§ 21. Countability axioms (character, weight of a space). Separability (density of a space).
21.1. Definition. A space X satisfies the first countability axiom at a point x
(or has countable character at the point x), if there exists a countable family B of neighborhoods
of the point X such that every neighborhood of the point x contains at least one neighborhood
from B.
A space X satisfies the first countability axiom (or the character of the space X
is countable), if X satisfies the first countability axiom at all points x ∈ X.
Metrizable spaces satisfy the first countability axiom. The Sorgenfrey line
satisfies the first countability axiom. The line with the Zariski topology does not satisfy the first countability axiom.
21.2. Definition. If a space has a countable base, it is said to satisfy the second countability axiom (or the weight of the space X is countable).
If a space satisfies the second countability axiom, then it satisfies the first
countability axiom.
The line with the standard topology satisfies the second countability axiom. An uncountable discrete space does not satisfy the second countability axiom.
21.3. Theorem. A subspace of a space satisfying the first countability axiom satisfies the first countability axiom. A countable product of spaces satisfying the first countability axiom satisfies the first countability axiom.
A subspace of a space satisfying the second countability axiom satisfies the second countability axiom. A countable product of spaces satisfying the second countability axiom satisfies the second countability axiom.
Proof. Let us prove the second statement.
Let B — be a countable base of X, Y ⊂ X. It is easy to see that the family {X ∩ U : U ∈ B} — is a base of Y .
The Tychonoff product topology is initial with respect to the projections, and its subbase is formed by preimages of elements of the bases of the factors under the projections. Since the number of factors is countable, the subbase has countably many elements, and the base it generates is countable.
21.4. Definition. A subset Z of a topological space X is called everywhere
dense in X, if Cl(Z) = X.
Problem. A subset Z of a topological space X is everywhere dense in X if and only
if Z ∩ O 6= ∅ for any nonempty open subset O of the space X.
21.5. Definition. A space X is called separable (or the density of the space X is countable), if it contains a countable everywhere dense subset.
If X satisfies the second countability axiom, then it is separable (the everywhere dense subset will be
the subset of points taken one from each element of the countable base).
21.6. Theorem. A countable product of separable spaces is separable.
Proof. Let Dn — be a countable dense subset of the spaces Xn, dn ∈ Dn — a
fixed point, n ∈ N. Set An = D1 × · · · × Dn × {dn+1} × . . . , n ∈ N. It is easily
verified that the family S
n∈N An is countable and everywhere dense in the product.
21.7. Theorem. A metrizable space satisfies the second countability axiom if and
only if it is separable.
Proof. Necessity is obvious. Let us prove sufficiency. Let ρ — be a metric on
X generating its topology, S — a countable everywhere dense subset of X. Let us show that
a base of the topology of X is formed by the family of open balls
B = {Oq(s), s ∈ S, q ∈ Q, q > 0}.
35
The family B — is a subfamily of the base of the topology. Let O — be an open subset of X. To complete the proof it suffices to show that for every point x ∈ O there exists U ∈ B
such that x ∈ U ⊂ O.
There exists q ∈ Q such that Oq(x) ⊂ O. Take s ∈ S ∩ Oq
2
(x) and the ball Oq
2
(s). Then x ∈
Oq
2
(s) ⊂ Oq(x).
§ 22. Separation axioms.
22.1. A topological space X is a T0-space or satisfies the Kolmogorov
separation axiom, if for any two distinct points x and y of the space X at
least one of them has a neighborhood not containing the other point.
A set of more than two points with the antidiscrete topology (for example the “glued-together two-point set” (Example 8.2.2 of Lecture 2)) is an example of a space not satisfying the axiom
T0.
22.2. A space X is called a T1-space, if for any distinct points x and
y of the space X there exist a neighborhood Ox not containing the point y, and a neighborhood Oy not
containing the point x.
The “connected two-point set”, i.e. the space X = {a, b}, in which the sets ∅, {a}, X are open
(Example 8.2.2 of Lecture 2), is a T0-space that is not a T1-space.
