Part IB — Metric and Topological Spaces
Based on lectures by J. Rasmussen
Notes taken by Dexter Chua
Easter 2015
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Metrics
Definition and examples. Limits and continuity. Open sets and neighbourhoods.
Characterizing limits and continuity using neighbourhoods and open sets. [3]
Topology
Definition of a topology. Metric topologies. Further examples. Neighbourhoods, closed
sets, convergence and continuity. Hausdorff spaces. Homeomorphisms. Topologi
cal and nontopological properties. Completeness. Subspace, quotient and product
top ologies. [3]
Connectedness
Definition using open sets and integervalued functions. Examples, including inter
vals. Components. The continuous image of a connected space is connected. Path
connectedness. Pathconnected spaces are connected but not conversely. Connected
op en sets in Euclidean space are pathconnected. [3]
Compactness
Definition using open covers. Examples: finite sets and [0, 1]. Closed subsets of
compact spaces are compact. Compact subsets of a Hausdorff space must b e closed.
The compact subsets of the real line. Continuous images of compact sets are compact.
Quotient spaces. Continuous realvalued functions on a compact space are bounded
and attain their bounds. The product of two compact spaces is compact. The compact
subsets of Euclidean space. Sequential compactness. [3]
Contents
0 Introduction
1 Metric spaces
1.1 Definitions
1.2 Examples of metric spaces
1.3 Norms
1.4 Open and closed subsets
2 Top ological spaces
2.1 Definitions
2.2 Sequences
2.3 Closed sets
2.4 Closure and interior
2.4.1 Closure
2.4.2 Interior
2.5 New topologies from old
2.5.1 Subspace topology
2.5.2 Product topology
2.5.3 Quotient topology
3 Connectivity
3.1 Connectivity
3.2 Path connectivity
3.2.1 Higher connectivity*
3.3 Components
3.3.1 Path components
3.3.2 Connected components
4 Compactness
4.1 Compactness
4.2 Products and quotients
4.2.1 Products
4.2.2 Quotients
4.3 Sequential compactness
4.4 Completeness
0 Introduction
The course Metric and Topological Spaces is divided into two parts. The first is
on metric spaces, and the second is on topological spaces (duh).
In Analysis, we studied real numbers a lot. We defined many properties such
as convergence of sequences and continuity of functions. For example, if (
x
n
) is
a sequence in R, x
n
→ x means
(∀ε > 0)(∃N)(∀n > N) x
n
− x < ε.
Similarly, a function f is continuous at x
0
if
(∀ε > 0)(∃δ > 0)(∀x) x − x
0
 < δ ⇒ f(x) − f(x
0
) < ε.
However, the definition of convergence doesn’t really rely on
x
n
being real
numbers, except when calculating values of
x
n
− x
. But what does
x
n
− x
really mean? It is the distance between
x
n
and
x
. To define convergence, we
don’t really need notions like subtraction and absolute values. We simply need a
(sensible) notion of distance between points.
Given a set
X
, we can define a metric (“distance function”)
d
:
X × X → R
,
where
d
(
x, y
) is the distance between the points
x
and
y
. Then we say a sequence
(x
n
) in X converges to x if
(∀ε > 0)(∃N)(∀n > N) d(x
n
, x) < ε.
Similarly, a function f : X → X is continuous if
(∀ε > 0)(∃δ > 0)(∀x) d(x, x
0
) < δ ⇒ d(f(x), f(x
0
)) < ε.
Of course, we will need the metric
d
to satisfy certain conditions such as being
nonnegative, and we will explore these technical details in the first part of the
course.
As people studied metric spaces, it soon became evident that metrics are not
really that useful a notion. Given a set
X
, it is possible to find many different
metrics that completely agree on which sequences converge and what functions
are continuous.
It turns out that it is not the metric that determines (most of) the properties
of the space. Instead, it is the open sets induced by the metric (intuitively, an
open set is a subset of
X
without “boundary”, like an open interval). Metrics
that induce the same open sets will have almost identical properties (apart from
the actual distance itself).
The idea of a topological space is to just keep the notion of open sets and
abandon metric spaces, and this turns out to be a really good idea. The second
part of the course is the study of these topological spaces and defining a lot of
interesting properties just in terms of open sets.
