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

66

If

z

A

then

z

∈ h

U

q

i

A

for every

q

Q

2

, thus

f

(

z

) = 0, because obviously

U

q

w

1.

If

z

B

then

z /

∈ h

U

q

i

A

for every

q

Q

2

, thus

f

(

z

) = 1, because

U

q

v

U

0

.

It remains to prove that

f

is continuous.

Let

D

(

x

) =

{

1

} ∪

n

q

Q

2

z

∈h

U

q

i

A

o

.

To show that f is continuous, we first prove two smaller results:
(a)

x

µ

1

h

U

r

i

A

f

(

x

)

r

.

We have

x

µ

1

h

U

r

i

A

⇒ ∀

s > r

:

x

∈ h

U

s

i

A

, so

D

(

x

) contains all

rationals greater than

r

. Thus

f

(

x

)

r

by definition of

f

.

(b)

x /

∈ h

U

r

i

A

f

(

x

)

r

.

We have

x /

∈ h

U

r

i

A

⇒ ∀

s < r

:

x /

∈ h

U

s

i

A

. So

D

(

x

) contains no rational less

than

r

. Thus

f

(

x

)

r

.

Let

x

0

S

and let ]

c

;

d

[ be an open real interval containing

f

(

x

). We will find

a neighborhood

T

of

x

0

such that

h

f

i

T

]

c

;

d

[.

Choose

p, q

Q

such that

c < p < f

(

x

0

)

< q < d

.

Let

T

=

h

U

q

i

A

\

µ

1

h

U

p

i

A

.

Then since

f

(

x

0

)

< q

, we have that (b) implies vacuously that

x

∈ h

U

q

i

A

.

Since

f

(

x

0

)

> p

, (a) implies

x

0

/

∈ h

U

p

i

A

.

Hence

x

0

T

. Then

T

is a neighborhood of

x

0

because

T

is open.

Finally, let

x

T

.

Then

x

∈ h

U

q

i

A

µ

1

h

U

q

i

A

. So

f

(

x

)

q

by (a).

Also

x /

µ

1

h

U

p

i

A

, so

x /

∈ h

U

p

i

A

and

f

(

x

)

p

by (b).

Thus:

f

(

x

)

[

p

;

q

]

]

c

;

d

[.

Therefore

f

is continuous.

Claim A:

f

(

x

)

> q

x /

∈ h

µ

1

i

h

U

q

i

A

Claim B:

f

(

x

)

< q

x

∈ h

U

q

i

A

Proof of claim A: If

f

(

x

)

> q

then then there must be some gap between

q

and

D

(

x

); in particular, there exists some

q

0

such that

q < q

0

< f

(

x

). But

q

0

< f

(

x

)

x /

∈ h

U

q

i

A

x /

µ

1

h

U

q

i

A

(using that

h

U

r

i

X

is open).

Proof of claim B: If

f

(

x

)

< q

then there exists

q

0

D

(

x

) such that

f

(

x

)

< q

0

<

q

, in which case

q

D

(

x

), so

x

∈ h

U

q

i

A

.

To show that

f

is continuous, it’s enough to prove that preimages of ]

a

; 1] and

[0;

a

[ are open.

Suppose

f

(

x

)

]

a

; 1]. Pick some

q

with

a < q < f

(

x

). We claim that the

open set

W

=

X

\

f

1

h

U

q

i

A

is a neighborhood of

x

that is mapped by

f

into

]

a

; 1]. First, by (A),

f

(

x

)

> q

x

W

, so

W

is a neighborhood of

x

. If

y

is

any point of

W

, then

f

(

y

) must be

q > a

; otherwise, if

f

(

y

)

< q

, then, by (B)

y

∈ h

U

q

i

A

f

1

h

U

q

i

A

.

Suppose

x

f

1

[0;

b

[ that is

f

(

x

)

< b

and pick

q

such that

f

(

x

)

< q < b

. By

(B)

x

∈ h

U

q

i

A

. We claim that the neighborhood

h

U

q

i

A

is mapped by

f

into

[0;

b

[. Suppose

y

is any point of

h

U

q

i

A

. Then

q

D

(

y

), so

f

(

y

)

q < b

.

Theorem

2392

.

(from [

1

]) If

µ

is a normal quasi-uniformity on a topolog-

ical space

ν

, then for any nonempty subset

A

Ob

ν

and entourage

U

up

µ

there exists a continuous function

f

: Ob

ν

[0; 1] such that

A

v

f

1

{

0

} v

f

1

[0; 1[

v

ν

1

ν

1

h

U

i

A

.

Proof.

Choose inductively a sequence of entourages (

U

n

)

n

=0

such that

U

0

=

U

and

U

n

+1

U

n

+1

v

U

n

.

Denote

l

r

= max

n

n

N

r

n

=1

o

.

Define

U

r

=

U

r

lr

l

r

. . .

U

r

1

1