition that dom g is a

subset of dom f).

Direct implication is obvious. Reverse implication is already proved

in the lemma (that dom g is a subset of dom f is taken into account).

End of proof.

Now let reprise the full form of the main theorem again and prove it:

Theorem. Let f and g are binary relations. Then the following

statements are equivalent:

1. g = f|[dom g].

2. Exists a set K such that g = f|[K].

3. gg-1 = fg-1 and g is a subset of f.

4. gg-1 = gf^-1 and g is a subset of f.

5. gg-1 = fg-1 and dom g is a subset of dom f.

6. gg-1 = gf^-1 and dom g is a subset of dom f.

Proof of the main theorem (full form).

Equivalences (3)<=>(4) and (5)<=>(6) can be proved by inverting the

formulas taking in account that (qp)-1 = p-1q-1 for any binary

relations p and q.

Implications (3)=>(5) and (4)=>(6) are obvious.

(5)=>(1) is exactly the lemma proved above.

(1)=>(3) is obvious (right multiply both sides of the equality (1)

with g-1).

(1)=>(2) is obvious. (Take K = dom g.)

To prove (2)=>(1) let assume that g = f|[K]. Then dom g is

intersection of dom f and the set K; from this follows g = f|[dom g].

From proved implications follows that all six statements are

equivalent. End of proof.

----

Sep 17, 2008, 7:03:54 AM9/17/08

to

On Sep 17, 12:29 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> This is a synopsis of a pair of papers by

> Victor Porton <por...@narod.ru>

>

> regarding a (partial) order on binary relations f subset XxY

> between two sets X and Y, that is an order on P(XxY) coarser than

> the subset order.

> This is a synopsis of a pair of papers by

> Victor Porton <por...@narod.ru>

>

> regarding a (partial) order on binary relations f subset XxY

> between two sets X and Y, that is an order on P(XxY) coarser than

> the subset order.

http://www.mathematics21.org/misc-math.html

^^^

William Elliot, I don't understand your purpose of re-publishing (with

a little additional info) info from my site it the newsgroup.

(However thanks for your attention to my work.)

Also note that you have violated the implied copyright of my work re-

publishing its large portions. (However I'm glad, not irritated, and

am not going to sue you.)

For the future I explicitly allow you to quote me (in reasonable

limits).

Also: there are no l[...] but I[...] in my texts. You messed the

letters L and i.

Note that my main works are not "Misc math" but Algebraic General

Topology:

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

> Notation. f o g = fg = { (x,y) | some t with (x,t) in f, (t,y) in g }

>

> Note. This composition for relations is

> the reverse of composition for functions.

You have messed the order (should be first g then f):

f o g = fg = { (x,y) | some t with (x,t) in g, (t,y) in f }

Glad to hear you further, William Elliot.

Sep 17, 2008, 7:14:05 AM9/17/08

to

On Sep 17, 2:03 pm, Victor Porton <por...@narod.ru> wrote:

> On Sep 17, 12:29 pm, William Elliot <ma...@hevanet.remove.com> wrote:

>

> > This is a synopsis of a pair of papers by

> > Victor Porton <por...@narod.ru>

>

> > regarding a (partial) order on binary relations f subset XxY

> > between two sets X and Y, that is an order on P(XxY) coarser than

> > the subset order.

>

> http://www.mathematics21.org/misc-math.html

> ^^^

> On Sep 17, 12:29 pm, William Elliot <ma...@hevanet.remove.com> wrote:

>

> > This is a synopsis of a pair of papers by

> > Victor Porton <por...@narod.ru>

>

> > regarding a (partial) order on binary relations f subset XxY

> > between two sets X and Y, that is an order on P(XxY) coarser than

> > the subset order.

>

> http://www.mathematics21.org/misc-math.html

> ^^^

Does it make sense to attempt to publish "Theorem about limiting

binary relations" (http://www.mathematics21.org/misc/limiting-binary-

relations-theorem.html) in a referated journal?

It seems being too simple theorem for publishing but indeed I suspect

that it may be a new result. To publish it officially or just to leave

as it (having an unofficial Web page about this theorem)?

Anyway I expect to use this theorem as a special case for proving

certain conjectures in my research on Algebraic General Topology

(http://www.mathematics21.org/algebraic-general-topology.html). So it

could anyway be published by me in a (future) book.

Sep 18, 2008, 3:36:27 AM9/18/08

to

Most of this appears to me to be similar to exercises in relation

algebras

and cylindric algebras. Before submitting your work for publication,

you

might look at some basic texts in universal algebra that deal with

relation

algebras. I recall only advanced work that Alfred Tarski did with

cylindric

algebras, but he might have written (or referred to) a more basic text

that

covers some of the same material.

Gerhard Paseman, 2008.09.18

Sep 18, 2008, 4:58:59 AM9/18/08

to

On Wed, 17 Sep 2008, Victor Porton wrote:

> > On Sep 17, 12:29 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> >

> > > This is a synopsis of a pair of papers by

> > > Victor Porton <por...@narod.ru>

> >

> > > regarding a (partial) order on binary relations f subset XxY

> > > between two sets X and Y, that is an order on P(XxY) coarser than

> > > the subset order.