1 Metric spaces
1.1 Definitions
As mentioned in the introduction, given a set
X
, it is often helpful to have a
notion of distance between points. This distance function is known as the metric.
Definition (Metric space). A metric space is a pair (
X, d
X
) where
X
is a set
(the space) and
d
X
is a function
d
X
:
X × X → R
(the metric) such that for all
x, y, z,
– d
X
(x, y) ≥ 0 (nonnegativity)
– d
X
(x, y) = 0 iff x = y (identity of indiscernibles)
– d
X
(x, y) = d
X
(y, x) (symmetry)
– d
X
(x, z) ≤ d
X
(x, y) + d
X
(y, z) (triangle inequality)
We will have two quick examples of metrics before going into other important
definitions. We will come to more examples afterwards.
Example.
– (Euclidean “usual” metric) Let X = R
n
. Let
d(v, w) = v − w =
v
u
u
t
n
X
i=1
(v
i
− w
i
)
2
.
This is the usual notion of distance we have in the
R
n
vector space. It is
not difficult to show that this is indeed a metric (the fourth axiom follows
from the CauchySchwarz inequality).
– (Discrete metric) Let X be a set, and
d
X
(x, y) =
(
1 x 6= y
0 x = y
To show this is indeed a metric, we have to show it satisfies all the axioms.
The first three axioms are trivially satisfied. How about the fourth? We
can prove this by exhaustion.
Since the distance function can only return 0 or 1,
d
(
x, z
) can be 0 or 1,
while
d
(
x, y
) +
d
(
y, z
) can be 0, 1 or 2. For the fourth axiom to fail, we
must have RHS
<
LHS. This can only happen if the right hand side is 0.
But for the right hand side to be 0, we must have
x
=
y
=
z
. So the left
hand side is also 0. So the fourth axiom is always satisfied.
Given a metric space (
X, d
), we can generate another metric space by picking
out a subset of X and reusing the same metric.
Definition (Metric subspace). Let (
X, d
X
) be a metric space, and
Y ⊆ X
.
Then (
Y, d
Y
) is a metric space, where
d
Y
(
a, b
) =
d
X
(
a, b
), and said to be a
subspace of X.
Example.
S
n
=
{
v
∈ R
n+1
:

v

= 1
}
, the
n
dimensional sphere, is a subspace
of R
n+1
.
Finally, as promised, we come to the definition of convergent sequences and
continuous functions.
Definition (Convergent sequences). Let (
x
n
) be a sequence in a metric space
(
X, d
X
). We say (
x
n
) converges to
x ∈ X
, written
x
n
→ x
, if
d
(
x
n
, x
)
→
0 (as a
real sequence). Equivalently,
(∀ε > 0)(∃N )(∀n > N ) d(x
n
, x) < ε.
Example.
–
Let (v
n
) be a sequence in
R
k
with the Euclidean metric. Write v
n
=
(
v
1
n
, ··· , v
k
n
), and v = (
v
1
, ··· , v
k
)
∈ R
k
. Then v
n
→
v iff (
v
i
n
)
→ v
i
for
all i.
–
Let
X
have the discrete metric, and suppose
x
n
→ x
. Pick
ε
=
1
2
.
Then there is some
N
such that
d
(
x
n
, x
)
<
1
2
whenever
n > N
. But if
d
(
x
n
, x
)
<
1
2
, we must have
d
(
x
n
, x
) = 0. So
x
n
=
x
. Hence if
x
n
→ x
,
then eventually all x
n
are equal to x.
Similar to what we did in Analysis, we can show that limits are unique (if
exist).
Proposition. If (
X, d
) is a metric space, (
x
n
) is a sequence in
X
such that
x
n
→ x, x
n
→ x
0
, then x = x
0
.
Proof.
For any
ε >
0, we know that there exists
N
such that
d
(
x
n
, x
)
< ε/
2 if
n > N . Similarly, there exists some N
0
such that d(x
n
, x
0
) < ε/2 if n > N
0
.
