Download Review of basic topology concepts

Chapter I
Topology Preliminaries
In this chapter we discuss (with various degrees of depth) several fundamental
topological concepts. The fact that the material is quite extensive is based on the
point of view that any competent mathematician - regardless of expertise area should know at least “this much topology,” and this chapter is thought to be the
“last push” in the attempt of reaching this goal. In particular, Section 2 offers an
exposition that is (unfortunately) seldom covered in many topology texts.
1. Review of basic topology concepts
In this lecture we review some basic notions from topology, the main goal being
to set up the language. Except for one result (Urysohn Lemma) there will be no
proofs.
Definitions. A topology on a (non-empty) set X is a family T of subsets of
X, which are called open sets, with the following properties:
(top1 ): both the empty set ∅ and the total set X are open;
(top2 ): an arbitrary union of open sets is open;
(top3 ): a finite intersection of open sets is open.
In this case the system (X, T ) is called a topological space.
If (X, T ) is a topological space and x ∈ X is an element in X, a subset N ⊂ X
is called a neighborhood of x if there exists some open set D such that x ∈ D ⊂ N .
A collection N of neighborhoods of x is called a basic system of neighborhoods
of x, if for any neighborhood M of x, there exists some neighborhood N in N such
that x ∈ N ⊂ M .
A collection V of neighborhoods of x is called a fundamental system of neighborhoods of x if for any neighborhood M of x there exists a finite sequence V1 , V2 , . . . , Vn
of neighborhoods in V such that x ∈ V1 ∩ V2 ∩ · · · ∩ Vn ⊂ M .
A topology is said to have the Hausdorff property if:
(h) for any x, y ∈ X with x 6= y, there exist open sets U 3 x and V 3 y such
that U ∩ V = ∅.
If (X, T ) is a topological space, a subset F ⊂ X will be called closed, if its
complement X r F is open. The following properties are easily derived from the
definition:
(c1 ) both the empty set ∅ and the total set X are closed;
(c2 ) an arbitrary intersection of closed sets is closed;
(c3 ) a finite union of closed sets is closed.
Examples 1.1. A. The standard topology T on R is defined as
T = D ⊂ R : D is a union of open intervals. .
The term “open interval” is suggestive of the fact that every such interval is an
open set. Likewise, every closed interval is closed in this topology. For a point
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CHAPTER I: TOPOLOGY PRELIMINARIES
x ∈ R, the collection of open intervals Jx = (x − ε, x + ε) : ε > 0 constitutes a
basic system of neighborhood of x.
B. The standard topology T on C is defined as
T = D ⊂ R : D is a union of open disks. .
Here an open disk is a set of the form
Dρ (ζ) = ξ ∈ C : |ξ − ζ| < ρ},
where ζ ∈ C is the center, and ρ > 0 is the radius. The term “open disk” is
suggestive of the fact that every such
disk is an open
set. For a number ζ ∈ C,
the collection of open disks Dζ = Dε (ζ) : ε > 0 constitutes a basic system of
neighborhood of ζ.
C. The standard topology on C can also be defined using open rectangles, which
are sets of the form
ε
δ
Rεδ (ζ) = ξ ∈ C : |Re ξ − Re ζ| < , |Im ξ − Im ζ| <
.
2
2
Here ζ ∈ C is the center, and ε, δ > 0 are the width and height of the rectangle.
Every such rectangle is an open set, and moreover, the topology defined above is
also given by
T = D ⊂ R : D is a union of open rectangles. .
For a number ζ ∈ C, the collection of open squares Vζ = Rεε (ζ) : ε > 0
constitutes a basic system of neighborhood of ζ.
Using the above properties of open/closed sets, one can perform the following
constructions. Let (X, T ) be a topological space and A ⊂ X be an arbitrary subset.
Consider the set Int(A) to be the union of all open sets D with D ⊂ A and consider
the set A to be the intersection of all closed sets F with F ⊃ A. The set Int(A)
◦
(sometimes denoted simply by A) is called the interior of A, while the set A is
called the closure of A. The properties of these constructions are summarized in
the following:
Proposition 1.1. Let (X, T ) be a topological space, and let A be an arbitrary
subset of X.
