§ 12. Closure, interior, and boundary of a subset.
12.1. Definition. Let X be a space and M ⊂ X. The intersection of all closed sets containing M
\
{F : F is closed in X and M ⊂ F},
which is closed according to condition (3’) after Definition 8.1, is called the closure of the set M
in X and is denoted by ClX(M) or Cl(M) (in the literature the notation
M is also common).
Cl(M) is the smallest closed subset of the space X containing M.
The union of all open sets contained in M
[
{U : U is open in X and U ⊂ M},
which is open according to condition (3) of Definition 8.1, is called the interior of the set M in X
and is denoted by IntX(M) or Int(M).
Int(M) is the largest open subset of the space X contained in M.
Always
Int(M) ⊂ M ⊂ Cl(M).
A subset M is open (respectively, closed) in X if and only if
Int(M) = M (respectively, Cl(M) = M).
12.2. Proposition. Let X be a space and M ⊂ X. Then
Cl(M) = X \ Int(X \ M).
The proof uses De Morgan's laws.
Cl(M) = \
{F : F is closed in X and M ⊂ F} =
\
{X \ U : U is open in X and U ⊂ X \ M} =
= X \
[
{U : U is open in X and U ⊂ X \ M} = X \ Int(X \ M).
12.3. Definition. Let X be a space and M ⊂ X. The set Cl(M) \ Int(M) is called the boundary of the set M. Notation Bd(M).
12.4. Neighborhoods. An arbitrary open set containing the set M ⊂ X
is called a neighborhood of the set M in the space X.
Neighborhoods of the set M are usually denoted by the symbol OM (Ox is a neighborhood of the point x,
ε-balls Oε(x) centered at the point x are neighborhoods of the point x in the metric topology). The intersection
of a finite number of neighborhoods of the set M is a neighborhood of it.
If a point x ∈ X has a neighborhood consisting of a single point, then x is called an isolated point of the space X. A space X is discrete if and only if all of
its points are isolated.
12.5. Definition. A point x ∈ M is called an interior point of the set M ⊂ X if
there exists a neighborhood Ox ⊂ M.
A point x ∈ X is called a point of contact (adherent point) of the set M ⊂ X if Ox ∩ M 6= ∅ for
every neighborhood Ox of the point x.
A point x ∈ X is called a limit point of the set M ⊂ X if every neighborhood
Ox of it contains a point of M different from x.
A point x ∈ X is called a boundary point of the set M if it is a point of contact of M
but not an interior point.
12.6. Proposition. The set of all interior points (points of contact, boundary
points) of the set M ⊂ X coincides with Int(M) (Cl(M), Bd(M) respectively).
20
We give the proof only of the first statement. It follows from Definition 12.1 that
every interior point of the set M belongs to Int(M). Conversely, if x ∈ Int(M), then
Ox = Int(M) ⊂ M.
12.7. Definition. Let (X, ρ) be a metric space and ξ = (xn)n∈N a sequence of points xn ∈ X. The sequence ξ is said to converge to a point x ∈ X if for
every ε > 0 there exists such n0 ∈ N that
xn ∈ Oε(x) for all n ≥ n0.
12.8. Remark. If (X, ρ) is a metric space, τρ is the metric topology, and
M ⊂ X, then x ∈ Cl(M) if and only if there exists a sequence (xn)n∈N
of points of the set M converging to x.
§ 13. Continuous mappings of topological spaces.
13.1. Definition. A mapping f : X → Y of a space X into a space Y is called
continuous at a point x ∈ X if for any neighborhood Oy of the point y = f(x) there is a
neighborhood Ox of the point x such that f(Ox) ⊂ Oy.
A mapping f : X → Y is called continuous if it is continuous at every point
x ∈ X.
13.2. Proposition. For a mapping f : X → Y the following conditions are equivalent:
1) f is continuous;
2) the preimage f
−1
(U) of every open set U in Y is open in X;
3) the preimage f
−1
(F) of every closed set F in Y is closed in X;
4) f
Cl(Z)
⊂ Cl
f(Z)
for every set Z ⊂ X.
Proof. Let us check the following chain of implications: 1) ⇒ 4) ⇒ 3) ⇒ 2) ⇒ 1).
