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MTH 313 — Knots and Surfaces
Fedor Duzhin
Week 2: Topological spaces
1 Topological spaces
Recall that a set A ⊂ R is called open if for each point x ∈ A there is an open interval (a, b)
such that x ∈ (a, b) ⊂ A. Obviously, the union of any number of open sets is open and the
intersection of finitely many open sets is open.
DEFINITION 1 Let X be a set. A topology on X is a collection of its subsets T, each called
an open set, such that the following conditions hold:
(i) ∅ and X are open;
(ii) The union of any collection of open sets is open;
(iii) The intersection of finitely many open sets is open.
The set X together with topology T is called a topological space.
EXAMPLE 2 Consider any set X. Then {∅, X} is a topology. It’s called the trivial topology
on X. It is the smallest topology on X.
♥
EXAMPLE 3 Consider any set X. Then the set of all its subsets 2X is a topology. It’s called
the discrete topology on X. It is the largest topology on X.
♥
Thus any set X admits at least two different topologies (unless it’s one element or empty)
— the trivial and the discrete ones. Those are two different topological spaces.
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EXAMPLE 4 Given any set X, the finite complement topology on X is defined by saying that
open sets are those whose complement is finite.
♥
It’s not hard to prove Example 4 directly, but it becomes more obvious once we formulate
the definition of a closed set.
DEFINITION 5 Let X be a topological space. A set A ⊂ X is called closed if X − A is open.
Now applying the fact that complement takes union to intersection and intersection to
union, we can reformulate the definition of a topological space via closed sets. Specifically,
to check that a certain collection of to-be-closed sets defines a topology on X (where open
sets are complements to closed ones), we need the following:
(i) X and ∅ are closed;
(ii) The intersection of any collection of closed sets is closed;
(iii) The union of finitely many closed sets is closed.
Now Example 4 becomes obvious. Indeed, instead of specifying open sets, we can just
say that closed sets are finite ones.
EXAMPLE 6 Closed sets of Zariski topology on Rn are algebraic sets defined as follows. Let
S be any set of polynomials of n variables. Then
A(S) = {x ∈ Rn : f(x) = 0 ∀f ∈ S}
is the corresponding algebraic set, a closed set of Zariski topology.
To see that this is, indeed, a topology, we need to check all the axioms. Let’s do it one
by one.
(i) We need to check that Rn and ∅ are closed. We have Rn = A(∅) and ∅ = A(x1 , x1 − 1)
and hence it’s true.
(ii) We need to check that the intersection of algebraic sets is also an algebraic set. Let
{Sα } be a collection of sets of polynomials, where α belongs to some index set I.
Obviously, we have then
!
\
[
A(Sα ) = A
Sα
α∈I
and hence we are done here.
α∈I
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(iii) We need to check that the union of finitely many algebraic sets is an algebraic set.
This is harder and requires some abstract algebra. If you haven’t done it or don’t
remember, just skip this part.
First, notice that it’s enough to show that the union of two closed sets is closed.
Further, given any set of polynomials S, let (S) be the ideal in the ring of polynomials
generated by the set S. Obviously, A(S) = A((S)) and thus any closed set is the null set
of some ideal. For two ideals I and J, we have A(I) ∪ A(J) = A(IJ) and hence we are
done too.
♥
The Zariski topology is also well-defined on Qn , Cn , Znm , or, generally, on Rn when
R is any commutative ring. It is the topology used by the branch of mathematics called
Algebraic Geometry.
EXAMPLE 7 The standard topology on Rn is defined as follows. First, recall that there is
Euclidean distance in Rn . Specifically,
p
d(x, y) = kx − yk = (x1 − y1 )2 + · · · + (xn − yn )2 ,
where x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ). Further, given x ∈ Rn and r > 0, the open ball
of radius r centred at x is
B(x, r) = {y ∈ Rn : kx − yk < r}.
Finally, a set A ⊂ Rn is open if for each x ∈ A there is ε > 0 such that B(x, ε) ⊂ A.