A. (Properties of the interior)
(i) The set Int(A) is open and Int(A) ⊂ A.
(ii) If D is an open set such that D ⊂ A, then D ⊂ Int(A).
(iii) x belongs to Int(A) if and only if A is a neighborhood of x.
(iv) A is open if and only if A = Int(A).
B. (Properties of the closure)
(i) The set A is closed and A ⊃ A.
(ii) If F is a closed set with F ⊃ A, then F ⊃ A.
(iii) A point x belongs to A, if and only if, A ∩ N 6= ∅ for any neighborhood
N of x.
(iv) A is closed if and only if A = A.
C. (Relationship between interior and closure) Int(X r A) = X r A and
X r A = X r Int(A).
Remark 1.1. When we work on R, equipped with the standard topology, for
a subset A ⊂ R, one has the following properties:
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(i) if sup A < ∞, then sup A ∈ A;
(ii) if inf A > −∞, then inf A ∈ A.
Indeed, for property (i), if the quantity s = sup A is finite, then for any neighborhood N of s, there exists some ε > 0, such that (s − ε, s + ε) ⊂ N , and by the
definition of the supremum, there exists some x ∈ A with s − ε < x ≤ s, which
clearly gives x ∈ N , thus proving that N ∩ A 6= ∅. The proof of (ii) is identical.
Exercise 1 ♦ . Let X be some non-empty set, and let P(X) denote the collection
of all subsets of X. Let cl : P(X) → P(X) be a map with the following properties:
(0) cl(∅) = ∅;
(1) cl(A) ⊃ A, ∀ A ∈ P(X);
(2) cl ◦ cl = cl;
(3) cl(A ∪ B) = cl(A) ∪ cl(B), ∀ A, B ∈ P(X).
Prove there exists a unique topology T on X, with respect to which we have the
equality cl(A) = A, ∀ A ∈ P(X).
Hints: Prove first the implication A ⊂ B ⇒ cl(A) ⊂ cl(B). Define then the collection
T = {A ⊂ X : cl(X r A) = X r A}.
A map cl with the above properties is called a closure operator.
Definition. Suppose (X, T ) is a topological space. Assume A ⊂ X is a subset
of X. On A we can introduce a natural topology, sometimes denoted by T |A which
consists of all subsets of A of the form A ∩ U with U open set in X. This topology
is called the relative (or induced ) topology.
Remark 1.2. If A is already open in the topology T , then a subset V ⊂ A
is open in the induced topology if and only if V is open in the topology T (this
follows from the fact that the intersection of any two open sets in T is again an
open set in T .
Definitions. Suppose (X, T ) and (Y, S) are topological spaces.
A. Given some point x ∈ X, a map f : X −→ Y is said to be continuous at x,
if for any neighborhood N of f (x) in the topology S (on Y ), the set
f −1 (N ) = {x ∈ X | f (x) ∈ N }
is a neighborhood of x in the topology T (on X).
B. If f : X → Y is continuous at every point in X, then f is said to be
continuous.
C. A continuous map f : X → Y is said to be a homeomorphism, if
• f is bijective, and
• the inverse map f −1 : Y → X is also continuous.
Continuity is “well behaved” with respect to compositions:
Proposition 1.2. Suppose (X, T ), (Y, S), and (Z, Z are topological spaces,
f
g
and X −→ Y −→ Z are two functions.
(i) If f is continuous at a point x ∈ X, and if g is continuous at f (x), then
g ◦ f is continuous at x.
(ii) If f and g are (globally) continuous, then so is g ◦ f .
The identity map on a topological space is always continuous.
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In terms of open/closed sets, the characterization of continuity is given by the
following.
Proposition 1.3. If (X, T ), (Y, S) are topological spaces and f : (X, T ) →
(Y, S) is a map, then the following are equivalent:
(i) f is continuous.
(ii) Whenever U ⊂ Y is an open set, it follows that f −1 (U ) is also an open
set (in X).
(iii) Whenever F ⊂ Y is a closed set, it follows that f −1 (F ) is also a closed
set (in X).
We conclude this section with a useful technical result.
