7. The Brouwer–Tietze–Urysohn extension theorem for functions. An example of a regular non-normal space. Peano curve. Tychonoff spaces. Metrizable spaces.

Lecture

23.2 The linear space of continuous bounded functions on X with the norm ||f|| =

sup

|f(x)| : x ∈ X

(see Example 9.4.5 of Lecture 2) is denoted C

∗

(X). The metric on C

∗

(X) generated by this norm (see Proposition 10.5 of Lecture 2) is called the metric of uniform

convergence. It is complete (i.e. any fundamental (Cauchy) sequence of points of the space

∗

(X) converges to some point of C

∗

(X)). This fact will be proved later. Let us state this

statement in terms of convergence of functions.

A sequence of functions fn ∈ C

∗

(X), n ∈ N, converges uniformly to a function f ∈ C

∗

(X),

if for an arbitrary ε > 0 there exists such n0 ∈ N that

||f(x) − fn(x)|| < ε for all n ≥ n0.

A sequence of functions fn ∈ C

∗

(X), n ∈ N, is called fundamental (Cauchy) if for an

arbitrary ε > 0 there exists such n0 ∈ N that

||fm(x) − fn(x)|| < ε for all m, n ≥ n0.

Theorem. A fundamental sequence of functions fn ∈ C

∗

(X), n ∈ N, converges uniformly

to a function f ∈ C

∗

(X).

23.3. The Brouwer–Tietze–Urysohn theorem. Let F be a closed subset of a normal space X and ϕ : F → R a continuous bounded function. Then there exists a continuous function ψ : X → R such that

ψ|F = ϕ and ||ψ|| = sup

|ψ(x)| : x ∈ X

= ||ϕ|| = sup

|ϕ(x)| : x ∈ F

Proof. Set ϕ0 = ϕ and µ0 = ||ϕ0||.

We assume µ0 > 0. Otherwise set ψ ≡ 0. Let

P0 =

x ∈ F : ϕ0(x) ≤ −µ0/3

, Q0 =

x ∈ F : ϕ0(x) ≥ µ0/3

The sets P0 and Q0 are closed in X and disjoint. By Urysohn's Lemma, replacing the segment [0, 1]

with [−µ0/3, µ0/3], there exists a continuous function ψ0 : X → [−µ0/3, µ0/3] such that

ψ0(P0) = −µ0/3, ψ0(Q0) = µ0/3.

Set ϕ1 = ϕ0 − ψ0 : F → R. The function ϕ1 is continuous and ||ϕ1|| = µ1 ≤ 2µ0/3.

Now set

P1 =

x ∈ F : ϕ1(x) ≤ −µ1/3

, Q1 =

x ∈ F : ϕ1(x) ≥ µ1/3

We construct a continuous function ψ1 : X → [−µ1/3, µ1/3] such that

ψ1(P1) = −µ1/3, ψ1(Q1) = µ1/3.

Set ϕ2 = ϕ1 − ψ1 and so on.

We obtain a sequence ϕ0, ϕ1, . . . , ϕn, . . . of continuous functions on F and a sequence ψ0, ψ1, . . . , ψn, . . . of continuous functions on X such that ϕn+1 = ϕn − ψn, ||ψn|| ≤

µn/3, ||ϕn+1|| = µn+1 ≤ 2µn/3. Hence,

||ϕn|| ≤ (2/3)nµ0, ||ψn|| ≤ (2/3)nµ0/3.

Set sn = ψ0 + . . . + ψn. The sequence of functions sn, continuous on X, is a

fundamental (Cauchy) sequence in C

∗

(X). Indeed, for m > n we have

||sm − sn|| = ||ψn+1 + . . . + ψm|| ≤ ||ψn+1|| + . . . + ||ψm|| ≤ Xm

k=n+1

k µ0

X∞

k=n+1

k µ0

n+1 µ0

X∞

k=0

n+1

µ0.