♥
Thus building blocks of the standard topology on Rn are open balls and any set can be
expressed as a union of open balls. This observation generalizes as follows.
DEFINITION 8 Let X be a set. A collection B of subsets of X is called a basis for a topology
on X if the following conditions hold:
(i) For each x ∈ X there is B ∈ B such that x ∈ B;
(ii) For any B1 , B2 ∈ B, suppose that x ∈ B1 ∩ B2 . Then there is B3 ∈ B such that
x ∈ B3 ⊂ B1 ∩ B2 .
Consider a set X with a basis B. The general method of constructing a topology from
the basis mimics constructing the standard topology on Rn from open balls.
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DEFINITION 9 Given a set X with a basis B, we say that a set A ⊂ X is open if for each
x ∈ A there is B ∈ B such that x ∈ B ⊂ A.
EXERCISE 10 Prove that the collection of open sets constructed in Definition 9 is, indeed, a
topology.
EXERCISE 11 Prove that given a set X and a basis B, a subset A is open if and only if it is
the union of basis elements.
Now we can reformulate the definition of standard topology on Rn using the notion of
a basis. In particular, Example 7 implicitly uses the basis consisting of all open balls. A
basis of a topology is not unique. In particular, the whole topology might be considered as
a basis (not a very interesting one though).
THEOREM 12 Let X be a set and let B1 and B2 be bases for two, possibly different,
topologies T1 and T2 on X. Then T1 = T2 if and only if each B1 ∈ B1 is open in T2
and each B2 ∈ B2 is open in T1 .
EXERCISE 13 Prove Theorem 12.
EXAMPLE 14 The following are also bases for the standard topology on R2 :
(i) Open balls B(x, r) such that x = (x1 , x2 ), where x1 , x2 , r are rational numbers.
(ii) Open rectangles (a, b) × (c, d).
(iii) Open balls for a different distance function, where
EXAMPLE 15 For each n ∈ Z, let
B(n) =
d(x, y) = |x1 − y1 | + |x2 − y2 |.
{n},
n odd,
{n − 1, n, n + 1}, n even.
♥
The collection of B(n) for all n ∈ Z is a basis for a topology on Z. This topology is called
the digital line topology and we call Z equipped with this topology the digital line.
♥
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EXAMPLE 16 Recall that an (infinite) arithmetic progression A(a, b) ⊂ Z is
A(a, b) = {. . . , a − 3b, a − 2b, a − b, a, a + b, a + 2b, a + 3b, . . . } = {a + mb : m ∈ Z},
where a ∈ Z and b ∈ Z>0 are some fixed integers. The set of infinite arithmetic progressions is a basis for a topology in Z. Let’s call it arithmetic progression topology.
♥
The following example is due to Harry Furstenberg (1955). It shows how classical
Euclid’s theorem can be proved using topology.
THEOREM 17 There are infinitely many prime numbers.
PROOF Consider Z equipped with the arithmetic progression topology. For each arithmetic
progression A(a, b), we have
Z \ A(a, b) = A(a + 1, b) ∪ A(a + 2, b) ∪ · · · ∪ A(a + b − 1, b)
and hence basis elements are also closed.
Further, the only integers not divisible by any prime number are −1 and 1. Thus we
have
[
Z \ {−1, 1} =
A(0, p).
prime p
If there were only finitely many primes p, then the right hand side would be a closed set as
a finite union of closed sets. But then {−1, 1} would be open, which is not true because it
doesn’t contain an infinite arithmetic progression. This contradiction concludes the proof.:)
Given two topologies on the same set, how do we compare them?
DEFINITION 18 Let X be a set, let T1 and T2 be topologies on X, and assume that T1 ⊂ T2
(that is, any set open in T1 is also open in T2 ). We say then T1 is coarser than T2 while
T2 is finer than T1 .