> >

> > http://www.mathematics21.org/misc-math.html

>

> > On Sep 17, 12:29 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> >

> > > This is a synopsis of a pair of papers by

> > > Victor Porton <por...@narod.ru>

> >

> > > regarding a (partial) order on binary relations f subset XxY

> > > between two sets X and Y, that is an order on P(XxY) coarser than

> > > the subset order.

> >

> > http://www.mathematics21.org/misc-math.html

>

> Does it make sense to attempt to publish "Theorem about limiting

> binary relations" (http://www.mathematics21.org/misc/limiting-binary-

> relations-theorem.html) in a referated journal?

>

I have found an astonishingly simple and direct way of proving the major> binary relations" (http://www.mathematics21.org/misc/limiting-binary-

> relations-theorem.html) in a referated journal?

>

lemma of that paper.

First I show for all f,g

x in dom f /\ dom g ==> f*(x) x g*(x) subset gf^-1

Hence immediately

x in dom f /\ dom g ==> f*(x) x g*(x) /\ gf^-1 = f*(x) x g*(x)

Consequently

ff^-1 = gf^-1 ==> for all x in dom f /\ dom g, f*(x) = g*(x)

which shows

ff^-1 = gf^-1, dom f subset dom g ==> for all x in dom f, f*(x) = g*(x)

and finally

ff^-1 = gf^-1, dom f subset dom g ==> f <= g

Sep 19, 2008, 3:04:59 AM9/19/08

to

On Thu, 18 Sep 2008, William Elliot wrote:

> On Wed, 17 Sep 2008, Victor Porton wrote:

> > > On Sep 17, 12:29 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> > >

> > > > This is a synopsis of a pair of papers by

> > > > Victor Porton <por...@narod.ru>

> > >

> > > > regarding a (partial) order on binary relations f subset XxY

> > > > between two sets X and Y, that is an order on P(XxY) coarser than

> > > > the subset order.

> > >

> > > http://www.mathematics21.org/misc-math.html

> >

> > Does it make sense to attempt to publish "Theorem about limiting

> > binary relations" (http://www.mathematics21.org/misc/limiting-binary-

> > relations-theorem.html) in a referated journal?

> >

> I have found an astonishingly simple and direct way of proving the major

> lemma of that paper.

>

> First I show for all f,g

> x in dom f /\ dom g ==> f*(x) x g*(x) subset gf^-1

>

> Hence immediately

> x in dom f /\ dom g ==> f*(x) x g*(x) /\ gf^-1 = f*(x) x g*(x)

>

These two are correct.> On Wed, 17 Sep 2008, Victor Porton wrote:

> > > On Sep 17, 12:29 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> > >

> > > > This is a synopsis of a pair of papers by

> > > > Victor Porton <por...@narod.ru>

> > >

> > > > regarding a (partial) order on binary relations f subset XxY

> > > > between two sets X and Y, that is an order on P(XxY) coarser than

> > > > the subset order.

> > >

> > > http://www.mathematics21.org/misc-math.html

> >

> > Does it make sense to attempt to publish "Theorem about limiting

> > binary relations" (http://www.mathematics21.org/misc/limiting-binary-

> > relations-theorem.html) in a referated journal?

> >

> I have found an astonishingly simple and direct way of proving the major

> lemma of that paper.

>

> First I show for all f,g

> x in dom f /\ dom g ==> f*(x) x g*(x) subset gf^-1

>

> Hence immediately

> x in dom f /\ dom g ==> f*(x) x g*(x) /\ gf^-1 = f*(x) x g*(x)

>

> Consequently

> ff^-1 = gf^-1 ==> for all x in dom f /\ dom g, f*(x) = g*(x)

>

I have overstated my hand and retract this claim.

Sep 19, 2008, 5:40:53 AM9/19/08

to

On Sep 18, 11:58 am, William Elliot <ma...@hevanet.remove.com> wrote:

> I have found an astonishingly simple and direct way of proving the major

> lemma of that paper.

>

> First I show for all f,g

> x in dom f /\ dom g ==> f*(x) x g*(x) subset gf^-1

>

> Hence immediately

> x in dom f /\ dom g ==> f*(x) x g*(x) /\ gf^-1 = f*(x) x g*(x)

>

> Consequently

> ff^-1 = gf^-1 ==> for all x in dom f /\ dom g, f*(x) = g*(x)

>

> which shows

> ff^-1 = gf^-1, dom f subset dom g ==> for all x in dom f, f*(x) = g*(x)

> and finally

> ff^-1 = gf^-1, dom f subset dom g ==> f <= g

> I have found an astonishingly simple and direct way of proving the major

> lemma of that paper.

