Download 6 | Continuous Functions

6 | Continuous Functions
Let X , Y be topological spaces. Recall that a function � : X → Y is continuous if for every open
set U ⊆ Y the set � −1 (U) ⊆ X is open. In this chapter we study some properties of continuous
functions. We also introduce the notion of a homeomorphism that plays a central role in topology: from
the topological perspective interesting properties of spaces are the properties that are preserved by
homeomorphisms.
6.1 Proposition. Let X , Y be topological spaces. A function � : X → Y is continuous if and only if for
every closed set A ⊆ Y the set � −1 (A) ⊆ X is closed.
Proof. Assume that � : X → Y is a continuous function and let A ⊆ Y be a closed set. We have
� −1 (A) = X r � −1 (Y r A)
The set Y r A is open in Y so by continuity of � the set � −1 (Y r A) ⊆ X is open in X . It follows that
� −1 (A) is closed in X .
Conversely, assume that � : X → Y is a function such that for every closed set A ⊆ Y the set � −1 (A) ⊆ X
is closed. Let U ⊆ Y be an open set. We have
� −1 (U) = X r � −1 (Y r U)
The set Y r U is closed in Y so by assumption the set � −1 (Y r U) is closed in X . If follows that
� −1 (U) is open in X . Therefore � is a continuous function.
For metric spaces continuous functions are precisely the functions that preserve convergence of
sequences:
6.2 Proposition. Let (X � �) be a metric space, let Y be a topological space, and let � : X → Y be a
function. The following conditions are equivalent:
1) � is continuous.
39
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40
2) If {�� } ⊆ X is a sequence and �� → � for some � ∈ X then �(�� ) → �(�).
X
�1
�2
Proof. 1) ⇒ 2) Exercise.
�
�
Y
�(�1 )
�(�2 )
�(�)
2) ⇒ 1) Let A ⊆ Y be a closed set. We will show that the set � −1 (A) is closed in X . By Proposition
5.8 it suffices to show that if {�� } ⊆ � −1 (A) is a sequence and �� → � then � ∈ � −1 (A).
If �� → � then by assumption we have �(�� ) → �(�). Since {�(�� )} ⊆ A and A is a closed set, thus by
Proposition 5.8 we obtain that �(�) ∈ A, and so � ∈ � −1 (A).
The implication 1) ⇒ 2) in Proposition 6.2 holds for maps between general topological spaces:
6.3 Proposition. Let � : X → Y be a continuous function of topological spaces. If {�� } ⊆ X is a
sequence and �� → � for some � ∈ X then �(�� ) → �(�).
Proof. Exercise.
6.4 Example. We will show that the implication 2) ⇒ 1) in Proposition 6.2 is not true if X is a general
topological space. Let X be the space defined in Example 5.16: X = R with the topology
T = {U ⊆ R | U = ? or U = (R r S) for some countable set S ⊆ R}
Recall that if {�� } is a sequence in X then �� → � if and only if there exists N > 0 such that �� = �
for all � > N. Let � : X → X be a function given by
�
0 if � ∈ (0� 1)
�(�) =
1 if � �∈ (0� 1)
This function is not continuous since the set {0} is closed in X and the set (0� 1) = � −1 ({0}) is not
closed in X . On the other hand let {�� } ⊆ X be a sequence and let �� → �. There is N > 0 such that
�� = � for � > N, so �(�� ) = �(�) for all � > N and so �(�� ) → �(�).
6.5 Proposition. If � : X → Y and � : Y → Z are continuous functions then the function �� : X → Z is
also continuous.
Proof. Exercise.
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41
Frequently functions � : X → Y are constructed by gluing together several functions defined on
subspaces of X . The next two facts are useful for verifying that functions obtained in this way are
continuous.
6.6 Open Pasting
Lemma. Let X , Y be topological spaces and let {U� }�∈I be a family of open sets in
�
X such that �∈I U� = X . Assume that for � ∈ I we have a continuous function �� : U� → Y such that
�� (�) = �� (�) if � ∈ U� ∩ U� . Then the function � : X → Y given by �(�) = �� (�) for � ∈ U� is continuous.