Hence if n > max(N, N
0
), then
0 ≤ d(x, x
0
)
≤ d(x, x
n
) + d(x
n
, x
0
)
= d(x
n
, x) + d(x
n
, x
0
)
≤ ε.
So 0 ≤ d(x, x
0
) ≤ ε for all ε > 0. So d(x, x
0
) = 0, and x = x
0
.
Note that to prove the above proposition, we used all of the four axioms. In
the first line, we used nonnegativity to say 0
≤ d
(
x, x
0
). In the second line, we
used triangle inequality. In the third line, we used symmetry to swap
d
(
x, x
n
)
with
d
(
x
n
, x
). Finally, we used the identity of indiscernibles to conclude that
x = x
0
.
To define a continuous function, here, we opt to use the sequence definition.
We will later show that this is indeed equivalent to the
ε − δ
definition, as well
as a few other more useful definitions.
Definition (Continuous function). Let (
X, d
X
) and (
Y, d
Y
) be metric spaces,
and
f
:
X → Y
. We say
f
is continuous if
f
(
x
n
)
→ f
(
x
) (in
Y
) whenever
x
n
→ x (in X).
Example. Let
X
=
R
with the Euclidean metric. Let
Y
=
R
with the discrete
metric. Then
f
:
X → Y
that maps
f
(
x
) =
x
is not continuous. This is since
1/n → 0 in the Euclidean metric, but not in the discrete metric.
On the other hand,
g
:
Y → X
by
g
(
x
) =
x
is continuous, since a sequence
in Y that converges is eventually constant.
1.2 Examples of metric spaces
In this section, we will give four different examples of metrics, where the first
two are metrics on
R
2
. There is an obvious generalization to
R
n
, but we will
look at R
2
specifically for the sake of simplicity.
Example (Manhattan metric). Let X = R
2
, and define the metric as
d(x, y) = d((x
1
, x
2
), (y
1
, y
2
)) = x
1
− y
1
 + x
2
− y
2
.
The first three axioms are again trivial. To prove the triangle inequality, we have
d(x, y) + d(y, z) = x
1
− y
1
 + x
2
− y
2
 + y
1
− z
1
 + y
2
− z
2

≥ x
1
− z
1
 + x
2
− z
2

= d(x, z),
using the triangle inequality for R.
This metric represents the distance you have to walk from one point to
another if you are only allowed to move horizontally and vertically (and not
diagonally).
Example (British railway metric). Let X = R
2
. We define
d(x, y) =
(
x − y if x = ky
x + y otherwise
To explain the name of this metric, think of Britain with London as the origin.
Since the railway system is
stupid
less than ideal, all trains go through London.
For example, if you want to go from Oxford to Cambridge (and obviously not
the other way round), you first go from Oxford to London, then London to
Cambridge. So the distance traveled is the distance from London to Oxford plus
the distance from London to Cambridge.
The exception is when the two destinations lie along the same line, in which
case, you can directly take the train from one to the other without going through
London, and hence the “if x = ky” clause.
Example (
p
adic metric). Let
p ∈ Z
be a prime number. We first define the
norm
n
p
to be
p
−k
, where
k
is the highest power of
p
that divides
n
. If
n
= 0,
we let n
p
= 0. For example, 20
2
= 2
2
· 5
2
= 2
−2
.
Now take
X
=
Z
, and let
d
p
(
a, b
) =
a−b
p
. The first three axioms are trivial,
and the triangle inequality can be proved by making some numbertheoretical
arguments about divisibility.
This metric has rather interesting properties. With respect to
d
2
, we have
1
,
2
,
4
,
8
,
16
,
32
, ··· →
0, while 1
,
2
,
3
,
4
, ···
does not converge. We can also use it
to prove certain numbertheoretical results, but we will not go into details here.
Example (Uniform metric). Let
X
=
C
[0
,
1] be the set of all continuous
functions on [0, 1]. Then define
d(f, g) = max
x∈[0,1]
f(x) − g(x).
The maximum always exists since continuous functions on [0
,
1] are bounded
and attain their bounds.