>

> First I show for all f,g

> x in dom f /\ dom g ==> f*(x) x g*(x) subset gf^-1

>

> Hence immediately

> x in dom f /\ dom g ==> f*(x) x g*(x) /\ gf^-1 = f*(x) x g*(x)

>

> Consequently

> ff^-1 = gf^-1 ==> for all x in dom f /\ dom g, f*(x) = g*(x)

>

> which shows

> ff^-1 = gf^-1, dom f subset dom g ==> for all x in dom f, f*(x) = g*(x)

> and finally

> ff^-1 = gf^-1, dom f subset dom g ==> f <= g

"f*(x) x g*(x) subset gf^-1"

What you mean by "x"?

Sep 19, 2008, 6:43:12 AM9/19/08

to

The cross product

. . AxB = { (x,y) | x in A, y in B }

However, if you read my other recent post, you will see that I retracked

that proof. Is this a counter example to the lemma? N is positive

integers.

f = { (n,m) in NxN | n <= m }

g = NxN; f subset g; dom f = dom g

ff^-1 = NxN = gf^-1

Sep 19, 2008, 7:36:19 AM9/19/08

to

First, you have messed f and g. I use f where you use g and vice

versa.

ff^-1 = {(x,y) | exists t in N such that (t,x) in f and (t,y) in f} =

{(x,y) | exists t in N such that t<=x and t<=y} =

{(x,y) | x, y in N} = NxN.

gf^-1 = {(x,y) | exists t in N such (t,x) in f and (t,y) in g} =

{(x,y) | exists t in N such t<=x, y in N} =

{(x,y) | x, y in N} = NxN

Hmm... yes, this seems to be a counterexample to my lemma... ~-)

Where is the error then? I will look at this when will have a free

time.

Now, you may discuss where is the error in this newsgroup.

http://www.mathematics21.org/misc/limiting-binary-relations-theorem.html

Sep 19, 2008, 1:48:34 PM9/19/08

to

It seems that mistakeous statement at

http://www.mathematics21.org/misc/limiting-binary-relations-theorem.html

is claiming

"what accordingly noted above is equivalent to g*g*-1 = f*g*-1."

So counterexample by William Elliot wins. My mistake.

Now the mistake is noted at

http://www.mathematics21.org/misc/limiting-binary-relations-theorem.html

This does not undetermines my other articles however.

--

Victor Porton - http://www.mathematics21.org

Sep 20, 2008, 3:13:00 AM9/20/08

to

On Fri, 19 Sep 2008, Victor Porton wrote:

> On Sep 19, 2:36 pm, Victor Porton <por...@narod.ru> wrote:

> > On Sep 19, 1:43 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> > > However, if you read my other recent post, you will see that I retracked

> > > that proof. Is this a counter example to the lemma? N is positive

> > > integers.

> >

> > > f = { (n,m) in NxN | n <= m }

> > > g = NxN; f subset g; dom f = dom g

> >

> > > ff^-1 = NxN = gf^-1

> >

> > First, you have messed f and g. I use f where you use g and vice

> > versa.

> >

> > ff^-1 = {(x,y) | exists t in N such that (t,x) in f and (t,y) in f} =

> > {(x,y) | exists t in N such that t<=x and t<=y} =

> > {(x,y) | x, y in N} = NxN.

> >

> > gf^-1 = {(x,y) | exists t in N such (t,x) in f and (t,y) in g} =

> > {(x,y) | exists t in N such t<=x, y in N} =

> > {(x,y) | x, y in N} = NxN

> >

> > Hmm... yes, this seems to be a counterexample to my lemma... ~-)

> > On Sep 19, 1:43 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> > > However, if you read my other recent post, you will see that I retracked

> > > that proof. Is this a counter example to the lemma? N is positive

> > > integers.

> >

> > > f = { (n,m) in NxN | n <= m }

> > > g = NxN; f subset g; dom f = dom g

> >

> > > ff^-1 = NxN = gf^-1

> >

> > First, you have messed f and g. I use f where you use g and vice

> > versa.

> >

> > ff^-1 = {(x,y) | exists t in N such that (t,x) in f and (t,y) in f} =

> > {(x,y) | exists t in N such that t<=x and t<=y} =

> > {(x,y) | x, y in N} = NxN.

> >

> > gf^-1 = {(x,y) | exists t in N such (t,x) in f and (t,y) in g} =

> > {(x,y) | exists t in N such t<=x, y in N} =

> > {(x,y) | x, y in N} = NxN

> >

> > Hmm... yes, this seems to be a counterexample to my lemma... ~-)

> It seems that mistakeous statement at

> http://www.mathematics21.org/misc/limiting-binary-relations-theorem.html

> is claiming "what accordingly noted above is equivalent to g*g*-1 =

> f*g*-1."

>

Correct.

> So counterexample by William Elliot wins. My mistake.

>

> Now the mistake is noted at

> http://www.mathematics21.org/misc/limiting-binary-relations-theorem.html

>

Here's another simpler counter example.

{ (0,0), (0,1), (1,1) }

and

{ (0,0), (0,1), (1,1), (1,0) }

or

{ (0,0), (0,1), (1,1), (1,0), (2,2) }

The latter with (2,2) is to have counter example

where one domain is proper subset of the other.