1) ⇒ 4). Take x ∈ Cl(Z). We must show that f(x) ∈ Cl
f(Z)
, i.e. that f(x) is a point of
contact of the set f(Z). By the continuity of f, for an arbitrary neighborhood Of(x)
there exists a neighborhood Ox such that f(Ox) ⊂ Of(x). Since x ∈ Cl(Z), we have Ox ∩ Z 6= ∅.
Take a point z ∈ Ox∩Z. Then f(z) ∈ f(Ox∩Z) ⊂ f(Ox)∩f(Z) ⊂ Of(x)∩f(Z). Hence,
Of(x) ∩ f(Z) 6= ∅, i.e. f(x) ∈ Cl
f(Z)
.
4) ⇒ 3). Suppose that there exists a closed set F ⊂ Y whose preimage f
−1
(F) is not closed in X. There exists a point
x ∈ Cl
f
−1
(F)
\ f
−1
(F). (13.1)
From 4) it follows that f(x) ∈ Cl
f
f
−1
(F)
⊂ Cl(F) = F. On the other hand, f(x) ∈/ F according to
(13.1). We obtain a contradiction.
The implication 3) ⇒ 2) follows from the equality f
−1
(Y \U) = X \f
−1
(U). Finally, 2) ⇒ 1). Take
a point x ∈ X and a neighborhood Of(x). The set f
−1
Of(x)
is open by 2). Since
x ∈ f
−1
Of(x)
, the set f
−1
Of(x)
is indeed a neighborhood Ox, for which f(Ox) =
f
f
−1
Of(x)
⊂ Of(x).
The following statement supplements Proposition 13.2.
13.3. Proposition. For the continuity of a mapping f : X → Y it suffices that
the preimages f
−1
(U) of the elements U of some subbase B of the space Y be open.
Proof. Let x ∈ X, and Of(x) be an arbitrary neighborhood. There exist
elements U1, . . . , Uk ∈ B such that f(x) ∈
T
k
i=1
Ui ⊂ Of(x). The sets f
−1
(Ui) are open by hypothesis.
Setting Ox =
T
k
i=1
f
−1
(Ui), we have f(Ox) ⊂ Of(x).
13.4. Rules for constructing continuous mappings. Let X, Y, Z be topological
spaces.
1. The constant mapping f = consty0
: X → Y, sending the whole space X to a point
y0 ∈ Y, is continuous. The preimage f
−1
(U) of any open set U ⊂ Y is either empty or equal to
X.
21
2. Let M be a subset of the space (X, T ). Denote by iM : M → X the mapping
of identical embedding, assigning to a point x ∈ M itself in X. The embedding
mapping iM : M → X of the subspace (M, T |M) into (X, T ) is continuous.
3. Theorem on the composite function. The composition g ◦ f : X → Z of continuous mappings f :
X → Y and g : Y → Z is continuous.
4. The restriction f|M of a continuous mapping f : X → Y to a subspace M ⊂ X
is continuous.
The statement follows from items 2 and 3, and the equality f|M = f ◦ iM.
5. A mapping f : X → Y is continuous if X can be represented as a union of open
sets X =
S
α∈A Oα such that f|Oα is continuous for every α ∈ A.
The continuity of the mapping at every point of the space X is obvious.
6. Let the space X be the union of a finite number of its closed subsets Fi
, i = 1, . . . , k, and let f : X → Y be a mapping such that f|Fi is continuous for each
i. Then the mapping f is continuous.
Take an arbitrary closed set Φ ⊂ Y. Then
f
−1
(Φ) = S
k
i=1
f|Fi
−1
(Φ).
Indeed, the inclusion ⊃ is obvious. Now let x ∈ f
−1
(Φ). Then x belongs to some Fi and, hence, x ∈
f|Fi
−1
(Φ).
From condition 3) of Proposition 13.2 and the continuity of the mappings f|Fi
it follows that the sets
f|Fi
−1
(Φ) are closed. Hence f
−1
(Φ) is also closed. Therefore f is continuous according to Proposition 13.2.
7. To an arbitrary topological space X one can assign its discrete duplicate Xd, i.e. the space on the same set X, endowed with the discrete topology (see item
8.1). Then the identity mapping id : Xd → X is continuous.