Now let
F
:
C
[0
,
1]
→ R
be defined by
F
(
f
) =
f
(
1
2
). Then this is continuous
with respect to the uniform metric on C[0, 1] and the usual metric on R:
Let
f
n
→ f
in the uniform metric. Then we have to show that
F
(
f
n
)
→ F
(
f
),
i.e. f
n
(
1
2
) → f (
1
2
). This is easy, since we have
0 ≤ F (f
n
) − F (f) = f
n
(
1
2
) − f(
1
2
) ≤ max f
n
(x) − f(x) → 0.
So f
n
(
1
2
) − f(
1
2
) → 0. So f
n
(
1
2
) → f (
1
2
).
1.3 Norms
There are several notions on vector spaces that are closely related to metrics.
We’ll look at norms and inner products of vector spaces, and show that they all
naturally induce a metric on the space.
First of all, we define the norm. This can be thought of as the “length” of a
vector in the vector space.
Definition (Norm). Let
V
be a real vector space. A norm on
V
is a function
k · k : V → R such that
– kvk ≥ 0 for all v ∈ V
– kvk = 0 if and only if v = 0.
– kλvk = λkvk
– kv + wk ≤ kvk + kwk.
Example. Let
V
=
R
n
. There are several possible norms we can define on
R
n
:
kvk
1
=
n
X
i=1
v
i

kvk
2
=
v
u
u
t
n
X
i=1
v
2
i
kvk
∞
= max{v
i
 : 1 ≤ i ≤ n}.
In general, we can define the norm
kvk
p
=
n
X
i=1
v
i

p
!
1/p
.
for any 1 ≤ p ≤ ∞, and kvk
∞
is the limit as p → ∞.
Proof that these are indeed norms is left as an exercise for the reader (in the
example sheets).
A norm naturally induces a metric on V :
Lemma. If k · k is a norm on V , then
d(v, w) = kv − wk
defines a metric on V .
Proof.
(i) d(v, w) = kv − wk ≥ 0 by the definition of the norm.
(ii) d(v, w) = 0 ⇔ kv − wk = 0 ⇔ v − w = 0 ⇔ v = w.
(iii) d(w, v) = kw − vk = k(−1)(v − w)k =  − 1kv − wk = d(v, w).
(iv) d(u, v) + d(v, w) = ku − vk+ kv − w k ≥ ku − wk = d(u, w).
Example. We have the following norms on C[0, 1]:
kfk
1
=
Z
1
0
f(x) dx
kfk
2
=
s
Z
1
0
f(x)
2
dx
kfk
∞
= max
x∈[0,1]
f(x)
The first two are known as the
L
1
and
L
2
norms. The last is called the uniform
norm, since it induces the uniform metric.
It is easy to show that these are indeed norms. The only slightly tricky part
is to show that kf k = 0 iff f = 0, which we obtain via the following lemma.
Lemma. Let
f ∈ C
[0
,
1] satisfy
f
(
x
)
≥
0 for all
x ∈
[0
,
1]. If
f
(
x
) is not
constantly 0, then
R
1
0
f(x) dx > 0.
Proof.
Pick
x
0
∈
[0
,
1] with
f
(
x
0
) =
a >
0. Then since
f
is continuous, there is
a
δ
such that
f
(
x
)
−f
(
x
0
)
 < a/
2 if
x −x
0
 < δ
. So
f
(
x
)
 > a/
2 in this region.
Take
g(x) =
(
a/2 x −x
0
 < δ
0 otherwise
Then f (x) ≥ g(x) for all x ∈ [0, 1]. So
Z
1
0
f(x) dx ≥
Z
1
0
g(x) dx =
a
2
· (2δ) > 0.
Example. Let X = C[0, 1], and let
d
1
(f, g) = kf − gk
1
=
Z
1
0
f(x) − g(x) dx.
Define the sequence
f
n
=
(
1 − nx x ∈ [0,
1
n
]
0 x ≥
1
n
.
f
x
1
n
1
Then
kfk
1
=
1
2
·
1
n
· 1 =
1
2n
→ 0
as n → ∞. So f
n
→ 0 in (X, d
1
) where 0(x) = 0.
On the other hand,
kf
n
k
∞
= max
x∈[0,1]
kf(x)k = 1.