> This does not undetermines my other articles however.

>

Would you like me to proof read, review another of your papers?

Let's start with Set Theoretic Filters, the lattice of filters.

It is known that the set of filters for a set S is a complete down

lattice, that is a meet complete semi-lattice. Does this paper go

beyond that?

Sep 20, 2008, 5:07:48 AM9/20/08

to

Certainly I would like you to proof read these.

> It is known that the set of filters for a set S is a complete down

> lattice, that is a meet complete semi-lattice. Does this paper go

> beyond that?

The set of filters for a set S is a complete lattice, distributive

lattice, atomistic lattice.

The paper goes far beyond that.

Now in the free time I'm writing the book "Filters on Posets" but it

will probably take yet long time.

Sep 20, 2008, 5:17:08 AM9/20/08

to

On Sep 20, 12:07 pm, Victor Porton <por...@narod.ru> wrote:

> On Sep 20, 10:13 am, William Elliot <ma...@hevanet.remove.com> wrote:

>

> > On Fri, 19 Sep 2008, Victor Porton wrote:

> > > This does not undetermines my other articles however.

>

> > Would you like me to proof read, review another of your papers?

> > Let's start with Set Theoretic Filters, the lattice of filters.

>

> Certainly I would like you to proof read these.

>

> > It is known that the set of filters for a set S is a complete down

> > lattice, that is a meet complete semi-lattice. Does this paper go

> > beyond that?

>

> The set of filters for a set S is a complete lattice, distributive

> lattice, atomistic lattice.

> On Sep 20, 10:13 am, William Elliot <ma...@hevanet.remove.com> wrote:

>

> > On Fri, 19 Sep 2008, Victor Porton wrote:

> > > This does not undetermines my other articles however.

>

> > Would you like me to proof read, review another of your papers?

> > Let's start with Set Theoretic Filters, the lattice of filters.

>

> Certainly I would like you to proof read these.

>

> > It is known that the set of filters for a set S is a complete down

> > lattice, that is a meet complete semi-lattice. Does this paper go

> > beyond that?

>

> The set of filters for a set S is a complete lattice, distributive

> lattice, atomistic lattice.

A correction: Atomistic is the reverse lattice to the lattice of

filters on a set S. I'm not sure about the direct (non-reverse)

lattice whether it is atomistic.

Sep 20, 2008, 6:26:29 AM9/20/08

to

On Sat, 20 Sep 2008, Victor Porton wrote:

> On Sep 20, 12:07 pm, Victor Porton <por...@narod.ru> wrote:

> > On Sep 20, 10:13 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > > On Fri, 19 Sep 2008, Victor Porton wrote:

> >

> > > Would you like me to proof read, review another of your papers?

> > > Let's start with Set Theoretic Filters, the lattice of filters.

> >

> > Certainly I would like you to proof read these.

> >

> > > It is known that the set of filters for a set S is a complete down

> > > lattice, that is a meet complete semi-lattice. Does this paper go

> > > beyond that?

> >

> > The set of filters for a set S is a complete lattice, distributive

> > lattice, atomistic lattice.

>

It is not complete because it has no top element.> On Sep 20, 12:07 pm, Victor Porton <por...@narod.ru> wrote:

> > On Sep 20, 10:13 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > > On Fri, 19 Sep 2008, Victor Porton wrote:

> >

> > > Would you like me to proof read, review another of your papers?

> > > Let's start with Set Theoretic Filters, the lattice of filters.

> >

> > Certainly I would like you to proof read these.

> >

> > > It is known that the set of filters for a set S is a complete down

> > > lattice, that is a meet complete semi-lattice. Does this paper go

> > > beyond that?

> >

> > The set of filters for a set S is a complete lattice, distributive

> > lattice, atomistic lattice.

>

Being complete down lattice, it's a bounded complete lattice

ie, every nonnul bounded set has a sup. It's also chain complete.

What do you mean it's distrubuitive? Consider N and the

principal filters F_A generated by the set A = {0}, {1}, {0,1}

(F_0,1 inf F_0) sup (F_0,1 inf F_1) = F_0,1 sup F_0,1 = F_0,1

/=

F_0,1 inf (F_0 sup F_1) because F_0 sup F_1 doesn't exist.

What are the atoms? They are not the princpals filters because

for b not in A, F_(A \/ {b}) proper subset F_{b}.

> A correction: Atomistic is the reverse lattice to the lattice of

> filters on a set S. I'm not sure about the direct (non-reverse)

> lattice whether it is atomistic.

>

The subset order ideals of S?

Anyway, what formats have you for that paper?

TeX and some other, what did you call it, that's easier

to read in raw file form than TeX? Can you convert them

into the web reable formate used for your two papers on

relations? Anyway as I can't do the pdf, post, email or

put on web a version that I can read and I'll consider what

I can for it.

