proof of Urysohn’s lemma

proof of Urysohn’s lemma

First we construct a family

U

p

of

open sets

of

X

indexed by

the

rationals

such that if

p

<

q

, then

U

p

¯

⊆

U

q

. These
are the sets we will use to define our

continuous function

.

Let

P

=

ℚ

∩

[

0

,

1

]

. Since

P

is

countable

, we can use

induction

(or recursive definition if you prefer) to define the sets

U

p

. List the elements of

P

is an

infinite

sequence

in some way;
let us assume that

1

and

0

are the first two elements of this
sequence. Now, define

U

1

=

X

\

D

(the

complement

of

D

in

X

). Since

C

is a

closed set

of

X

contained

in

U

1

, by

normality

of

X

we can choose an open set

U

0

such that

C

⊆

U

0

and

U

0

¯

⊆

U

1

.

In general, let

P

n

denote the set consisting of the first

n

rationals in our sequence. Suppose that

U

p

is defined for all

p

∈

P

n

and

if

⁢

p

<

q

,

then

⁢

U

p

¯

⊆

U

q

.

(1)

Let

r

be the next rational number in the sequence. Consider

P

n

+

1

=

P

n

∪

{

r

}

. It is a finite subset of

[

0

,

1

]

so it inherits
the usual

ordering

<

of

ℝ

. In such a set, every

element

(other than the smallest or largest) has an immediate

predecessor

and

successor

. We know that

0

is the smallest element and

1

the
largest of

P

n

+

1

so

r

cannot be either of these. Thus

r

has
an immediate predecessor

p

and an immediate successor

q

in

P

n

+

1

. The sets

U

p

and

U

q

are already defined by the

inductive hypothesis

so using the normality of

X

, there exists an
open set

U

r

of

X

such that

U

p

¯

⊆

U

r

⁢

and

⁢

U

r

¯

⊆

U

q

.

We now show that (1) holds for every pair of elements in

P

n

+

1

. If
both elements are in

P

n

, then (1) is true by the inductive
hypothesis. If one is

r

and the other

s

∈

P

n

, then if

s

≤

p

we have

U

s

¯

⊆

U

p

¯

⊆

U

r

and if

s

≥

q

we have

U

r

¯

⊆

U

q

⊆

U

s

.

Thus (1) holds for ever pair of elements in

P

n

+

1

and therefore by
induction,

U

p

is defined for all

p

∈

P

.

We have defined

U

p

for all rationals in

[

0

,

1

]

. Extend this

definition

to every rational

p

∈

ℝ

by defining

U

p

=

∅

if

⁢

p

<

0

U

p

=

X

if

⁢

p

>

1

.

Then it is easy to check that (1) still holds.

Now, given

x

∈

X

, define

ℚ

⁢

(

x

)

=

{

p

:

x

∈

U

p

}

. This
set contains no number less than

0

and contains every number

greater

than

1

by the definition of

U

p

for

p

<

0

and

p

>

1

. Thus

ℚ

⁢

(

x

)

is bounded below and its

infimum

is an element in

[

0

,

1

]

. Define

f

⁢

(

x

)

=

inf

⁢

ℚ

⁢

(

x

)

.

Finally we show that this

function

f

we have defined

satisfies

the
conditions of lemma. If

x

∈

C

, then

x

∈

U

p

for all

p

≥

0

so

ℚ

⁢

(

x

)

equals the set of all nonnegative rationals and

f

⁢

(

x

)

=

0

. If

x

∈

D

, then

x

∉

U

p

for

p

≤

1

so

ℚ

⁢

(

x

)

equals all the rationals greater than 1 and

f

⁢

(

x

)

=

1

.

To show that

f

is

continuous

, we first prove two smaller results:

(a)

x

∈

U

r

¯

⇒

f

⁢

(

x

)

≤

r

Proof

. If

x

∈

U

r

¯

, then

x

∈

U

s

for all

s

>

r

so

ℚ

⁢

(

x

)

contains all rationals greater than

r

. Thus

f

⁢

(

x

)

≤

r

by definition of

f

.

(b)

x

∉

U

r

⇒

f

⁢

(

x

)

≥

r

.

Proof

. If

x

∉

U

r

, then

x

∉

U

s

for all

s

<

r

so

ℚ

⁢

(

x

)

contains no rational less than

r

. Thus

f

⁢

(

x

)

≥

r

.

Let

x

0

∈

X

and let

(

c

,

d

)

be an

open interval

of

ℝ

containing

f

⁢

(

x

)

. We will find a

neighborhood

U

of

x

0

such that

f

⁢

(

U

)

⊆

(

c

,

d

)

. Choose

p

,

q

∈

ℚ

such that

c

<

p

<

f

⁢

(

x

0

)

<

q

<

d

.

Let

U

=

U

q

\

U

p

¯

. Then since

f

⁢

(

x

0

)

<

q

, (b)
implies that

x

∈

U

q

and since

f

⁢

(

x

0

)

>

p

, (a) implies that

x

0

∉

U

p

¯

. Hence

x

0

∈

U

.

Finally, let

x

∈

U

. Then

x

∈

U

q

⊆

U

q

¯

, so

f

⁢

(

x

)

≤

q

by (a). Also,

x

∉

U

p

¯

so

x

∉

U

p

and

f

⁢

(

x

)

≥

p

by (b). Thus

f

⁢

(

x

)

∈

[

p

,

q

]

⊆

(

c

,

d

)

as desired. Therefore

f

is continuous and we are done.

Title

proof of Urysohn’s lemma

Canonical

name

ProofOfUrysohnsLemma

Date of creation

2013-03-22 13:09:23

Last modified on

2013-03-22 13:09:23

Owner

scanez (1021)

Last modified by

scanez (1021)

Numerical id

4

Author

scanez (1021)

Entry type

Proof

Classification

msc 54D15

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