So f
n
6→ 0 in the uniform metric.
So the function (
C
[0
,
1]
, d
1
)
→
(
C
[0
,
1]
, d
∞
) that maps
f 7→ f
is not continu
ous. This is similar to the case that the identity function from the usual metric
of
R
to the discrete metric of
R
is not continuous. However, the discrete metric
is a silly metric, but d
1
is a genuine useful metric here.
Using the same example, we can show that the function
G
: (
C
[0
,
1]
, d
1
)
→
(R, usual) with G(f ) = f(0) is not continuous.
We’ll now define the inner product of a real vector space. This is a general
ization of the notion of the “dot product”.
Definition (Inner product). Let
V
be a real vector space. An inner product on
V is a function h·, ·i : V × V → R such that
(i) hv, vi ≥ 0 for all v ∈ V
(ii) hv, vi = 0 if and only if v = 0.
(iii) hv, wi = hw, vi.
(iv) hv
1
+ λv
2
, w) = hv
1
, wi + λhv
2
, wi.
Example.
(i) Let V = R
n
. Then
hv, wi =
n
X
i=1
v
i
w
i
is an inner product.
(ii) Let V = C[0, 1]. Then
hf, gi =
Z
1
0
f(x)g(x) dx
is an inner product.
We just showed that norms induce metrics. The proof was completely trivial
as the definitions were almost the same. Now we want to prove that inner
products induce norms. However, this is slightly less trivial. To do so, we need
the CauchySchwarz inequality.
Theorem (CauchySchwarz inequality). If h·, ·i is an inner product, then
hv, wi
2
≤ hv, vihw, wi.
Proof. For any x, we have
hv + xw, v + xwi = hv, vi + 2xhv, wi+ x
2
hw, wi ≥ 0.
Seen as a quadratic in
x
, since it is always nonnegative, it can have at most one
real root. So
(2hv, wi)
2
− 4hv, vihw, wi ≤ 0.
So the result follows.
With this, we can show that inner products induce norms (and hence metrics).
Lemma. If h·, ·i is an inner product on V , then
kvk =
p
hv, vi
is a norm.
Proof.
(i) kvk =
p
hv, vi ≥ 0.
(ii) kvk = 0 ⇔ hv, vi = 0 ⇔ v = 0.
(iii) kλvk =
p
hλv, λvi =
p
λ
2
hv, vi = λkvk.
(iv)
(kvk + kwk)
2
= kvk
2
+ 2kvkkwk+ kwk
2
≥ hv, vi + 2hv, wi + hw, wi
= kv + wk
2
1.4 Open and closed subsets
In this section, we will study open and closed subsets of metric spaces. We will
then proceed to prove certain key properties of open and closed subsets, and
show that they are all we need to define continuity. This will lead to the next
chapter (and the rest of the course), where we completely abandon the idea of
metrics and only talk about open and closed subsets.
To define open and closed subsets, we first need the notion of balls.
Definition (Open and closed balls). Let (
X, d
) be a metric space. For any
x ∈ X, r ∈ R,
B
r
(x) = {y ∈ X : d(y, x) < r}
is the open ball centered at x.
¯
B
r
(x) = {y ∈ X : d(y, x) ≤ r}
is the closed ball centered at x.
Example.
(i) When X = R, B
r
(x) = (x − r, x + r).
¯
B
r
(x) = [x − r, x + r].
(ii) When X = R
2
,
(a)
If
d
is the metric induced by the
k
v
k
1
=
v
1

+
v
2

, then an open ball
is a rotated square.
v
1
v
2
(b)
If
d
is the metric induced by the
k
v
k
2
=
p
v
2
1
+ v
2
2
, then an open ball
is an actual disk.
v
1
v
2
(c)
If
d
is the metric induced by the
k
v
k
∞
=
max{v
1
, v
2
}
, then an
open ball is a square.
v
1
v
2
Definition (Open subset).
U ⊆ X
is an open subset if for every
x ∈ U
,
∃δ >
0
such that B
δ
(x) ⊆ U .
C ⊆ X is a closed subset if X \ C ⊆ X is open.