According to item 23.2 the sequence sn converges uniformly to a continuous function ψ ≡

P∞

n=0

ψn. We have

||ψ|| ≤ ||ψ − sn|| + ||sn|| ≤ ||ψ − sn|| +

X∞

n=0

n µ0

= ||ψ − sn|| + µ0

for any n ∈ N. Since limn→∞ ||ψ − sn|| = 0, we have ||ψ|| ≤ µ0 = ||ϕ||. Further,

ϕ − sn = ϕ0 − ψ0 − ψ1 − . . . − ψn = ϕ1 − ψ1 − . . . − ψn = ϕn − ψn = ϕn+1.

Hence, ||ϕ − sn|| ≤

n+1

µ0. Therefore limn→∞ ||ϕ − sn|| = 0, ψ|F = ϕ and ||ϕ|| = ||ψ||.

23.4. Corollary. The statement of Theorem 23.3 also holds for unbounded functions.

Proof. Indeed, let ϕ : F → R be an unbounded function. Consider the

function ϕ0 : F →

−

, defined as follows:

ϕ0(x) = arctg

ϕ(x)

According to Theorem 23.3 the function ϕ0 extends to a continuous function ψ0 : X →

−

Set F0 = ψ

−1

−

. By Urysohn's Lemma there exists a function ψ1 : X → [0, 1],

such that ψ1(F0) = 0 and ψ1(F) = 1. Define the function (prove its continuity independently)

ψ : X → R, setting

ψ(x) = tg

ψ1(x) · ψ0(x)

This will be the desired extension of the function ϕ.

23.5. Example (of a regular non-normal space).

The Sorgenfrey line X (Example 8.5.3 of Lecture 2) is an example of a normal space X whose square is not normal.

1. The space X is normal. It is easy to see that X is a T1-space (for example, because

every interval [a, b) is both open and closed in X). Let F, T be disjoint

closed subsets of X. For any point x ∈ F (y ∈ T) take an arbitrary clopen

neighborhood of it of the form [x, ax) ([y, by)), not intersecting T (F resp.). Set

OF =

[

{[x, ax) : x ∈ F}, OT =

[

{[y, by) : y ∈ T}.

It is easy to see that OF ∩ OT = ∅.

2. The square X ×X is not normal. Indeed, consider the anti-diagonal, i.e. the set

Y of all points of the form (a, −a), a ∈ R. Since the sets [a, b) × [−a, c) are neighborhoods

of the point (a, −a) in X ×X, every point y ∈ Y is isolated in Y, i.e. Y is a discrete subspace

of the product X × X. At the same time, Y is closed in X × X. Therefore every set Z ⊂ Y

is closed in X × X.

The set Y has the cardinality of the continuum c. For each set Z ∈ 2

Y consider

the continuous function ϕZ : Y → [0, 1], equal to 0 on Z and 1 on Y \ Z. The number of such functions

has cardinality 2

. If the space X × X were normal, then every function ϕZ

could be extended to a continuous function fZ : X × X → [0, 1] by the Brouwer–

Tietze–Urysohn theorem. But the space X ×X has a countable dense set D of points of the form (x1, x2),

where xi ∈ Q are rational numbers. According to Proposition 22.4 of Lecture 5 every continuous

mapping f : X ×X → [0, 1] is uniquely determined by its values on the countable dense

set D. But the set [0, 1]D of all (not necessarily continuous) mappings g : D → [0, 1]

has cardinality |(2N)

N| = |2

N·N| = |2

N| = c. This contradiction (Theorem 5.1 of Lecture 1) shows

that the square X is not normal.

23.6. Remark The square of the Sorgenfrey line is an example of a separable space containing a non-separable subspace (for example, the anti-diagonal).

§ 24. Peano curve.

In Proposition 5.3 a surjective mapping f : 2N → I = [0, 1] is defined. To a point of the product (in)n∈N there corresponds a point t ∈ I, with binary expansion t = 0, i1i2 . . . in . . ..