X
U3
U1
U2
�
Y
Proof. Let V ⊆ Y be an open set. We will show that the set � −1 (V ) ⊆ X is open. Since
we have
�
�
� −1 (V ) =
� −1 (V ) ∩ U� =
��−1 (V )
�∈I
�∈I
�
�∈I
U� = X
Since �� : U� → Y is a continuous function the set ��−1 (V ) is open in U� . Also, since U� is open in X
thus by Exercise 5.7 we obtain that the set ��−1 (V ) is open in X . Thus � −1 (V ) is an open set.
6.7 Closed Pasting Lemma. Let X , Y be topological spaces and let A1 � A2 ⊆ X be closed sets such
that A1 ∪ A2 = X . Assume that for � = 1� 2 we have a continuous function �� : A� → Y such that
�1 (�) = �2 (�) if � ∈ A1 ∩ A2 . Then the function � : X → Y given by �(�) = �� (�) for � ∈ A� is continuous.
Proof. Exercise.
6.8 Example. Let � : R → R be the absolute value function, �(�) = |�|. On the set A1 = (−∞� 0] this
function is given by �|A1 (�) = −�, and on A2 = [0� +∞) it is given by �|A2 (�) = �. Since both �|A1 and
�|A2 are continuous functions and A1 � A2 are closed sets in R thus by the Closed Pasting Lemma 6.7
we obtain that � : R → R is continuous.
6.9 Note. �Lemma 6.7 holds if instead of two closed sets we take any finite number of sets A1 � � � � � A�
such that ��=1 A� = X . On the other hand the statement of the lemma does not hold in general if the
collection of sets {A� } is infinite.
6.10 Definition. A homeomorphism is a continuous function � : X → Y such that � is a bijection and
the inverse function � −1 : Y → X is continuous.
6.11 Proposition. 1) For any topological space the identify function idX : X → X given by idX (�) = �
is a homeomorphism.
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42
2) If � : X → Y and � : Y → Z are homeomorphisms then the function �� : X → Z is also a
homeomorphism.
3) If � : X → Y is a homeomorphism then the inverse function � −1 : Y → X is also a homeomorphism.
4) If � : X → Y is a homeomorphism and Z ⊆ X then the function �|Z : Z → �(Z ) is also a homeomorphism.
Proof. Exercise.
6.12 Note. If � : X → Y is a continuous bijection then � need not be a homeomorphism since the
inverse function � −1 may be not continuous. For example, let X = {�1 � �2 } be a space with the discrete
topology and let Y = {�1 � �2 } be a space with the antidiscrete topology. Let � : X → Y be given by
�(�� ) = �� . The function � is continuous but � −1 is not continuous since the set {�1 } is open in X , but
the set (� −1 )−1 ({�1 }) = {�1 } is not open in Y .
6.13 Proposition. Let � : X → Y be a continuous bijection. The following conditions are equivalent:
(i) The function � is a homeomorphism.
(ii) For each open set U ⊆ X the set �(U) ⊆ Y is open.
(iii) For each closed set A ⊆ X the set �(A) ⊆ Y is closed.
Proof. Exercise.
6.14 Example. Recall that S 1 denotes the unit circle:
S 1 = {(�1 � �2 ) ∈ R2 | �12 + �2 = 1}
The function � : [0� 1) → S 1 given by �(�) = (cos 2π�� sin 2π�) is a continuous bijection, but it is not a
homeomorphism since the set U = [0� 12 ) is open in [0� 1), but �(U) is not open in S 1 .
0
U
1
2
1
�
�(U)
�( 12 )
�(0)
6.15 Definition. We say that topological spaces X , Y are homeomorphic if there exists a homeomorphism
� : X → Y . In such case we write: X ∼
= Y.
6.16 Note. Notice that if X ∼
= Y and Y ∼
= Z then X ∼
= Z.
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6.17 Example. For any � < � and � < � the open intervals (�� �)� (�� �) ⊆ R are homeomorphic. To
see this take e.g. the function � : (�� �) → (�� �) defined by
�
�
�
�
�−�
�� − ��
�(�) =
�+
�
�−�
�−�
)
� (�
=
This function is a continuous bijection. Its inverse function � −1 : (�� �) →
�
(�� �) is given by
�
�
�
�
�
�−�
�� − ��
−1
� (�) =
�+
�
�
�−�
�−�
so it is also continuous. By the same argument for any � < � and � < � the closed intervals
[�� �]� [�� �] ⊆ R are homeomorphic.
