Download 1 A crash course in point set topology

1
A crash course in point set topology
In this handout we review elements of point set topology we’ll need throughout the course. The
notion of a topology on a space is an abstraction of properties of open sets in Rn . Let us set some
notation first. Recall that the norm ||x|| of a vector x = (x1 , . . . , xn ) ∈ Rn is defined by
!1/2
||x|| :=
X
x2i
.
i
It satisfies the triangle inequality:
||x + y|| ≤ ||x|| + ||y||.
An open ball Br (x) of radius r centered at a point x ∈ Rn is defined by
Br (x) := {y ∈ Rn | ||x − y|| < r}
Definition 1.1. A subset V of Rn is open if for any x ∈ V there is r > 0 such that Br (x) ⊂ V .
Exercise 1.1. For any r > 0 and any x ∈ Rn the open ball Br (x) is open in Rn .
Remark 1.2. The empty set ∅ is an open subset of Rn .
Open subsets of Rn have the following properties:
1. the empty set and all of Rn are open;
2. if U and V are open then their intersection U ∩ V is open;
S
3. if {Uα }α∈A is a collection of open sets then their union Uα is open.
Exercise 1.2. Prove properties 2 and 3 of open sets above.
Continuity of a function f : Rn → Rm can be phrased in terms of open sets. First recall the
calculus definition: f : Rn → Rm is continuous if for any x ∈ Rn and any ε > 0 there is δ > 0 such
that if ||x − y|| < δ then ||f (x) − f (y)|| < for all y ∈ Rn .
Exercise 1.3. Prove that f : Rn → Rm is continuous if and only if for any open set V ⊂ Rm the
preimage f −1 (V ) is open in Rn .
The latter condition will be the basis for defining continuity of functions between topological
spaces. We now abstract the above observations about open sets in Rn .
Definition 1.3. A topological space is a set X together with a collection T of subsets of X (“a
topology on X”) such that
1. the empty set and all of X are in T ;
2. if U and V are in T then their intersection U ∩ V is in T ;
S
3. if {Uα }α∈A is a collection of sets in T then their union Uα is in T ..
1
The sets in T are referred to as open sets; the pair (X, T ) is called a topological space. One
often says “X is a topological space” with the collection T understood.
Definition 1.4. Let (X, T ) be a topological space. A subset C ⊂ X is closed if and only if
X r C ∈ T , i.e., the complement of C is open.
Example 1.5 (The standard topology on Rn ). We declare the collection T of open sets to be the
open sets in the sense of definition 1.1.
Example 1.6. Let X be a subset of Rn . We define a topology T on X by U ∈ T if and only if
there is an open set Ũ ⊂ Rn such that
U = X ∩ Ũ .
Check that T is a topology.
Solution: ∅ = X ∩ ∅, X = X ∩ Rn . If U, V ∈ T then there are open sets Ũ , Ṽ ⊂ Rn with U = X ∩ Ũ , V = X ∩ Ṽ . Then U ∩ V = (X ∩ Ũ ) ∩ (X ∩ Ṽ ) =
X ∩ (Ũ ∩ Ṽ ). But Ũ ∩ Ṽ is open in Rn . Hence U ∩ V ∈ T . Similarly, if {Uα } ⊂ T then ∪Uα ∈ T .
The example above can be generalized.
Definition 1.7. Let (X, T ) be a topological space, Y ⊂ X a subset. The subspace topology TY on
Y is defined by
A ∈ TY ⇔ there is à ∈ T with A = à ∩ Y.
Exercise 1.4. Check that TY defined above is indeed a topology on Y .
Definition 1.8 (Continuous maps). . Let (X, T ) and (X, T 0 ) be two topological spaces. A map
f : X → X 0 is continuous if for any U 0 ∈ T 0 the preimage f −1 (U 0 ) is in T . That is, the preimage
of an open set is open.
Example 1.9. Consider R with the standard topology. The map
1 x≥0
f (x) =
0 x<0
is not continuous by the standard calculus definition. And indeed, f −1 ((1/2, 3/2)) = [0, ∞), which
is not open in R. So f is not continuous by our definition either.
Definition 1.10. A continuous map f : (X, T ) → (X 0 , T 0 ) of topological spaces is a homeomorphism iff
1. f is a bijection and
2. f −1 is continuous.
Example 1.11. The map f : R → R, f (x) = x3 is a homeomorphism.
Example 1.12. Give [0, 2π) ⊂ R the subspace topology. Give complex numbers C the topology of
R2 and give S 1 := {eiθ ∈ C | θ ∈ R} the subspace topology. The map f : [0, 2π) → S 1 , f (θ) = eiθ
is a continuous bijection, but it is not a homeomorphism. Why not?
2
Definition 1.13. A topological space X is Hausdorff if for any two distinct points x and y in X
there exist open sets U and V such that x ∈ U , y ∈ V and U ∩ V = ∅.
Example 1.14. Rn with the standard topology is Hausdorff: if x and y are distinct points in Rn ,
let r = ||x − y||/2. Then the intersection Br (x) ∩ Br (y) of open balls is empty.
Not all topologies are Hausdorff. Here is a useful example.
Example 1.15. Let X be a line with two origins. Technically speaking
X = (R r {0}) ∪ {a, b}
where a, b are two points (or symbols). Define a topology on X as follows: if U ⊂ R is open in the
standard topology and 0 6∈ U , then U is open in X. If U ⊂ R is open and 0 ∈ U then the sets
(U r {0}) ∪ {a} and (U r {0}) ∪ {b} are both open in X. Declare this to be the complete list of
open sets in X. Check that this is a topology. It is not Hausdorff: for any two open sets U and V
in X with a ∈ U and b ∈ V the intersection U ∩ V is never empty.