24.1. Proposition. f : 2N → I = [0, 1] is a continuous mapping.

Proof. For a point (ik)k∈N ∈ 2

N and a neighborhood O = (t −

2n , t +

2n ) of its image t,

the set U =

{pr−1

(ik) : k = 1, . . . , n + 1} is an open neighborhood of (ik)k∈N in 2

N, and

f(U) ⊂ O. It follows that f is continuous at an arbitrary point (ik)k∈N ∈ 2

N, and hence on all of

24.2. Theorem (Peano curve). There exists a continuous mapping of the segment onto the square.

Proof. Since 2

N ×2

N is homeomorphic to 2

N, the square f ×f of the mapping f is

a surjection of the set 2

N (homeomorphic to the Cantor set C) onto the square I

. Denote

the surjection of C onto I

2 by F.

Regarding the Cantor set C as a subset of the segment I, and the square as a subset of R

, let us extend the mapping F to the segment. The mapping F is defined at the endpoints x < y

of the interval J and of the intervals Ji1...ik

. For t ∈ Ji1...ik

(or J) set F˜(t) = y−t

y−x

F(x) + t−x

y−x

F(y).

Then F˜ = F. The continuity of the restriction of F˜ to the segments that are the closure of

the interval J and of the intervals Ji1...ik

is obvious. Hence, F˜ is continuous at the points of the complement of C.

Let x ∈ C. For the ε-neighborhood Oε(F˜(x)) of its image, which is a convex subset of R

, there exists a segment Ii1...ik

such that F(C ∩Ii1...ik

) ⊂ Oε(F˜(x)) (since F is a continuous mapping). Then the images of all intervals Ji1...ik...im, contained in Ii1...ik

, under the mapping F˜ are contained in Oε(F˜(x)) (the image of each interval is an interval with endpoints at points

belonging to Oε(F˜(x)). If x is not an endpoint of the segment Ii1...ik

, then its interior is a neighborhood of the point x in I, whose image belongs to Oε(F˜(x)), and the continuity of F˜ at this

point is proved.

Let x be an endpoint of the segment Ii1...ik

. Then it is an endpoint of the segment I

, which is the closure of the interval adjacent to the segment Ii1...ik

. Since F˜ is continuous on I

, there exists a neighborhood of x in I

(in particular, a half-interval [x, t) or (t, x]), whose image under the mapping F˜

is contained in Oε(F˜(x)). Thus the interior of the set [x, t) ∪ Ii1...ik or (t, x] ∪ Ii1...ik is a neighborhood of the point x and its image belongs to Oε(F˜(x)). The continuity of the mapping F˜

at this point is also proved.

§ 25. Tychonoff spaces.

25.1. Definition. A T1-space is called completely regular or Tychonoff if

for every point x ∈ X and every closed set F not containing it there exists a

continuous function f : X → [0, 1] such that f(x) = 1, f(F) = 0.

By Urysohn's Lemma 23.1 a normal space is Tychonoff, which, in turn,

is regular.

25.2. Proposition. Every subspace of a Tychonoff space is a Tychonoff space.

The product of Tychonoff spaces is a Tychonoff space.

Proof. Let us prove the second statement. Let X =

α∈A Xα.

For an arbitrary neighborhood Ox of a point x ∈ X let us find a function f : X → [0, 1] such that

f(x) = 1, f(X \ Ox) = 0. By the definition of the product topology there exist a finite

set of indices α1, . . . , αk and open sets Vi ⊂ Xα, such that

x = (xα) ∈

{pr−1

αi

(Vi) : i = 1, . . . , k} ⊂ Ox.

Set gi

: Xαi → [0, 1] such that gi(xαi

) = 1, gi(X \ Vi) = 0, i = 1, . . . , k.

fi = gi ◦ prαi

: X → [0, 1], i = 1, . . . , k.

Then

f = f1 · . . . · fk : X → [0, 1]

is the required function.