A space X is a T1-space if and only if all one-point subsets of X are closed. Among the topologies on a set X in which X is a T1-space,
there is a smallest one. This is the cofinite topology (Example 8.2.4 of Lecture 2).
22.3. A space X is called Hausdorff or a T2-space, if for every
pair of distinct points of X there exist disjoint neighborhoods of them.
An infinite set equipped with the cofinite topology (Example 8.2.4 of Lecture
2), is a non-Hausdorff T1-space.
22.4. Proposition. Let fi
: X → Y, i = 1, 2, — be continuous mappings of a space X into a Hausdorff space Y . Suppose that f1
Z = f2
Z for some everywhere
dense subset Z of X. Then f1 = f2.
Proof. Suppose that for some point x ∈ X the points f1(x) and f2(x) are distinct. Since the space Y is Hausdorff, there exist disjoint neighborhoods
Ofi(x), i = 1, 2. From the continuity of fi it follows that there exist neighborhoods Oix, i = 1, 2,
such that fi
Oix
⊂ Ofi(x).
Set Ox = O1x ∩ O2x. Since Z is everywhere dense in X, there exists a point z ∈ Z ∩ Ox.
Then f1(z) = f2(z) ∈ Of1(x)∩Of2(x). But this contradicts the fact that the sets Of1(x) and Of2(x)
do not intersect.
The axioms T0, T1, T2 form an increasing chain of strength and give ever narrower, as the given examples show, classes of spaces.
22.5. A space X is called a T3-space, if for every point x and every
closed set F not containing it there exist disjoint neighborhoods Ox and OF.
The axiom T3 does not even imply T0. This is shown by the example of a set of more than two points with the
antidiscrete topology.
22.6. A space simultaneously satisfying the axioms T0 and T3 is called regular.
Every regular space X is Hausdorff. Indeed, let x, y ∈ X and x 6= y. For
one of the points, say x, there exists a neighborhood Ox not containing y. Taking disjoint neighborhoods of the point x and the closed set X \ Ox, we enclose x and y in disjoint neighborhoods.
An example of a Hausdorff non-regular space is the real line, whose base of topology is formed by all sets of the form U and U \ K, where U — is an interval of the real
line, K = {
1
n
: n ∈ N} (Example 8.8.5 of Lecture 2). Prove that the set K is closed in this
space and cannot be separated from zero by disjoint neighborhoods.
22.7. Proposition. Every subspace of a regular space (respectively
a T0-, T1-, T2-space) is regular (respectively a T0-, T1-, T2-space).
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22.8. Proposition. A product of Ti-spaces, i = 0, 1, 2, 3, is a Ti-space. In
particular, a product of regular spaces is regular.
Proof. Let us verify the last statement. Regular spaces are spaces simultaneously satisfying the axioms T0 and T3. Let X =
Q
α∈A Xα and the factors Xα be regular. Take distinct points x, y ∈ X. There exists α such that prα(x) 6= prα(y).
Since Xα ∈ T0, one of the points prα(x) and prα(y) has a neighborhood not containing the other
point. Suppose that the open set V ⊂ Xα contains prα(x) and does not contain prα(y).
Then x ∈ pr−1
α (V ) and y /∈ pr−1
α (V ). Thus, X is a T0-space.
To verify the axiom T3, we need for an arbitrary neighborhood Ox of the point x ∈ X to find
a neighborhood Ux such that Cl(Ux) ⊂ Ox. 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. (22.1)
Since Xαi satisfies the axiom T3, there exists a neighborhood Oxαi
, such that Cl(Oxαi
) ⊂ Vi
.
Set
Ux =
\
{pr−1
αi
(Oxαi
) : i = 1, . . . , k}.
Then Cl(Ux) ⊂
T
{pr−1
αi
(Cl(Oxαi
)) : i = 1, . . . , k} ⊂ T
{pr−1
αi
(Vi) : i = 1, . . . , k} ⊂ (22.1) ⊂ Ox.
22.9. A space X is called a T4-space, if for any disjoint pair of
sets F and T closed in X there exist disjoint neighborhoods OF and OT.