Definition 1.16 (Product topology). Let (X, T ), (X 0 , T 0 ) be two topological spaces. The product
topology on X × X 0 is defined as follows: A set W ⊂ X × X 0 is open if for any (a, b) ∈ W there
are open sets U ⊂ X, V ⊂ X 0 such that
(a, b) ∈ U × V ⊂ W.
Exercise 1.5. Check that a product topology is a topology.
paracompactness and partitions of unity
We next defined paracompactness and partitions of unity. We then state a theorem to the effect
that there are partitions of unity on paracompact manifolds.
Definition 1.17. The closure S̄ of a subset S of a topological space X is the intersection of all
closed subsets of X containing S:
\
S̄ = {C ⊂ X | C closed, S ⊂ C}.
Since arbitrary intersections of closed sets are closed, the closure is a closed set.
Definition 1.18. The support supp f of a continuous function f : X → R is the closure of the set
of points where the function is nonzero:
supp f = {x ∈ X | f (x) 6= 0}.
An open cover (also called an open covering) of a topological space X is a collection of open
sets {Uα } such that
[
Uα = X.
Let {Uα }α∈A and {Vβ }β∈B be two open covers of a space X. The cover {Vβ } is a refinement of
{Uα } if for each Vβ there is a Uα with Vβ ⊂ Uα . That is, there is a map h : B → A with
Vβ ⊂ Uh(β) .
3
A cover {Uα } of X is locally finite if for each point x ∈ X there is an open set W containing x so
that
W ∩ Uα 6= ∅
only for finitely many indices α.
Example 1.19. The cover {n, n + 2}n∈Z is a locally finite cover of R, while {(− n1 , n1 )}n∈Z is not a
locally finite cover of the interval (−1, 1).
Definition 1.20 (Partition of unity). Let {Uα } be an open cover of a space X. A partition of
unity subordinate to the cover is a collection of (continuous) functions
{fβ : X → [0, 1]}β
such that
1. there is a locally finite refinement {Vβ } of {Uα } such that supp fβ ⊂ Vβ and
P
2.
fβ (x) = 1 for all x ∈ X.
Note that condition (1) guarantees that
P for any given point x ∈ X there are only finitely many
indices β with fβ (x) 6= 0; hence the sum
fβ (x) makes sense.
Definition 1.21. A Hausdorff topological space is paracompact if every open cover has a locally
finite refinement.
The key result that we will use, but won’t prove, is:
Theorem 1.22. Let M be a paracompact manifold. Then every open cover of M has a smooth
(C ∞ ) partition of unity subordinate to the cover.
This leave us with a question: do paracompact manifolds exist? A traditional way to deal with
the question is to consider the so called “second-countable” manifolds.
Definition 1.23. A topological space X is second countable if there is a countable collection of
open sets {Ui } such that any open set in X is a union of Ui ’s.
Example 1.24. Rn is second countable: the collection {Ui } consists of open balls with rational
radii centered at points with rational coordinates.
Theorem 1.25. Any second countable Hausdorff C ∞ manifold is paracompact.
We will not prove this theorem either.
compactness
A concept related to the notion of paracompactness is that of compactness.
Definition 1.26. A topological space X is compact if for any open cover {Uα } there exist finitely
many indices α1 , . . . αn such that
n
[
X=
Uαi .
i=1
4
It should not be hard to convince yourself that a compact space is paracompact. Facts
1. The interval [0, 1] is compact. This fact is known as the Bolzano-Weierstrass theorem.
2. If X and Y are compact, then so is their product X × Y .
3. If f : X → Y is continuous and X is compact then f (X) is compact. This is an easy fact.
Prove it.
4. If X is a Hausdorff topological space and K ⊂ X is a compact subset (i.e., K is compact in
the subspace topology) then K is closed.
5. If X is compact and K ⊂ X is closed, then K is compact. Again, this is easy.
6. If X is compact, Y is Hausdorff and f : X → Y is a continuous bijection, then f is a
homeomorphism.
connectedness
Definition 1.27. A topological space X is connected if for any two open subsets U, V of X
(U ∩ V = ∅ and U ∪ V = X) ⇒ U = ∅ or V = ∅.
In other words, a topological space is connected if it cannot be written as a disjoint union of two
open sets.
Facts (also known as theorems without proof):
1. The interval [0, 1] is connected.
2. If f : X → Y is continuous and X is connected then f (X) is connected. This is fairly easy.
Prove it.
S
3. If {Yα } is a collection of connected subsets of Y and no two Yα ’s are disjoint, then Yα is
connected.
4. The relation “points p and q belong to a connected subset of X” is an equivalence relation.
The equivalence classes of this relation are called the connected components of the space X.
S
Remark 1.28. Facts (1–3) show that R = n∈N [−n, n] is connected.
Example 1.29. Connected components of R r {0} are (−∞, 0) and (0, ∞).
Exercise 1.6. A space X is path-connected if for any two points p, q ∈ X there is a continuous
map (a path) f : [0, 1] → X with f (0) = p and f (1) = q. Show that a path-connected space is
connected.
Solution: Suppose X is not connected. Then there exist two open sets U, V ⊂ X with U ∩V = ∅ and U ∪V = X. Since f is continuous, f −1 (U )
and f −1 (V ) are open. It is easy to see that f −1 (U ) ∩ f −1 (V ) = ∅ and that f −1 (U ) ∪ f −1 (V ) = [0, 1]. But [0, 1] is connected. Contradiction.
5