Sep 20, 2008, 6:47:09 AM9/20/08

to

On Sep 20, 1:26 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> On Sat, 20 Sep 2008, Victor Porton wrote:

> > On Sep 20, 12:07 pm, Victor Porton <por...@narod.ru> wrote:

> > > On Sep 20, 10:13 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > > > On Fri, 19 Sep 2008, Victor Porton wrote:

>

> > > > Would you like me to proof read, review another of your papers?

> > > > Let's start with Set Theoretic Filters, the lattice of filters.

>

> > > Certainly I would like you to proof read these.

>

> > > > It is known that the set of filters for a set S is a complete down

> > > > lattice, that is a meet complete semi-lattice. Does this paper go

> > > > beyond that?

>

> > > The set of filters for a set S is a complete lattice, distributive

> > > lattice, atomistic lattice.

>

> It is not complete because it has no top element.

> Being complete down lattice, it's a bounded complete lattice

> ie, every nonnul bounded set has a sup. It's also chain complete.

> On Sat, 20 Sep 2008, Victor Porton wrote:

> > On Sep 20, 12:07 pm, Victor Porton <por...@narod.ru> wrote:

> > > On Sep 20, 10:13 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > > > On Fri, 19 Sep 2008, Victor Porton wrote:

>

> > > > Would you like me to proof read, review another of your papers?

> > > > Let's start with Set Theoretic Filters, the lattice of filters.

>

> > > Certainly I would like you to proof read these.

>

> > > > It is known that the set of filters for a set S is a complete down

> > > > lattice, that is a meet complete semi-lattice. Does this paper go

> > > > beyond that?

>

> > > The set of filters for a set S is a complete lattice, distributive

> > > lattice, atomistic lattice.

>

> It is not complete because it has no top element.

> Being complete down lattice, it's a bounded complete lattice

> ie, every nonnul bounded set has a sup. It's also chain complete.

Oh, well, I forgot to say that my definition of filter is different

than the definition of filters in some other's works.

See

http://www.mathematics21.org/binaries/set-filters.pdf

for the exact definition.

In my terminology it has top and bottom elements.

> What do you mean it's distrubuitive? Consider N and the

> principal filters F_A generated by the set A = {0}, {1}, {0,1}

>

> (F_0,1 inf F_0) sup (F_0,1 inf F_1) = F_0,1 sup F_0,1 = F_0,1

> /=

> F_0,1 inf (F_0 sup F_1) because F_0 sup F_1 doesn't exist.

The same note about different definition.

I skip about atoms of the lattice of filters. It is probably the same

problem with different definition.

> Anyway, what formats have you for that paper?

> TeX and some other, what did you call it, that's easier

> to read in raw file form than TeX? Can you convert them

> into the web reable formate used for your two papers on

> relations? Anyway as I can't do the pdf, post, email or

> put on web a version that I can read and I'll consider what

> I can for it.

I have sent you by email LaTeX file set-filters.tex. I can't offer you

something better.

Sep 20, 2008, 7:28:33 AM9/20/08

to

On Sat, 20 Sep 2008, Victor Porton wrote:

> > > > On Sep 20, 10:13 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > > > > On Fri, 19 Sep 2008, Victor Porton wrote:

> >

> > > > > Would you like me to proof read, review another of your papers?

> > > > > Let's start with Set Theoretic Filters, the lattice of filters.

> >

> > > > Certainly I would like you to proof read these.

> >

> > > > > It is known that the set of filters for a set S is a complete down

> > > > > lattice, that is a meet complete semi-lattice. Does this paper go

> > > > > beyond that?

> >

> > > > The set of filters for a set S is a complete lattice, distributive

> > > > lattice, atomistic lattice.

> >

> > It is not complete because it has no top element.

> > Being complete down lattice, it's a bounded complete lattice

> > ie, every nonnul bounded set has a sup. It's also chain complete.

>

> Oh, well, I forgot to say that my definition of filter is different

> than the definition of filters in some other's works.

> > > > On Sep 20, 10:13 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > > > > On Fri, 19 Sep 2008, Victor Porton wrote:

> >

> > > > > Would you like me to proof read, review another of your papers?

> > > > > Let's start with Set Theoretic Filters, the lattice of filters.

> >

> > > > Certainly I would like you to proof read these.

> >

> > > > > It is known that the set of filters for a set S is a complete down

> > > > > lattice, that is a meet complete semi-lattice. Does this paper go

> > > > > beyond that?

> >

> > > > The set of filters for a set S is a complete lattice, distributive

> > > > lattice, atomistic lattice.

> >

> > It is not complete because it has no top element.

> > Being complete down lattice, it's a bounded complete lattice

> > ie, every nonnul bounded set has a sup. It's also chain complete.

>

> Oh, well, I forgot to say that my definition of filter is different

> than the definition of filters in some other's works.

Then it needs a different name.

> See

> http://www.mathematics21.org/binaries/set-filters.pdf

> for the exact definition.

>

> In my terminology it has top and bottom elements.

>

> I skip about atoms of the lattice of filters. It is probably the same

> problem with different definition.

>

Atoms for filters of a set S are the filters A_x = {S, S\x}, x in S.

The atoms for the filter F_B are A_x, x not in B.