It is easy to verify that this condition is equivalent to the following: for every closed set F and every its neighborhood OF there exists a neighborhood O1F, such that Cl(O1F) ⊂ OF.
Another equivalent condition: any disjoint pair of closed sets can be enclosed in neighborhoods with disjoint closures.
The example of a set of more than two points with the antidiscrete topology shows that
T4 does not imply T0. The real line R, on which ∅, R and the infinite intervals of the form
(a, ∞) are open (Example 8.8.4 of Lecture 2), shows that the axiom T4 does not imply T3 either.
22.10. A space simultaneously satisfying the axioms T1 and T4 is called normal.
Every normal space is regular, since T3 follows from T1 and T4. At the same time,
as the “connected two-point set” shows, T0 plus T4 does not imply T3.
§ 23. Urysohn's lemma. The Brouwer–Tietze–Urysohn extension theorem for functions.
23.1. Urysohn's lemma. For any disjoint closed subsets F0 and F1
of a normal space X there exists a continuous function f : X → [0, 1], such that
f(x) = 0 for x ∈ F0 and f(x) = 1 for x ∈ F1.
Proof. The validity of Urysohn's lemma in the case F0 = ∅ or F1 = ∅ is obvious.
For every dyadic rational number r ∈ [0, 1] we define an open set Γr
such that
r < r0 =⇒ Cl(Γr) ⊂ Γr
0 , (23.1)
F0 ⊂ Γ0, F1 ⊂ X \ Γ1. (23.2)
There exists a neighborhood OF0, such that Cl(OF0) ∩ F1 = ∅. Set Γ0 = OF0 and Γ1 = X \ F1.
Next, there exists a neighborhood OCl(Γ0), such that Cl(OCl(Γ0)) ⊂ Γ1. Set Γ 1
2
= OCl(Γ0).
In a similar way we construct the sets Γ 1
4
and Γ 3
4
such that
Cl
Γ0
⊂ Γ 1
4
⊂ Cl
Γ 1
4
⊂ Γ 1
2
;
Cl
Γ 1
2
⊂ Γ 3
4
⊂ Cl
Γ 3
4
⊂ Γ1.
Continuing this process, we construct the desired family of sets Γr. Define the function f :
X → I by the formula
f(x) =
inf{r : x ∈ Γr} for x ∈ Γ1,
1 for x ∈ X \ Γ1.
Take numbers a, b from the interval (0, 1). Let us show that
f
−1
[0, a)
=
[
Γr : r < a
; (23.3)
f
−1
(b, 1]
=
[
X \ Cl(Γr) : r > b
. (23.4)
Equality (23.3) follows from the following property:
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f(x) < a ⇐⇒ there exists r such that r < a, and x ∈ Γr.
Next,
f(x) > b ⇐⇒ there exists r
0
such that b < r0
, and x /∈ Γr
0 . (23.5)
Conditions (23.1) and (23.5) imply that
f(x) > b ⇐⇒ there exists r such that b < r < r0
, and x /∈ Cl(Γr).
Thus, equality (23.4) is also verified. From (23.3) and (23.4), applying Proposition 13.3 of Lecture 3, follows the continuity of the function f.
The fulfilment of the condition f(x) = 0 for x ∈ F0 and f(x) = 1 for x ∈ F1 is obvious.
Assignment N 6
1. Prove that a space X is a T1-space if and only if all one-point subsets of X are closed.
Prove that among the topologies on a set X in which X is a T1-space,
there is a smallest one.
2. Prove that the line with the Zariski topology is a non-Hausdorff space.
3. Give an example of two distinct continuous mappings fi
: X → Y, i = 1, 2, of a space X into a non-Hausdorff space Y , coinciding on an everywhere dense subset
Z.
4. Prove that for continuous mappings f, g : X → Y into a Hausdorff space Y
the set of coincidence points {x ∈ X : f(x) = g(x)} is closed.
Can the Hausdorff condition on the image be dropped?
5. Prove that a space X is Hausdorff if and only if the diagonal
∆ = {(x, x) : x ∈ X} of the product is closed in X × X.
