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1. ON “EACH REGULAR PARATOPOLOGICAL GROUP IS COMPLETELY REGULAR” ARTICLE

64

Proposition

2386

.

A funcoid

ν

is normally quasi-uniformizable iff there exist

a quasi-proximity space (= reflexive and transitive funcoid)

µ

such that

µ

is normal

on

ν

and

ν

= Compl

µ

.

Proof.

Obvious

2380

and the fact that (

FCD

) is an isomorphism between

reflexive and transitive funcoids and reflexive and transitive reloids.

In other words, it is normally reloidazable or normally quasi-uniformizable when

(Compl

µ

)

(Compl

µ

)

1

v

(Compl

µ

)

1

µ

for suitable

µ

.

1.2. Urysohn’s lemma and friends.

For a detailed proof of Urysohn’s

lemma see also:

http://homepage.math.uiowa.edu/~jsimon/COURSES/M132Fall07/
UrysohnLemma_v5.pdf
https://proofwiki.org/wiki/Urysohn’s_Lemma
http://planetmath.org/proofofurysohnslemma

https://en.wikipedia.org/wiki/Proximity_space

says

that

“The

resulting

topology is always completely regular. This can be proven by imitating the usual
proofs of Urysohn’s lemma, using the last property of proximal neighborhoods to
create the infinite increasing chain used in proving the lemma.”

Below follows an alternative proof of Urysohn lemma.

The proof was based on

a conjecture proved false, see example 1444!

Lemma

2387

.

If

h

µ

iA  B

for a complete funcoid

µ

and

A

,

B

are filters on

relevant sets, then there exists

U

up

µ

such that

h

U

iA  B

.

Proof.

Prove that

n

h

U

iA

U

up

µ

o

is a filter base. That it is nonempty is obvious.

Let

X

,

Y ∈

n

h

U

iA

U

up

µ

o

. Then

X

=

h

U

X

iA

,

Y

=

h

U

Y

ih

A

i

. Because

µ

is complete,

we have (proposition 1227)

U

X

u

U

Y

up

µ

. Thus

X

,

Y w h

U

X

u

U

Y

iA ∈

n

h

U

iA

U

up

µ

o

.

Thus

h

µ

iA  B ⇔ B u h

µ

iA

=

⊥ ⇔ ∃

U

up

µ

:

B u h

U

iA

=

⊥ ⇔ ∃

U

up

µ

:

h

U

iA  B

.

Corollary

2388

.

If

h

µ

iA  h

µ

iB

for a complete funcoid

µ

and

A

,

B

are filters

on relevant sets, then there exists

U

up

µ

such that

h

U

iA  h

U

iB

.

Proof.

Applying the lemma twice we can obtain

P, Q

up

µ

such that

h

P

iA  h

Q

iB

. But because

µ

is complete, we have

U

=

P

u

Q

up

µ

, while

obviously

h

U

iA  h

U

iB

.

Lemma

2389

.

(assuming conjecture 1444) For every

U

up

µ

(where

µ

is a

T

4

topological space) such that

¬

A

U

U

1

B

there is

W

up

µ

such that

U

U

1

w

W

W

1

W

W

1

. For it holds

¬

A

W

W

1

B

. We can assume

that

h

W

i

X

is open for every set

X

.

Proof.

U

U

1

up(

µ

µ

1

)

up(

µ

µ

1

µ

µ

1

) (normality used). Thus

by the conjecture there exists

W

up

µ

such that

U

U

1

w

W

W

1

W

W

1

.

W

W

1

v

U

U

1

thus

¬

A

W

W

1

B

.

To prove that

h

W

i

X

is open for every set

X

, replace every

h

W

i

{

x

}

with an

open neighborhood

E

⊆ h

W

i

X

of

h

µ

i

{

x

}

(and note that union of open sets is

open). This new

W

holds all necessary properties.