The Tychonoff cube of weight κ ≥ ℵ0 is the space I

κ =

α∈A

Iα, where Iα = [0, 1] for each α ∈ A,

|A| = κ (the κ-th power of the segment, or the product of κ copies of the segment). The Tychonoff cube I

ℵ0

(the countable power of the segment) is called the Hilbert cube.

25.3. Lemma. Let a family of mappings fα : X → Iα = [0, 1], α ∈ A, |A| = κ, of a Tychonoff space X separate points and closed sets, i.e. satisfy the

condition:

for any point x ∈ X and any closed set T, x 6∈ T, there exists α ∈ A such

that fα(x) = 1, fα(T) = 0.

Then the diagonal product F = ∆α∈Afα : X →

α∈A

Iα of the mappings fα, α ∈ A, is

an embedding of X in the Tychonoff cube I

(i.e. a homeomorphism of X onto F(X)).

Proof. It is obvious that the mapping F is injective and continuous. To prove the continuity of the inverse mapping F

−1

: F(X) → X let us check its continuity at an

arbitrary point y ∈ F(X). Let F(x) = y (i.e. F

−1

(y) = x), and Ox a neighborhood of the point x.

By the condition on the family fα, α ∈ A, for x ∈ X and the closed set T = X \ Ox there exists

α0 ∈ A such that fα0

(x) = 1, fα0

(T) = 0. Then W = pr−1

α0

((0, 1]) ∩ F(X) is a neighborhood of the point y

in F(X), and

if F(t) ∈ W, then t ∈ Ox.

Hence F

−1

(W) ⊂ Ox.

From Proposition 25.2 (sufficiency) and Lemma 25.3 (necessity) we obtain.

25.4. Theorem. A space X is Tychonoff if and only if

it is embeddable in the Tychonoff cube I

for some κ.

§ 26. Metrizable spaces. Urysohn's theorem.

26.1. Theorem. Any metrizable space is normal.

Proof. Let ρ be a metric on the space X generating its topology, F, T

disjoint closed subsets of X.

Consider the non-negative functions ρF : X → R, ρF (x) = inf{ρ(x, t) : t ∈ F}, and ρT : X → R,

ρT (x) = inf{ρ(x, t) : t ∈ T}. They are continuous (see Example 13.6 of Lecture 3), and X \ F = ρ

−1

(0, +∞),

X\T = ρ

−1

(0, +∞). Let us prove the first equality (the second is proved similarly). If x ∈ F, then

ρF (x) = 0, if x ∈ X \F, then there exists ε > 0 such that Oε(x) ⊂ X \F, and, hence, ρF (x) ≥ ε > 0.

The function h(x) = ρF (x)

ρF (x)+ρT (x)

is well defined, continuous and F = h

−1

(0), T = h

−1

(1).

Then the sets h

−1

(−∞,

) and h

−1

(

, +∞) are disjoint neighborhoods of F and T respectively.

The satisfaction of axiom T1 is obvious.

The Sorgenfrey line is a normal non-metrizable space.

26.2. Definition. Let two metrics ρ1 and ρ2 be given on a set X. They are called

topologically equivalent if Tρ1 = Tρ2

26.3. Proposition. Every metric ρ1 on a set X is topologically equivalent

to a metric ρ2, in which the diameter of X is ≤ 1 (diam X = sup{ρ2(x, y) : x, y ∈ X} ≤ 1).

Proof. One can set, for example, ρ2(x, y) = min{ρ1(x, y), 1}, and note that

the topologies generated by these metrics coincide, since the bases of the topologies consist of open -balls, where < 1.

26.4. Proposition. The sum X = ⊕{Xα : α ∈ A} of metrizable spaces Xα, α ∈ A,

is metrizable.

Proof. Let ρα be a metric on Xα generating the topology of the space Xα.