The topology of the space Xd is generated by the metric of Example 10.2.3 (x 6= y ⇒ ρ(x, y) = 1). Thus
any space is a continuous image of a metrizable space.
8. Let T1 ≥ T2 be topologies on X. Then the identity mapping id : (X, T1) → (X, T2)
is continuous.
13.5. Continuous mappings of metric spaces. Let (X, ρ1) and (Y, ρ2) be
metric spaces. A mapping f : X → Y is called continuous at a point x0 ∈ X
in the Cauchy sense if for any ε > 0 there exists such δ > 0 that
f
Oδ(x0)
⊂ Oε
f(x0)
. (13.2)
A mapping f : X → Y is called continuous if it is continuous at every point x0 ∈ X.
From the definition of the metric topology it follows that any continuous mapping of metric spaces (X, ρ1) and (Y, ρ2) is a continuous mapping of topological
spaces (X, Tρ1
) and (Y, Tρ2
).
13.6. Example. For any subset M of a metric space (X, ρ) the mapping
ρM : X → R, defined by the formula ρM(x) = inf{ρ(x, a) : a ∈ M} (the distance from the point to
the set), is continuous.
For any points x, x0 ∈ X and a ∈ M, using the triangle inequalities, we have
ρ(x
0
, a) ≤ ρ(x, a) + ρ(x, x0
).
Hence,
ρM(x
0
) = inf{ρ(x
0
, a) : a ∈ M} ≤ ρM(x) + ρ(x, x0
), where ρM(x) = inf{ρ(x, a) : a ∈ M}, i.e.
ρ(x, x0
) ≥ ρM(x
0
) − ρM(x). (13.3)
Carrying out similar reasoning, using the inequality ρ(x, a) ≤ ρ(x
0
, a) + ρ(x, x0
), we have
ρ(x, x0
) ≥ ρM(x) − ρM(x
0
). (13.4)
From (13.3) and (13.4) it follows the inequality
ρ(x, x0
) ≥ |ρM(x) − ρM(x
0
)|,
which allows us to obviously complete the proof.
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13.7. Definition of continuity by Heine. Let X and Y be metric spaces.
A mapping f : X → Y is called continuous at a point x0 ∈ X if for every sequence ξ = (xn : n ∈ N) converging to the point x0, the sequence f(ξ) =
f(xn) : n ∈ N)
converges to the point f(x0).
13.8. Theorem. The definitions of continuity of a mapping f : X → Y at a point x0 ∈ X by
Cauchy and by Heine are equivalent.
Proof. Let f be continuous at the point x0 by Cauchy and let ξ = (xn : n ∈ N) be a
sequence converging to the point x0. Take an arbitrary ε > 0. There exists such
δ > 0 that condition (13.1) is satisfied. Since ξ converges to x0, there exists such n0 ∈ N that
xn ∈ Oδ(x0) for all n ≥ n0.
But then
f(xn) ∈ f
Oδ(x0)
⊂ Oε
f(x0)
for all n ≥ n0.
Hence, f(ξ) converges to f(x0).
Conversely, let f be continuous at the point x0 by Heine. Take δn = 1/n. If f is not
continuous at the point x0 by Cauchy, then for some ε > 0 and for every n there exists such a
point xn ∈ Oδn
(x0), that f(xn) ∈/ Oε
f(x0)
. Then the sequence ξ = (xn : n ∈ N) converges
to x0, but the sequence f(ξ) =
f(xn) : n ∈ N
does not converge to f(x0), since f(ξ) ∩
Oε
f(x0)
= ∅. We obtain a contradiction.
§ 14. Homeomorphism.
14.1. Definition. Let f : X → Y be a one-to-one mapping of a space
X onto a space Y. Suppose, in addition, that the mapping f and the mapping f
−1
inverse to it are continuous. Then f is called a homeomorphism, and the spaces X and Y are called homeomorphic.
Notation X ≈ Y .
The relation X ≈ Y is an equivalence relation on topological spaces.
14.2. Examples.
1. All segments of the real line are homeomorphic to the segment [0, 1].
2. All intervals of the real line are homeomorphic to the interval (0, 1).
3. The open ball in R
n centered at the origin and of radius 1 will be denoted by
Dn.
The open ball Dn is homeomorphic to the space R
n.