The atoms for the Flechet filter are A_x, x in S.

Sep 20, 2008, 7:57:08 AM9/20/08

to

On Sat, 20 Sep 2008, William Elliot wrote:

> Atoms for filters of a set S are the filters A_x = {S, S\x}, x in S.

> The atoms for the filter F_B are A_x, x not in B.

> The atoms for the Flechet filter are A_x, x in S.

>

sup{ A_x | x in S } = Flechet filter

Thus the set of filters isn't atomistic.

It's bounded complete, chain complete, with bottom.

The non-negative reals are bounded complete, with bottom

but not chain complete.

On the other hand doesn't F_B = sup { A_x | x not in B } ?

> > I have sent you by email LaTeX file set-filters.tex. I can't offer you

> > something better.

> >

It has not arrived. Did you forget to remove "remove" from my email

address before sending?

----

Sep 20, 2008, 9:57:03 AM9/20/08

to

On Sep 20, 2:28 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> On Sat, 20 Sep 2008, Victor Porton wrote:

> > > > > On Sep 20, 10:13 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > > > > > On Fri, 19 Sep 2008, Victor Porton wrote:

>

> > > > > > Would you like me to proof read, review another of your papers?

> > > > > > Let's start with Set Theoretic Filters, the lattice of filters.

>

> > > > > Certainly I would like you to proof read these.

>

> > > > > > It is known that the set of filters for a set S is a complete down

> > > > > > lattice, that is a meet complete semi-lattice. Does this paper go

> > > > > > beyond that?

>

> > > > > The set of filters for a set S is a complete lattice, distributive

> > > > > lattice, atomistic lattice.

>

> > > It is not complete because it has no top element.

> > > Being complete down lattice, it's a bounded complete lattice

> > > ie, every nonnul bounded set has a sup. It's also chain complete.

>

> > Oh, well, I forgot to say that my definition of filter is different

> > than the definition of filters in some other's works.

>

> Then it needs a different name.

> On Sat, 20 Sep 2008, Victor Porton wrote:

> > > > > On Sep 20, 10:13 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > > > > > On Fri, 19 Sep 2008, Victor Porton wrote:

>

> > > > > > Would you like me to proof read, review another of your papers?

> > > > > > Let's start with Set Theoretic Filters, the lattice of filters.

>

> > > > > Certainly I would like you to proof read these.

>

> > > > > > It is known that the set of filters for a set S is a complete down

> > > > > > lattice, that is a meet complete semi-lattice. Does this paper go

> > > > > > beyond that?

>

> > > > > The set of filters for a set S is a complete lattice, distributive

> > > > > lattice, atomistic lattice.

>

> > > It is not complete because it has no top element.

> > > Being complete down lattice, it's a bounded complete lattice

> > > ie, every nonnul bounded set has a sup. It's also chain complete.

>

> > Oh, well, I forgot to say that my definition of filter is different

> > than the definition of filters in some other's works.

>

> Then it needs a different name.

Some authors (I can't remember who) also like me allow empty sets as

member of filters.

> > See

> >http://www.mathematics21.org/binaries/set-filters.pdf

> > for the exact definition.

>

> > In my terminology it has top and bottom elements.

>

> > I skip about atoms of the lattice of filters. It is probably the same

> > problem with different definition.

>

> Atoms for filters of a set S are the filters A_x = {S, S\x}, x in S.

> The atoms for the filter F_B are A_x, x not in B.

> The atoms for the Flechet filter are A_x, x in S.

The order of filters in my article is reversed. So what I call atomic

filters correspond to ultrafilters in the traditional terminology.

Sep 20, 2008, 10:03:48 AM9/20/08

to

On Sep 20, 2:57 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> On Sat, 20 Sep 2008, William Elliot wrote:

> > Atoms for filters of a set S are the filters A_x = {S, S\x}, x in S.

> > The atoms for the filter F_B are A_x, x not in B.

> > The atoms for the Flechet filter are A_x, x in S.

> On Sat, 20 Sep 2008, William Elliot wrote:

> > Atoms for filters of a set S are the filters A_x = {S, S\x}, x in S.

> > The atoms for the filter F_B are A_x, x not in B.

> > The atoms for the Flechet filter are A_x, x in S.

I've forgot to say that I use non-standard terminology.

This annuls these problems.

See my other message in this thread for some more info about the

terminology I use.

Or better see

http://www.mathematics21.org/binaries/set-filters.pdf

> sup{ A_x | x in S } = Flechet filter

> Thus the set of filters isn't atomistic.

> It's bounded complete, chain complete, with bottom.

> The non-negative reals are bounded complete, with bottom

> but not chain complete.

>

> On the other hand doesn't F_B = sup { A_x | x not in B } ?

>

> > > I have sent you by email LaTeX file set-filters.tex. I can't offer you

> > > something better.

>

> It has not arrived. Did you forget to remove "remove" from my email

> address before sending?

I've not forgot to remove .remove. Now I sent it once more.