6.18 Note. In Chapter 7 we will show that an open interval (�� �) is not homeomorphic to a closed
interval [�� �].
6.19 Example. We will show that for any � < � the open interval (�� �) is homeomorphic to R. Since
(�� �) ∼
= (−1� 1) it will be enough to check that R ∼
= (−1� 1). Take the function � : R → (−1� 1) given by
�(�) =
�
1 + |�|
This function is a continuous bijection with the inverse function
� −1 : (−1� 1) → R is given by
� −1 (�) =
�
1 − |�|
Since � −1 is continuous we obtain that � is a homeomorphism.
1
� = �(�)
−1
6.20 Note. If spaces X and Y are homeomorphic then usually there are many homeomorphisms
� : X → Y . For example, the function � : (−1� 1) → R given by
� �
�(�) = tan π2 �
is another homeomorphism between the spaces (−1� 1) and R.
6.21 Example. We will show that for any point �0 ∈ S 1 there is a homeomorphism S 1 r {�0 } ∼
= R.
1
Denote by S(0�1)
⊆ R the circle of radius 1 with the center at the point (0� 1) ∈ R2 :
1
S(0�1)
:= {(�1 � �2 ) ∈ R2 | �12 + (�2 − 1)2 = 1}
1 r{(0� 2)}.
It is easy to check that for �0 ∈ S 1 the space S 1 r{�0 } is homeomorphic to the space X = S(0�1)
Likewise, it is easy to check that R is homeomorphic to the subspace Y ⊆ R2 that consists of all points
of the �-axis:
Y := {(�1 � 0) ∈ R2 | �1 ∈ R}
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It is then enough to show that X ∼
= Y . A homeomorphism � : X → Y can be constructed as follows. For
any point � ∈ X there is a unique line in R2 that passes through � and though the point (0� 2) ∈ R2 .
We define �(�) to be the point of intersection of this line with the �-axis:
�
(0, 2)
�(�)
The function � is called the stereographic projection.
In a similar way we can construct a stereographic projection in any dimension � ≥ 1 that gives a
homeomorphism between the space S � r {�0 } (i.e. the �-dimensional sphere with one point deleted)
and the space R� :
(0� 0� 2)
�
�(�)
Exercises to Chapter 6
E6.1 Exercise. Consider the set of rational numbers Q as a subspace of R. Show that Q is not
homeomorphic to a space with the discrete topology.
E6.2 Exercise. Prove Proposition 6.3.
E6.3 Exercise. Prove Proposition 6.5.
E6.4 Exercise. Prove Lemma 6.7.
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E6.5 Exercise. Prove Proposition 6.13.
E6.6 Exercise. Let X be a topological space and let �� � : X → R be continuous functions.
a) Show that the set
is closed in X .
A = {� ∈ X | �(�) ≥ �(�)}
b) Let � : X → R be a function given by �(�) = max{�(�)� �(�)}. Show that � is continuous.
E6.7 Exercise. Let �� � : R → R be continuous functions such that �(�) > �(�) for all � ∈ R. Define
subspaces X , Y of R2 as follows.
X := {(�� �) ∈ R2 | �(�) ≤ � ≤ �(�)}
Show that X ∼
= Y.
Y := {(�� �) ∈ R2 | 0 ≤ � ≤ 1}
E6.8 Exercise. Let �0 = (0� 0) ∈ R2 and let B(�0 � 1) ⊆ R2 be a closed ball defined by the Euclidean
metric �:
B(�0 � 1) = {� ∈ R2 | �(�� �0 ) ≤ 1}
Define subspaces X � Y ⊆ R2 as follows:
Show that X ∼
= Y.
X := R2 r {�0 }
Y := R2 r B(�0 � 1)
E6.9 Exercise. Let (X � �) be a metric space. A subspace Y ⊆ X is a retract of X if there exists a
continuous function � : X → Y such that �(�) = � for all � ∈ Y . Show that if Y ⊆ X is a retract of X
then Y is a closed in X .