Proof. For x = (x1, . . . , xn) ∈ Dn set
f(x) =
tg
π
2
kxk
·
x
kxk
, if x 6= 0,
0, if x = 0,
where ||x|| =
s
Pn
i=1
|xi
|
2. It is clear that f : Dn → R
n is a bijection.
The inverse mapping f
−1
: R
n → Dn is defined as follows (check independently that it is indeed inverse to f):
f
−1
(x) = 2
π
arctg
kxk
x
kxk
, if x 6= 0,
0, if x = 0.
The continuity of the mappings f and f
−1 is proved in the same way.
Let us check the continuity of the mapping f at the point x0 ∈ R
n, x0 6= 0. Let M > 0 and the neighborhood
Oδ
0 (x0) be such that
tg
π
2
kxk
kxk < M for any point x ∈ Oδ
0 (x0). For > 0 choose Oδ
00 (x0) so
that |
tg
π
2
kxk
kxk −
tg
π
2
kx0k
kx0k
| <
2||x0|| for x ∈ Oδ
00 (x0). Set δ = min{δ
0
,
2M , δ00}. Then for
x ∈ Oδ(x0) we have
||f(x) − f(x0)|| = ||tg
π
2
kxk
kxk
· x −
tg
π
2
kx0k
kx0k
· x0|| ≤
||tg
π
2
kxk
kxk
· x −
tg
π
2
kxk
kxk
· x0|| + ||tg
π
2
kxk
kxk
· x0 −
tg
π
2
kx0k
kx0k
· x0|| =
= |
tg
π
2
kxk
kxk
| · ||x − x0|| + |
tg
π
2
kxk
kxk
−
tg
π
2
kx0k
kx0k
| · ||x0|| < M
2M
+
2||x0|| · ||x0|| < .
23
The continuity of the mapping f at the point x0 has been established. (Check the continuity of the mapping f at the point
0 independently.)
4. S
n−1 =
(x1, . . . , xn) ∈ R
n : x
2
1 + . . . + x
2
n = 1
— the sphere in R
n of radius 1 centered at the origin
of coordinates.
The set En−1 =
(x1, . . . , xn) ∈ S
n−1
: xn = 0
— the equator of the sphere S
n−1
.
The equator divides the sphere S
n−1 into two hemispheres:
S
n−1
+ =
(x1, . . . , xn) ∈ S
n−1
: xn > 0
;
S
n−1
− =
(x1, . . . , xn) ∈ S
n−1
: xn < 0
.
The hemispheres S
n−1
+ and S
n−1
+ are homeomorphic.
5. The open ball Dn−1
is homeomorphic to the hemisphere S
n−1
+ .
The homeomorphism h : Dn−1 → S
n−1
+ is defined by the equality
h
x1, . . . , xn−1
=
x1 . . . , xn−1,
q
1 − x
2
1 − . . . − x
2
n−1
.
The inverse mapping h
−1
is given by the equality
h
−1
x1, . . . , xn−1, xn
=
x1, . . . , xn−1
.
Assignment N 3
1. Prove:
(1) ClA = A ∪ BdA;
(2) IntA = A \ BdA;
(3) X \ BdA = IntA ∪ Int(X \ A);
(4) BdA = ClA ∩ Cl(X \ A);
(5) A is closed if and only if BdA ⊂ A;
(6) A is open if and only if BdA ∩ A = ∅;
(7) Ad \ A = BdA \ A, where Ad is the set of limit points of A;
(8) A is closed if and only if Ad ⊂ A.
2. Are the following relations valid:
(1) if A ⊂ B, then IntA ⊂ IntB, ClA ⊂ ClB;
(2) Int(A ∩ B) = IntA ∩ IntB; Cl(A ∩ B) = ClA ∩ ClB; Bd(A ∩ B) = BdA ∩ BdB;
(3) Int(A ∪ B) = IntA ∪ IntB; Cl(A ∪ B) = ClA ∪ ClB; Bd(A ∪ B) = BdA ∪ BdB;
(4) Bd(A ∩ B) ⊂ BdA ∪ BdB;
(5) Bd(X \ A) = BdA, Bd(ClA) ⊂ BdA, Bd(IntA) ⊂ BdA;
(6) (A ∪ B)
d = Ad ∪ Bd
?