Sep 21, 2008, 2:44:38 AM9/21/08

to

On Sat, 20 Sep 2008, Victor Porton wrote:

> On Sep 20, 2:57 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> > On Sat, 20 Sep 2008, William Elliot wrote:

> > > Atoms for filters of a set S are the filters A_x = {S, S\x}, x in S.

> > > The atoms for the filter F_B are A_x, x not in B.

> > > The atoms for the Flechet filter are A_x, x in S.

>

> I've forgot to say that I use non-standard terminology.

> This annuls these problems.

>

> See my other message in this thread for some more info about the

> terminology I use.

>

> Or better see

> http://www.mathematics21.org/binaries/set-filters.pdf

> On Sep 20, 2:57 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> > On Sat, 20 Sep 2008, William Elliot wrote:

> > > Atoms for filters of a set S are the filters A_x = {S, S\x}, x in S.

> > > The atoms for the filter F_B are A_x, x not in B.

> > > The atoms for the Flechet filter are A_x, x in S.

>

> I've forgot to say that I use non-standard terminology.

> This annuls these problems.

>

> See my other message in this thread for some more info about the

> terminology I use.

>

> Or better see

> http://www.mathematics21.org/binaries/set-filters.pdf

I cannot read that. I asked adobe@pdf2txt to convert that

file to text. It did not work because all the spaces between words had

been removed. That is quite unusual for a conversion. Are you using some

special software that use strange character for space?

> > sup{ A_x | x in S } = Flechet filter

> > Thus the set of filters isn't atomistic.

> > It's bounded complete, chain complete, with bottom.

> > The non-negative reals are bounded complete, with bottom

> > but not chain complete.

> >

> > On the other hand doesn't F_B = sup { A_x | x not in B } ?

> >

> > > > I have sent you by email LaTeX file set-filters.tex. I can't offer

> > > > you something better.

> >

> > It has not arrived. Did you forget to remove "remove" from my email

> > address before sending?

>

> I've not forgot to remove .remove. Now I sent it once more.

>

It still has not arrived. Perhaps it got caught in the spam trap.

I'll check on that. On the other hand, a possible work around would be

the address ma...@rdrop.com .

----

Sep 21, 2008, 2:53:02 AM9/21/08

to

On Sat, 20 Sep 2008, Victor Porton wrote:

> >

> > > > It is not complete because it has no top element.

> > > > Being complete down lattice, it's a bounded complete lattice

> > > > ie, every nonnul bounded set has a sup. It's also chain complete.

> >

> > > Oh, well, I forgot to say that my definition of filter is different

> > > than the definition of filters in some other's works.

> >

> > Then it needs a different name.

>

> Some authors (I can't remember who) also like me allow empty sets as

> member of filters.

>

> > > See

> > >http://www.mathematics21.org/binaries/set-filters.pdf

> > > for the exact definition.

> >

That doesn't work as explained in other post.> >

> > > > It is not complete because it has no top element.

> > > > Being complete down lattice, it's a bounded complete lattice

> > > > ie, every nonnul bounded set has a sup. It's also chain complete.

> >

> > > Oh, well, I forgot to say that my definition of filter is different

> > > than the definition of filters in some other's works.

> >

> > Then it needs a different name.

>

> Some authors (I can't remember who) also like me allow empty sets as

> member of filters.

>

> > > See

> > >http://www.mathematics21.org/binaries/set-filters.pdf

> > > for the exact definition.

> >

> > > In my terminology it has top and bottom elements.

>

> The order of filters in my article is reversed. So what I call atomic

> filters correspond to ultrafilters in the traditional terminology.

>

> http://www.mathematics21.org/binaries/set-filters.pdf

>

From what you tell me, here's a synopsis of what I expect the paper

to cover. Have I missed any points?

--

We shall consider the set of filters F on a set S and the extended set F*

of filters which is F with the addition of the 'filter' generated by the

empty set.

The following three theorems are familiar to those who use filters.

The intersection of any set of filters is a filter.

The union of a chain of filters is a filter.

The intersection of the ultrafilters containing a filter,

is that filter.

Expressing these facts as order theory, F is a

complete down lattice, ie a meet complete semi-lattice

a bounded complete lattice

chain complete

and F* is

a complete anti-atomic lattice

for which the ultrafilters are anti-atoms.

In other words, if the subset order of F* is reversed, it is atomic.

If f,g in F, then f inf g = f /\ g = intersection f,g.

f sup g = { A /\ B | A in f, B in G }

provided that that set does not contain the empty set.

If it does, then f sup g doesn't exist in F, that is { f,g } is

unbounded. However f sup g will exist in F* as top element.

Let F_A be the principal filter generated by A. Then

F_A inf F_B = F_(A \/ B)

and

F_A sup F_B = F_(A /\ B)

(provided A /\ B /= nulset when within F).

It can easily be shown that F* is finitely distributive from

which one discerns that F is finitely bounded distributive.

The question if F* is infinitely or completely distributive

has been left unaddressed.

--

Anyway, what does the sequel paper that generalizes those

results to (partially) ordered sets concern itself about?