REMARK 19 The terms ’finer’ and ’coarser’ are taken from Adams and Franzosa. More
standard terms are ’weaker’ and ’stronger’. However, the notion of weaker/stronger has
exactly opposite meanings in topology and analysis, so it’s better to use finer/coarser. X
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EXERCISE 20 We have constructed the following topologies on Z: trivial, discrete, finite
complement, digital, arithmetic progression. Arrange them by the relation finer/coarser.
EXERCISE 21 We have constructed the following topologies on R: trivial, discrete, finite
complement, Zariski, standard. Arrange them by the relation finer/coarser.
EXERCISE 22 We have constructed the following topologies on R2 : trivial, discrete, finite
complement, Zariski, standard. Arrange them by the relation finer/coarser.
DEFINITION 23 A topological space X is called Hausdorff if for any two points x, y ∈ X
there are disjoint open sets U ∋ x and V ∋ y.
EXAMPLE 24 Any set X equipped with the discrete topology is Hausdorff. If X has at least
two elements, then the trivial topology on X is not Hausdorff.
♥
EXAMPLE 25 If a set X is infinite, then the finite complement topology on X is not Hausdorff.
♥
EXAMPLE 26 The standard topology on Rn is Hausdorff.
♥
EXAMPLE 27 The Zariski topology on Rn is not Hausdorff. In fact, the intersection of any
two nonempty Zariski-open sets is a nonempty Zariski-open set.
♥
DEFINITION 28 Let X be a topological space and let x ∈ X. Any open set U containing x
is called a neighbourhood of x.
2 Constructing new topologies
From now on, we are mainly going to work with Rn equipped with the standard topology
and we denote it just Rn . If, in a rare occasion, we need to consider some other topology
on Rn , we specify it explicitly. Other topological spaces here are are usually constructed
from Rn using operations described below. Recall that Rn is Hausdorff.
We’ll consider the following three constructions — subspace, quotient space, and product
space.
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Subspace
DEFINITION 29 Let (X, T) be a topological space and let A ⊂ X. Open sets of the subspace
topology on A are U ∩ A when U is open in X. The set A equipped with this topology is
called a subspace of X.
EXAMPLE 30 Consider the real line R and the subspace topology on [0, 2]. Then the interval
[0, 1) is open in this topology since [0, 1) = (−1, 1) ∩ [0, 2]. Notice, however, that [0, 1) is not
open in R.
♥
EXERCISE 31 Let X be a topological space and let A ⊂ X. Prove that a set B ⊂ A is closed
in the subspace topology of A if and only if B = A ∩ C for some set C ⊂ X closed in X.
EXAMPLE 32 Consider R as the x-axis in R2 . Then the interval (0, 1) is open in R but neither
open nor closed in R2 .
♥
EXERCISE 33 What is the subspace topology of Z ⊂ R?
Quotient space
Recall that a plane model for a surface is a polygon together with instructions of how its
edges are to be glued. Gluing is formalized by the concept of quotient topology.
DEFINITION 34 Let X be a topological space, let Q be a set, and let a function p : X → Q
be onto. Define a subset U of Q to be open if and only if p−1 (U) is open. The resultant
collection of open sets in Q is called the quotient topology induced by p, the function p
is called a quotient map, and Q is called a quotient space.
Let’s see how this concept works for plane models of surfaces.
EXAMPLE 35 Consider the standard plane model aa−1 for the sphere shown in Figure 1.
How do we glue two edges labelled a and a−1 to each other? Let’s begin with the closed unit
disk X given by x 2 + y 2 ≤ 1, which is our plane model. According to the gluing instruction,
a point (x, y) on the boundary of the disk is supposed to be glued to the point (−x, y). Thus
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Figure 1: Gluing a sphere from its model aa−1 .
U2
U2
a
a
U1
Figure 2: Open sets in the sphere model aa−1 .
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an arbitrary point of the set Q is either (x, y) for x 2 + y 2 < 1 or a triple (−x, x, y) for
x 2 + y 2 = 1. The quotient map p : X → Q goes like this:
(x, y),
x 2 + y 2 < 1,
p(x, y) =
(−x, x, y), x 2 + y 2 = 1.