3. Let C : 2X → 2
X be an operator on the family of subsets of the set X, satisfying
the conditions:
(1) C(∅) = ∅;
(2) M ⊂ C(M);
(3) C(M ∪ N) = C(M) ∪ C(N);
(4) C(C(M)) = C(M).
Prove that on X there exists a topology such that Cl(M) = C(M) for all M ∈ 2
X.
4. Find the closures of the subsets of the lexicographically ordered square I
2
:
(1) C = {(x, 0) : 0 < x < 1};
(2) D = {(x, 1
2
) : 0 < x < 1};
(3) E = {(
1
2
, y) : 0 < y < 1}.
5. Let (X, ρ) be a metric space, M ⊂ X. Prove that the following conditions
are equivalent:
(1) x ∈ Cl(M);
(2) there exists a sequence (xn)n∈N of points of the set M converging to x;
(3) ρ(x, M) = inf{ρ(x, y) : y ∈ M} = 0.
6. Find the interior and boundary of the following subsets of R
2
:
(1) A = {(x, y) : y = 0};
(2) B = {(x, y) : x > 0, y 6= 0};
(3) C = A ∪ B;
24
(4) D = {(x, y) : x ∈ Q};
(5) E = {(x, y) : 0 < x2 − y
2 ≤ 1};
(6) F = {(x, y) : x 6= 0, y ≤
1
x
}.
7. Let A ⊂ Y ⊂ X. Prove that ClY (A) = ClX(A) ∩ Y . Is the analogous formula true
for the interiors of the set A in X and Y ?
8. Find the closure of the set {
1
n
: n ∈ N} on the line in the Zariski topology.
9. Prove that the following conditions on a mapping f : X → Y are equivalent:
(a) the mapping f is continuous;
(d) for any subset B ⊂ Y we have Cl(f
−1B) ⊂ f
−1
(ClB);
(e) for any subset B ⊂ Y we have f
−1
(IntB) ⊂ Int(f
−1B).
10. Prove that the identity mapping id : (X, T1) → (X, T2) is continuous if and
only if T1 ≥ T2 (the topology T1 on X is not weaker than the topology T2).
11. Let f : X → Y be a continuous mapping, M ⊂ X. Will the image of a limit point
of the set M be a limit point of the set f(M)?
12. Let Y be a linearly ordered space with the order topology, f, g :
X → Y continuous mappings.
a) Prove that the set {x ∈ X : f(x) ≤ g(x)} is closed in X.
b) Prove that the mapping h(x) = min{f(x), g(x)} is continuous.
13. Let X be a countable union of closed sets {Ai
: i ∈ N}, the restrictions
f|Ai
of a mapping f : X → Y to the subsets Ai being continuous, i ∈ N. Will the mapping f
be continuous?
14. A mapping of metric spaces f : X → Y is called an isometric embedding if for any points x, y ∈ X we have ρ(x, y) = ρ(f(x), f(y)). Prove that any
isometric embedding is continuous.
15. A mapping f of a metric space X into itself is called contracting (Lipschitz) if there exists 0 < α < 1 (respectively α > 0) such that for any points x, y ∈ X
we have ρ(f(x), f(y)) ≤ αρ(x, y). Prove that any contracting (any Lipschitz) mapping of a metric space X is continuous.
16. Prove that any continuous mapping of the segment [0, 1] into itself has a fixed
point.
Additional problems for Assignment N 3
17. State conditions on an operator I : 2X → 2
X such that there exists a topology on X for
which Int(M) = I(M) for all M ∈ 2
X.
18. The subset of the space `
2
, consisting of all points (x1, ..., xk, ...), for which 0 ≤
xk ≤ 1/2
k
, k ∈ N, is called the Hilbert cube. Prove that the Hilbert cube is a closed
subset of `
2
; the interior of the Hilbert cube in `
2 is the empty set.
19. List all the distinct sets that can be obtained from a single set,
applying to it successively the operations Cl and Int.
20. For which n ∈ N can one construct on the line n pairwise disjoint open sets having the same boundary?
21. Prove that on the space T (of countable transfinites T(ω1) and the first uncountable
transfinite ω1, Example 7.1 of Lecture 1) with the order topology Cl(T(ω1)) = T, and there does not
exist a sequence of points of T(ω1) converging to ω1.
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