As if we’ve not emphasized this enough, this is a very very important
definition.
We first prove that this is a sensible definition.
Lemma. The open ball
B
r
(
x
)
⊆ X
is an open subset, and the closed ball
¯
B
r
(x) ⊆ X is a closed subset.
Proof. Given y ∈ B
r
(x), we must find δ > 0 with B
δ
(y) ⊆ B
r
(x).
v
1
v
2
r
y
Since
y ∈ B
r
(
x
), we must have
a
=
d
(
y, x
)
< r
. Let
δ
=
r − a >
0. Then if
z ∈ B
δ
(y), then
d(z, x) ≤ d(z, y) + d(y, x) < (r − a) + a = r.
So z ∈ B
r
(x). So B
δ
(y) ⊆ B
r
(x) as desired.
The second statement is equivalent to
X \
¯
B
r
(
x
) =
{y ∈ X
:
d
(
y, x
)
> r}
is
open. The proof is very similar.
Note that openness is a property of a subset.
A ⊆ X
being open depends on
both
A
and
X
, not just
A
. For example, [0
,
1
2
) is not an open subset of
R
, but
is an open subset of [0
,
1] (since it is
B
1
2
(0)), both with the Euclidean metric.
However, we are often lazy and just say “open set” instead of “open subset”.
Example.
(i)
(0
,
1)
⊆ R
is open, while [0
,
1]
⊆ R
is closed. [0
,
1)
⊆ R
is neither closed
nor open.
(ii) Q ⊆ R
is neither open nor closed, since any open interval contains both
rational numbers and irrational numbers. So any open interval cannot be
a subset of Q or R \ Q.
(iii)
Let
X
= [
−
1
,
1]
\ {
0
}
with the Euclidean metric. Let
A
= [
−
1
,
0)
⊆ X
.
Then
A
is open since it is equal to
B
1
(
−
1).
A
is also closed since it is
equal to
¯
B
1
2
(−
1
2
).
Definition (Open neighborhood). If
x ∈ X
, an open neighborhood of
x
is an
open U ⊆ X with x ∈ U.
This is not really an interesting definition, but is simply a convenient short
hand for “open subset containing x”.
Lemma. If
U
is an open neighbourhood of
x
and
x
n
→ x
, then
∃N
such that
x
n
∈ U for all n > N.
Proof.
Since
U
is open, there exists some
δ >
0 such that
B
δ
(
x
)
⊆ U
. Since
x
n
→ x
,
∃N
such that
d
(
x
n
, x
)
< δ
for all
n > N
. This implies that
x
n
∈ B
δ
(
x
)
for all n > N. So x
n
∈ U for all n > N.
Definition (Limit point). Let
A ⊆ X
. Then
x ∈ X
is a limit point of
A
if there
is a sequence x
n
→ x such that x
n
∈ A for all n.
Intuitively, a limit point is a point we can get arbitrarily close to.
Example.
(i)
If
a ∈ A
, then
a
is a limit point of
A
, by taking the sequence
a, a, a, a, ···
.
(ii) If A = (0, 1) ⊆ R, then 0 is a limit point of A, e.g. take x
n
=
1
n
.
(iii) Every x ∈ R is a limit point of Q.
It is possible to characterize closed subsets by limit points. This is often a
convenient way of proving that sets are closed.
Proposition.
C ⊆ X
is a closed subset if and only if every limit point of
C
is
an element of C.
Proof.
(
⇒
) Suppose
C
is closed and
x
n
→ x
,
x
n
∈ C
. We have to show that
x ∈ C.
Since
C
is closed,
A
=
X \ C ⊆ X
is open. Suppose the contrary that
x 6∈ C
.
Then
x ∈ A
. Hence
A
is an open neighbourhood of
x
. Then by our previous
lemma, we know that there is some
N
such that
x
n
∈ A
for all
n ≥ N
. So
x
N
∈ A
. But we know that
x
N
∈ C
by assumption. This is a contradiction. So
we must have x ∈ C.
(⇐) Suppose that C is not closed. We have to find a limit point not in C.