Theorem 1.1 (Urysohn’s Lemma). Let (X, T ) be a topological Hausdorff space
with the following property:
(n) For any two disjoint closed sets A, B ⊂ X, there exist two disjoint open
sets U, V ⊂ X, such that U ⊃ A and V ⊃ B.
Then for any two disjoint closed sets A,
B ⊂ X, there exists a continuous function
f : X → [0, 1] such that f A = 0 and f B = 1.
Proof. We begin with a refinement of property (n):
(n0 ) For any disjoint closed sets A, B ⊂ X, there exist two open sets U, W ⊂ X,
such that A ⊂ U , U ⊂ W , and W ∩ B = ∅.
To prove (n0 ), we first apply (n) to find two disjoint open sets W, Z ⊂ X such that
(1)
W ⊃ A and Z ⊃ B.
Next we apply again (n) to the pair of closed sets A and X r W , and find two
disjoint open sets U, V ⊂ X such that
(2)
U ⊃ A and V ⊃ X r W.
On the one hand, using the fact that U ∩ V = ∅ and the fact that V is open, we
get the inclusion U ⊂ X r V . Using (2) this gives
U ⊂ X r V ⊂ W.
On the other hand, using the fact that W ∩ Z = ∅ and the fact that Z is open, we
get W ⊂ X r Z. But using (1) this will give
W ⊂ X r Z ⊂ X r B,
and we are done.
To prove the Theorem, start with two disjoint closed sets A, B ⊂ X. For every
integer n ≥ 0 we define the set Dn = { 2kn : k ∈ Z, 0 ≤ k ≤ 2n }, and we consider
D=
∞
[
Dn .
n=0
(Notice that Dn ⊂ Dn+1 , for all n ≥ 0.)
We are going to construct a family (Vt )t∈D of open sets in X with the following
properties
(i) V0 ⊃ A and V 1 ∩ B = ∅;
(ii) V t ⊂ Vs , for all t, s ∈ D with t < s.
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Let us start by constructing V0 and V1 . We use property (n0 ) to find open sets
U, W ⊂ X, with
A ⊂ U ⊂ U ⊂ W and W ∩ B = ∅,
and we simply take V0 = U and V1 = W .
The construction of the family (Vt )t∈D is carried on recursively. Assume, for
some integer n ≥ 0, we have constructed the sets (Vt )t∈Dn with property (i) and (ii)
(satisfied for t, s ∈ Dn ), and let us construct the next block of sets (Vt )t∈Dn+1 rDn .
We start off by observing that for every t ∈ Dn+1 r Dn , then the numbers
t± = t ±
1
2n+1
belong to Dn . Apply (n0 ) to the pair of disjoint closed sets V t− and X r Vt+ to
find two open sets U, W ⊂ X such that
V t− ⊂ U ⊂ U ⊂ W and W ∩ X r Vt+ = ∅.
Notice that the equality W ∩ (X r Vt+ ) = ∅, coupled with the inclusion U ⊂ W ,
gives U ∩ (X r Vt+ ), so we get U ⊂ Vt+ . We can then define Vt = U , and we will
obviously have the inclusions
(3)
V t− ⊂ Vt ⊂ V t ⊂ Vt+ .
Now the extended family (Vt )t∈Dn+1 will also satisfy property (ii), since for t, s ∈
Dn+1 with t < s, one of the following will hold:
• either t, s ∈ Dn , or
• t ∈ Dn , s ∈ Dn+1 r Dn , and t ≤ s− , or
• t ∈ Dn+1 r Dn , s ∈ Dn , and t+ ≤ s, or
• t, s ∈ Dn+1 r Dn , and t+ ≤ s− .
(In either case, one uses (3) combined with the inductive hypothesis.)
Having constructed the family (Vt )t∈D , with properties (i) and (ii), we define
the functions f : X → [0, 1] by
inf{t ∈ D : x ∈ Vt }, if x ∈ V1
f (x) =
1, if x 6∈ V1
(4)
Claim 1: The function f is equivalently defined by
0, if x ∈ V 0
f (x) =
sup{t ∈ D : x 6∈ V t }, if x 6∈ V 0
Let us denote by g : X → [0, 1] be the function defined by formula (4). Fix
some point x ∈ X. We break the proof in several cases
Case I: x ∈ V 0 .