6. Prove that for a continuous mapping f : X → Y of a space X into a Hausdorff
space Y the graph of the mapping Γf = {(x, f(x)) : x ∈ X} is closed in X × Y . Does the
closedness of the graph of a mapping imply its continuity?
7. Prove that the space of Example 8.5.5 of Lecture 2 is Hausdorff but not regular.
8. Prove that the fulfilment of the axioms T0 and T3 is equivalent to the fulfilment of the axioms T1 and T3.
9. Prove that in a regular space any point and a closed set not containing it have disjoint neighborhoods whose closures do not intersect.
Prove that the following conditions on a space X are equivalent:
a) X — is a normal space;
b) for every closed set F and every its neighborhood OF there exists a neighborhood O1F, such that Cl(O1F) ⊂ OF;
c) any disjoint pair of closed sets can be enclosed in neighborhoods with disjoint closures.
10. Prove that every subspace of a T0 (respectively T1, T2, regular) space is a T0 (respectively T1, T2, regular) space.
Prove that every closed subset of a normal space is a normal space.
11. Let topologies T1 ≤ T2 be given on a set X. If the space (X, T1) is Hausdorff
(regular, normal), what can be said about the separation properties of the space (X, T2)?
If the space (X, T2) is Hausdorff (regular, normal), what can be said about the
separation properties of the space (X, T1)?
12. Let f, g : X → R — be continuous functions. Prove that the functions f + g ((f + g)(x) =
f(x)+g(x)), fg ((fg)(x) = f(x)g(x)) and
1
f
(
1
f
(x) = 1
f(x)
), provided f(x) 6= 0, x ∈ X, are continuous.
13. Let f, g : X → R — be continuous functions. Prove that the functions max{f, g}, min{f, g}
and |f| = max{f, −f} are continuous.
14. Prove that any continuous function on the line with the Zariski topology is
constant.
15. Does there exist a continuous function on the plane R
2
, taking the value 0 on the coordinate
axes Ox and Oy and the value 1 on the graph of the hyperbola y = 1/x?
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Prove Urysohn's lemma using the Brouwer–Tietze–Urysohn theorem.
16. Shrinking lemma. Let u = {U1, . . . , Uk} — be an open cover of a normal space X. Prove that there exists an open cover v = {V1, . . . , Vk} of the space X,
such that Cl(Vi) ⊂ Ui
, i = 1, . . . , k.
17. Partition of unity. Let ϕ : X → R — be a continuous function. The open set
Uϕ =
x ∈ X : ϕ(x) 6= 0
is called the support of the function ϕ and is denoted supp(ϕ).
Let u = {U1, . . . , Uk} — be an open cover of X. A family of continuous functions ϕi
: X →
[0, 1], i = 1, . . . , k, is called a partition of unity subordinate to the cover u, if
Cl(supp(ϕi)) ⊂ Ui
, i = 1, . . . , k,
and X
k
i=1
ϕi = 1.
Prove that for every finite open cover u = {U1, . . . , Uk} of a normal
space X there exists a partition of unity {ϕ1, . . . , ϕk}, subordinate to the cover u.
Additional problems for Assignment N 6
18. Prove that in every finite T0-space
a) there exists an isolated point;
b) the set of isolated points is everywhere dense.
19. Prove that every regular space satisfying the second countability axiom is normal.
Prove that every countable regular space is normal.
20. Prove that a linearly ordered space with the interval topology is normal.
21. Which of the separation axioms are preserved in the image under continuous mappings?
22. Does there exist a regular space X, containing more than two points, on which
every continuous function f : X → R is constant?
23. Describe all continuous functions on the spaces of Example 8.5 of Lecture 2.
24. Prove that any uniformly continuous function on the interval (0, 1) ⊂ R can
be extended to R.
25. Let A — be a closed subset of a metric space (X, ρ), f : A → I —
a continuous mapping. Prove that the mapping
g(x) = (
inf{f(a) + ρ(x,a)
ρ(x,A) − 1 : a ∈ A} x ∈ X \ A
f(x) x ∈ A
is a continuous extension of f to X.
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