What about topology on Q? By definition of the quotient topology, there are two options
for an open set U in Q:
(i) U lies in the open disk x 2 + y 2 < 1 entirely as U1 in Figure 2;
(ii) U intersects the circle x 2 + y 2 = 1 along symmetric intervals as U2 in Figure 1.
♥
EXERCISE 36 What do we get if we begin with a closed interval, say [0, 1] and glue its
endpoints together?
Recall that an equivalence relation ∼ on a set X is a reflexive (which means a ∼ a),
symmetric (a ∼ b ⇔ b ∼ a), and transitive (a ∼ b, b ∼ c Ñ a ∼ c) binary relation. It
partitions X into disjoint equivalence classes.
We can now formulate the definition of the quotient topology as follows. Let’s begin
with a topological space X and assume that there is an equivalence relation ∼ on X. Let
Q = X/ ∼ be the set of equivalence classes and let p : X → Q be the function that sends
each x ∈ X to its equivalence class.
EXERCISE 37 Prove that this construction is same as the one in Definition 34.
EXAMPLE 38 Let’s apply this approach to the standard plane model of a torus shown in
Figure 3. The polygon X is the square given by 0 ≤ x, y ≤ 1. The equivalence relation on
X is defined by (x, 0) ∼ (x, 1) and (0, y) ∼ (1, y).
♥
EXERCISE 39 Let X be the unit sphere in R3 given by x 2 = 1, where x = (x1 , x2 , x3 ) and x 2
means the dot product x · x. Consider the equivalence relation x ∼ −x. What is X/ ∼?
Product of spaces
Given two sets X and Y , recall that their product X × Y is the set of pairs (x, y), where
x ∈ X and y ∈ Y . If X and Y are topological spaces, is there a natural way to construct a
topology on X × Y ?
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a
b
b
a
Figure 3: Plane model aba−1b−1 .
EXERCISE 40 Is it possible to say that open sets of X × Y are products U × V , where U is
open in X and V is open in Y ?
DEFINITION 41 Let X and Y be topological spaces. Then the product topology on X × Y
is generated by the basis
B = {U × V : U is open in X and V is open in Y}.
EXAMPLE 42 R × R is R2 .
♥
EXAMPLE 43 Let S 1 be the circle obtained by gluing [0, 1]/0 ∼ 1. Then S 1 × S 1 is the square
[0, 1] × [0, 1] with gluing (x, 0) ∼ (x, 1) and (0, y) ∼ (1, y), which is the torus.
♥
Configuration spaces
A configuration space is a topological space that describes possible states of a physical
system. In this course we’ll consider configuration spaces associated with robots.
EXAMPLE 44 Suppose that a robot arm has three joints that are able to rotate by 360◦. Then
the space of positions of each joint is topologically the circle S 1 and hence the configuration
♥
space of the robot arm is the product S 1 × S 1 × S 1 .
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EXAMPLE 45 Consider three identical vehicles moving in a disk D ⊂ R2 . Their positions
are points on the product space D × D × D. Further, the vehicles are not supposed to
collide, so the only available points are (P, Q, R) ∈ D × D × D such that P 6= Q, Q 6= R, and
P 6= R. These triples of mutually distinct points form a subspace of the product D × D × D.
Specifically, let ∆ ⊂ D × D × D be those points where P = Q, Q = R, or P = R. Then the
configuration space of triples of vehicles is D × D × D − ∆.
Further, assume that, additionally, vehicles are identical and if we switch them, we won’t
notice the difference. It gives us the following equivalence relation on the configuration
space:
(P, Q, R) ∼ (P, R, Q) ∼ (Q, P, R) ∼ (Q, R, P) ∼ (Q, P, R) ∼ (Q, R, P)
and configuration space of triples of identical vehicles is (D × D × D − ∆)/ ∼.
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♥