Since
C
is not closed,
A
is not open. So
∃x ∈ A
such that
B
δ
(
x
)
6⊆ A
for all
δ > 0. This means that B
δ
(x) ∩ C 6= ∅ for all δ > 0.
So pick
x
n
∈ B
1
n
(
x
)
∩C
for each
n >
0. Then
x
n
∈ C
,
d
(
x
n
, x
) =
1
n
→
0. So
x
n
→ x. So x is a limit point of C which is not in C.
Finally, we get to the Really Important Result
TM
that tells us metrics are
useless.
Proposition (Characterization of continuity). Let (
X, d
x
) and (
Y, d
y
) be metric
spaces, and f : X → Y . The following conditions are equivalent:
(i) f is continuous
(ii) If x
n
→ x, then f (x
n
) → f (x) (which is the definition of continuity)
(iii) For any closed subset C ⊆ Y , f
−1
(C) is closed in X.
(iv) For any open subset U ⊆ Y , f
−1
(U) is open in X.
(v)
For any
x ∈ X
and
ε >
0,
∃δ >
0 such that
f
(
B
δ
(
x
))
⊆ B
ε
(
f
(
x
)).
Alternatively, d
x
(x, z) < δ ⇒ d
y
(f(x), f(z)) < ε.
Proof.
– 1 ⇔ 2: by definition
–
2
⇒
3: Suppose
C ⊆ Y
is closed. We want to show that
f
−1
(
C
) is closed.
So let x
n
→ x, where x
n
∈ f
−1
(C).
We know that
f
(
x
n
)
→ f
(
x
) by (2) and
f
(
x
n
)
∈ C
. So
f
(
x
) is a limit
point of
C
. Since
C
is closed,
f
(
x
)
∈ C
. So
x ∈ f
−1
(
C
). So every limit
point of f
−1
(C) is in f
−1
(C). So f
−1
(C) is closed.
–
3
⇒
4: If
U ⊆ Y
is open, then
Y \ U
is closed in Y. So
f
−1
(
Y \ U
) =
X \ f
−1
(U) is closed in X. So f
−1
(U) ⊆ X is open.
–
4
⇒
5: Given
x ∈ X, ε >
0,
B
ε
(
f
(
x
)) is open in
Y
. By (4), we know
f
−1
(
B
ε
(
f
(
x
))) =
A
is open in
X
. Since
x ∈ A
,
∃δ >
0 with
B
δ
(
x
)
⊆ A
.
So
f(B
δ
(x)) ⊆ f (A) = f(f
−1
(B
ε
(f(x)))) = B
ε
(f(x))
–
5
⇒
2: Suppose
x
n
→ x
. Given
ε >
0,
∃δ >
0 such that
f
(
B
δ
(
x
))
⊆
B
ε
(
f
(
x
)). Since
x
n
→ x
,
∃N
such that
x
n
∈ B
δ
(
x
) for all
n > N
. Then
f(x
n
) ∈ f (B
δ
(x)) ⊆ B
ε
(f(x)) for all n > N . So f(x
n
) → f (x).
The third and fourth condition can allow us to immediately decide if a subset
is open or closed in some cases.
Example. Let f : R
3
→ R be defined as
f(x
1
, x
2
, x
3
) = x
2
1
+ x
4
2
x
6
3
+ x
8
1
x
2
3
.
Then this is continuous. So
{
x
∈ R
3
:
f
(x)
≤
1
}
=
f
−1
((
−∞,
1]) is closed in
R
3
.
Before we end, we prove some key properties of open subsets. These will be
used as the defining properties of open subsets in the next chapter.
Lemma.
(i) ∅ and X are open subsets of X.
(ii) Suppose V
α
⊆ X is open for all α ∈ A. Then U =
[
α∈A
V
α
is open in X.
(iii) If V
1
, ··· , V
n
⊆ X are open, then so is V =
n
\
i=1
V
i
.
Proof.
(i) ∅
satisfies the definition of an open subset vacuously.
X
is open since for
any x, B
1
(x) ⊆ X.
(ii)
If
x ∈ U
, then
x ∈ V
α
for some
α
. Since
V
α
is open, there exists
δ >
0
such that B
δ
(x) ⊆ V
α
. So B
δ
(x) ⊆
[
α∈A
V
α
= U. So U is open.