According to Proposition 26.3 we may assume diam (Xα, ρα) ≤ 1. Define the function ρ :

X × X → R+ as follows:

ρ(x, y) =

ρα(x, y), if x, y ∈ Xα;

1, if x and y lie in different summands.

It is easy to check that ρ is a metric on X (here it is important that diam Xα ≤ 1). The open

ε-balls Oε(x) form a base of the metric topology on X. The preimage of any ε-ball Oε(x) under

the embedding mapping iα : Xα → X is open in Xα, α ∈ A (if ε ≤ 1 and x ∈ Xα0 , then i

−1

α (Oε(x)) = ∅,

for α 6= α

0 and i

−1

α0 (Oε(x)) = Oε(x), where the latter set is the open ε-ball of the point x

in the metric space (Xα0 , ρα0 ); if ε > 1, then Oε(x) = X and i

−1

α (Oε(x)) = Xα, for any

α ∈ A). It remains to note that any set open in X (in the final topology on X

with respect to the embeddings) is a union of ε-balls Oε(x).

26.5. Theorem. A countable product of metrizable spaces is metrizable.

Proof. Let X =

n∈N Xn. According to Proposition 26.3 the spaces Xn can

be endowed with metrics ρn such that

diam Xn ≤ 1, n ∈ N.

On X define a metric ρ, taking on an arbitrary pair of points x = (xn) and y = (yn) from

X the values ρ(x, y) = sup{

ρn(xn,yn)

: n ∈ N}. It is easy to check that ρ is a metric.

Denote by XM the product X with the metric topology Tρ. It suffices to prove that

the identity mapping id : X → XM is a homeomorphism.

Let us first prove that id is continuous. Take an arbitrary point x = (xn) ∈ XM and its ε-

neighborhood Oε(x), where ε > 0. Choose N so that 1

N < ε.

Define the neighborhood Ox ⊂ X of the point x as follows:

Ox =

n=1

pr−1

(Oε(xn)),

where Oε(xn) is the open ε-neighborhood centered at the point xn in the space Xn. Then for every

point y ∈ Ox we have ρ(x, y) = sup{

ρn(xn,yn)

: n ∈ N} < ε. Thus, Ox ⊂ Oε(x), i.e.

the mapping id is continuous.

To prove the continuity of the inverse mapping from XM onto X it suffices to prove

the continuity of its composition with an arbitrary projection prn (see Corollary 17.3 of Lecture 4).

For a point x = (xn) let Oε(xN ) be the ε-neighborhood of the point xN in the space XN (with the metric

ρN ). Then for any point y = (yn) from O ε

(x) we have

ρN (xN , yN )

I.e. under the composition the image of the neighborhood O ε

(x) is contained in Oε(xN ), and the composition is continuous.

Assignment N 7

1. Prove that in the Brouwer–Tietze–Urysohn theorem the extension of a function is

defined non-uniquely. Is it true that there are at least a continuum of extensions of a function?

2. In Example 17.8 of Lecture 4 a homeomorphism g of the Cantor set onto the countable

power of the discrete two-point set 2

N is defined. In Proposition 24.1 of Lecture 7 a mapping f

of the product 2

N onto the segment I = [0, 1] is defined.

Prove that the composition f ◦ g of the mappings g and f is a monotonically increasing function on the Cantor set, for which the preimages of dyadic-rational numbers of the interval

(0, 1) are two-point, and the preimages of the other points of I are one-point.

(Cantor staircase) Prove that there exists a unique extension of the function f ◦ g :

C → I to the segment I, which is a monotonically increasing function. Draw its graph.

3. Construct a continuous surjective mapping of the segment onto the finite-dimensional cube I

n, n ∈

N, the Hilbert cube I

ℵ0 (the countable power of the segment).

4 (Niemytzki plane). A base of the topology on the half-plane {(x, y) ∈ R

: y ≥ 0} is formed by

the open balls

Oε(x, y) ∩ R

, x ∈ R, y > 0, ε > 0,