 1. THE REST

40

Suppose

↑ {

(

x, x

)

} v h

f

×

(

A

)

f

i

(

g

\

U

); then

g

\

U

6 h

f

1

×

(

A

)

f

1

i{

(

x, x

)

}

;

U

@

h

f

1

×

(

A

)

f

1

i{

(

x, x

)

} v h

f

1

×

(

A

)

f

1

i

1

RLD

what is impossible.

Thus there exist

x

6

=

y

such that

{

(

x, y

)

} v

Cor

h

f

×

(

A

)

f

i

(

g

\

U

). Thus

{

(

x, y

)

} v h

f

×

(

A

)

f

i

g

.

Thus by the lemma

{

(

x, y

)

} v

1

RLD

what is impossible. So

U

up

g

.

We have up

h

f

×

(

A

)

f

i

1

RLD

up

g

;

h

f

×

(

A

)

f

i

1

RLD

w

g

.

Corollary

2109

.

Let

f

is a

T

1

-separable (the same as

T

2

for symmetric transi-

tive) compact funcoid and

g

is a uniform space (reflexive, symmetric, and transitive

endoreloid) such that (

FCD

)

g

=

f

. Then

g

=

h

f

×

(

A

)

f

i

1

RLD

.

An (incomplete) attempt to prove one more theorem follows:

Theorem

2110

.

Let

µ

and

ν

be uniform spaces, (

FCD

)

µ

be a compact funcoid.

Then a map

f

is a continuous map from (

FCD

)

µ

to (

FCD

)

ν

iff

f

is a (uniformly)

continuous map from

µ

to

ν

.

Proof.

FiXme

: errors in this proof.

http://math.stackexchange.com/questions/665202/bourbaki-

on-the-fact-that-continuous-function-on-a-compact-is-uniformly-
continuo/670956?iemail=1&noredirect=1#670956

We have

µ

=

h

(

FCD

)

µ

×

(

FCD

)

µ

i ↑

RLD

1

RLD

f

C

?

((

FCD

)

µ,

(

FCD

)

ν

). Then

f

×

(

A

)

f

C

?

((

FCD

)(

µ

×

(

A

)

µ

)

,

(

FCD

)(

ν

×

(

A

)

ν

))

(

f

×

(

A

)

f

)

(

FCD

)(

µ

×

(

A

)

µ

)

v

(

FCD

)(

ν

×

(

A

)

ν

)

(

f

×

(

A

)

f

)

For every

V

up(

ν

×

(

A

)

ν

) we have

h

g

1

i

V

∈ h

(

FCD

)(

µ

×

(

A

)

µ

)

i{

y

}

for some

y

.

h

g

1

i

V

∈ h

(

FCD

)

µ

×

(

A

)

(

FCD

)

µ

i ↑

RLD

1

RLD

= up

µ

h

g

ih

g

1

i

V

v

V

We need to prove

f

C(

µ, ν

) that is

p

up

ν

q

up

µ

:

h

f

i

q

v

p

. But this

follows from the above.

FiXme

: A space is compact if and only if it is both, complete and totally

bounded.

http://math.stackexchange.com/questions/1101995/

non-symmetric-version-of-compact-totally-bounded-complete