In particular, using (ii) we get x ∈ Vt , for all t ∈ D, with t > 0, and since
x ∈ V1 , we have
f (x) = inf{t ∈ D : x ∈ Vt } = inf{t ∈ D : t > 0} = 0 = g(x).
Case II: x 6∈ V1 .
Using (ii) we have x 6∈ V t , for all t ∈ D, with t < 1, and since x 6∈ V 0 , we have
g(x) = sup{t ∈ D : x 6∈ V t } = sup{t ∈ D : t < 1} = 1 = f (x).
Case III: x ∈ V1 r V 0 .
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CHAPTER I: TOPOLOGY PRELIMINARIES
By the definition of f (x) we know:
(5)
x 6∈ Vt , ∀ t ∈ D, with t < f (x).
(6)
∀ ε > 0, ∃ sε ∈ D, with f (x) ≤ sε < f (x) + ε, such that x ∈ Vsε .
By the definition of g(x) we know:
(7)
x ∈ V t , ∀ t ∈ D, with t > g(x);
(8)
∀ ε > 0, ∃ rε ∈ D, with g(x) ≥ rε > g(x) − ε, such that x 6∈ V rε .
Using (6) and (8) we see that we must have
sε ≥ rε , ∀ ε > 0.
(9)
Indeed, if there exists some ε > 0 for which we have sε < rε , then using (6) we
would have
x ∈ V sε ⊂ V sε ⊂ V r ε ⊂ V r ε ,
which contradicts (8).
Now the inequality (9) gives
f (x) + ε > g(x) − ε, ∀ ε > 0,
so we have in fact the inequality
f (x) ≥ g(x).
Suppose now this inequality is strict. Using (5) and (7) we will get
(10)
x ∈ V t and x 6∈ Vt , for all t ∈ D, with f (x) > t > g(x).
Using the fact that D is dense in [0, 1], we could then find at least two elements
t1 , t2 ∈ D such that
f (x) > t1 > t2 > g(x).
In this case (10) immediately creates a contradiction, since
x ∈ V t2 ⊂ Vt1 .
Claim 2: The function f is continuous.
Since any open set in R is a union of open intervals, it suffice to prove the
following two properties1
(usc): f −1 (∞, t) is open for all t ∈ R;
(lsc): f −1 (t, ∞) is open for all t ∈ R.
In order to prove property (usc) it suffices to prove the equality
[
(11)
f −1 (∞, t) =
Vs .
s∈D
s<t
Start with a point x ∈ f −1 (t, ∞) , which means that f (x) < t. Using (6), there
exists some s ∈ D with f (x) < s < t, such that x ∈ Vs , so x indeed belongs to
the right hand side of (11). Conversely, if x belongs to the right hand side of (11),
there exists some s < t such that
x ∈ Vs . By the definition of f (x), it follows that
f (x) ≤ s < t, so x ∈ f −1 (∞, t) .
1 The condition (usc) means that f is upper semi-continuous, while the condition (lsc)
means that f is lower semi-continuous.
§1. Review of basic topology concepts
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In order to prove property (lsc) it suffices to prove the equality
[
(12)
f −1 (t, ∞) =
(X r V r ).
r∈D
r>t
Start with a point x ∈ f −1 (t, ∞) , which means that f (x) > t. Using (8), there
exists some r ∈ D with f (x) > r > t, such that x 6∈ V r , that is, x ∈ X r V r , so x
indeed belongs to the right hand side of (12). Conversely, if x belongs to the right
hand side of (12), there exists some r > t such that x ∈ X r V s , i.e. x 6∈ V r By
the equivalent definition
of f (x) given by Claim 1, it follows that f (x) ≥ r > t, so
x ∈ f −1 (t, ∞) .
Having proven that f is continuous, let us finish the proof. Since A ⊂ V0 , by
the definition
of f , we get f A = 0. Since B ⊂ X r V1 , again by the definition of
f , we get f B = 1.
Definition. A Hausdorff space (X, T ) with property (n) is called normal.