----

Sep 21, 2008, 5:32:00 AM9/21/08

to

On Sep 21, 9:44 am, William Elliot <ma...@hevanet.remove.com> wrote:

> It still has not arrived. Perhaps it got caught in the spam trap.

> I'll check on that. On the other hand, a possible work around would be

> the address [ANTISPAM] .> It still has not arrived. Perhaps it got caught in the spam trap.

> I'll check on that. On the other hand, a possible work around would be

I've tried to send attached research.zip to this email.

Say me whether this arrives.

If not download it from http://www.mathematics21.org/research.zip

Sep 21, 2008, 5:42:01 AM9/21/08

to

On Sep 21, 9:53 am, William Elliot <ma...@hevanet.remove.com> wrote:

> From what you tell me, here's a synopsis of what I expect the paper

> to cover. Have I missed any points?

> From what you tell me, here's a synopsis of what I expect the paper

> to cover. Have I missed any points?

I quickly (having busy time) viewed and answered what you've write in

your review. Not sure that I have not missed anything. Don't look on

my below replies as authoritative. I suggest better to look inside the

paper itself:

http://www.mathematics21.org/binaries/set-filters.pdf

> We shall consider the set of filters F on a set S and the extended set F*

> of filters which is F with the addition of the 'filter' generated by the

> empty set.

Yes. I call a filter also the "extended" 'filter' generated by the

empty set.

> The following three theorems are familiar to those who use filters.

> The intersection of any set of filters is a filter.

> The union of a chain of filters is a filter.

> The intersection of the ultrafilters containing a filter,

> is that filter.

In my reversed order intersection of any set of filters is a filter

AND union of any set of filters is a filter.

> Expressing these facts as order theory, F is a

> complete down lattice, ie a meet complete semi-lattice

> a bounded complete lattice

> chain complete

> and F* is

> a complete anti-atomic lattice

> for which the ultrafilters are anti-atoms.

> In other words, if the subset order of F* is reversed, it is atomic.

Like this.

> If f,g in F, then f inf g = f /\ g = intersection f,g.

> f sup g = { A /\ B | A in f, B in G }

> provided that that set does not contain the empty set.

> If it does, then f sup g doesn't exist in F, that is { f,g } is

> unbounded. However f sup g will exist in F* as top element.

> The question if F* is infinitely or completely distributive

> has been left unaddressed.

No, it was addressed. F* is infinitely distribuitive for \/ over

infinite /\.

It is not infinitely distribuitive for /\ over infinite \/ (not said

in the article but easy to find a counterexample). Consequently F* is

not completely distributive.

There are yet several things/theorems which my paper "Set Theoretic

Filters" consider.

> Anyway, what does the sequel paper that generalizes those

> results to (partially) ordered sets concern itself about?

This will be not a paper but a book, a complete reference about

filters on posets and some info about more general lattices.

Sep 21, 2008, 7:42:46 AM9/21/08

to

On Sun, 21 Sep 2008, Victor Porton wrote:

> On Sep 21, 9:53 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > We shall consider the set of filters F on a set S and the extended set F*

> > of filters which is F with the addition of the 'filter' generated by the

> > empty set.

>

> Yes. I call a filter also the "extended" 'filter' generated by the

> empty set.

>

> > The following three theorems are familiar to those who use filters.

> > The intersection of any set of filters is a filter.

> > The union of a chain of filters is a filter.

> > The intersection of the ultrafilters containing a filter,

> > is that filter.

>

> In my reversed order intersection of any set of filters is a filter

> AND union of any set of filters is a filter.

No. What's being reversed? Anything other than for filters f,g,

f <= g when f subset g is being reversed to f <= g when g subset f?

Then f \/ g <= f,g <= f /\ g and sup and inf of sets of filters

is reversed. No. F_{0} \/ F_{1} is not an (extended) filter because

{0} /\ {1} = nulset and it lacks nulset.

Sep 21, 2008, 12:53:11 PM9/21/08

to

On Sep 21, 2:42 pm, William Elliot <ma...@hevanet.remove.com> wrote:

> On Sun, 21 Sep 2008, Victor Porton wrote:

> > On Sep 21, 9:53 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > In my reversed order intersection of any set of filters is a filter

> > AND union of any set of filters is a filter.

>

> No. What's being reversed? Anything other than for filters f,g,

>

> f <= g when f subset g is being reversed to f <= g when g subset f?

> On Sun, 21 Sep 2008, Victor Porton wrote:

> > On Sep 21, 9:53 am, William Elliot <ma...@hevanet.remove.com> wrote:

> > In my reversed order intersection of any set of filters is a filter

> > AND union of any set of filters is a filter.

>

> No. What's being reversed? Anything other than for filters f,g,

>

> f <= g when f subset g is being reversed to f <= g when g subset f?

Yes.

> Then f \/ g <= f,g <= f /\ g and sup and inf of sets of filters

> is reversed. No. F_{0} \/ F_{1} is not an (extended) filter because

> {0} /\ {1} = nulset and it lacks nulset.

sup and inf of sets of filters is reversed as well as <=.

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