(iii)
If
x ∈ V
, then
x ∈ V
i
for all
i
= 1
, ··· , n
. So
∃δ
i
>
0 with
B
δ
i
(
x
)
⊆ V
i
.
Take
δ
=
min{δ
1
, ··· , δ
n
}
. So
B
δ
(
x
)
⊆ V
i
for all
i
. So
B
δ
(
x
)
⊆ V
. So
V
is
open.
Note that we can take infinite unions or finite intersection, but not infinite
intersections. For example, the intersection of all (
−
1
n
,
1
n
) is
{
0
}
, which is not
open.
2 Topological spaces
2.1 Definitions
We have previously shown that a function
f
is continuous iff
f
−1
(
U
) is open
whenever
U
is open. Convergence can also be characterized using open sets only.
This suggests that we can dump the metric and just focus on the open sets.
What we will do is to define the topology of a space
X
to be the set of open
sets of
X
. Then this topology would define most of the structure or geometry of
X, and we don’t need to care about metrics.
Definition (Topological space). A topological space is a set
X
(the space)
together with a set U ⊆ P(X) (the topology) such that:
(i) ∅, X ∈ U
(ii) If V
α
∈ U for all α ∈ A, then
[
α∈A
V
α
∈ U.
(iii) If V
1
, ··· , V
n
∈ U, then
n
\
i=1
V
i
∈ U.
The elements of
X
are the points, and the elements of
U
are the open subsets of
X.
Definition (Induced topology). Let (
X, d
) be a metric space. Then the topology
induced by d is the set of all open sets of X under d.
Example. Let
X
=
R
n
and consider the two metrics
d
1
(x
,
y) =
k
x
−
y
k
1
and
d
∞
(x, y) = kx − yk
∞
. We will show that they induce the same topology.
Recall that the metrics are defined by
kvk =
n
X
i=1
v
i
, kvk
∞
= max
1≤i≤n
v
i
.
This implies that
kvk
∞
≤ kvk
1
≤ nkvk
∞
.
This in turn implies that
B
∞
r
(x) ⊇ B
1
r
(x) ⊇ B
∞
r/n
(x).
v
1
v
2
B
1
B
∞
To show that the metrics induce the same topology, suppose that
U
is open with
respect to
d
1
, and we show that it is open with respect to
d
∞
. Let
x ∈ U
. Since
U
is open with respect to
d
1
, there exists some
δ >
0 such that
B
1
δ
(
x
)
⊆ U
. So
B
∞
δ/n
(x) ⊆ B
1
δ
(x) ⊆ U . So U is open with respect to d
∞
.
The other direction is similar.
Example. Let
X
=
C
[0
,
1]. Let
d
1
(
f, g
) =
kf − gk
1
and
d
∞
(
f, g
) =
kf − gk
∞
.
Then they do not induce the same topology, since (
X, d
1
)
→
(
X, d
∞
) by
f 7→ f
is not continuous.
It is possible to have some other topologies that are not induced by metrics.
Example.
(i) Let X be any set.
(a) U = {∅, X} is the coarse topology on X.
(b) U
=
P
(
X
) is the discrete topology on
X
, since it is induced by the
discrete metric.
(c) U
=
{A ⊆ X
:
X \A is finite or A
=
∅}
is the cofinite topology on
X
.
(ii) Let X = R, and U = {(a, ∞) : a ∈ R} is the right order topology on R.
Now we can define continuous functions in terms of the topology only.
Definition (Continuous function). Let
f
:
X → Y
be a map of topological
spaces. Then f is continuous if f
−1
(U) is open in X whenever U is open in Y .
When the topologies are induced by metrics, the topological and metric
notions of continuous functions coincide, as we showed in the previous chapter.
Example.
(i) Any function f : X → Y is continuous if X has the discrete topology.
(ii) Any function f : X → Y is continuous if Y has the coarse topology.
(iii)
If
X
and
Y
both have cofinite topology, then
f
:
X → Y
is continuous iff
f
−1
({y}) is finite for every y ∈ Y .
Lemma. If
f
:
X → Y
and
g
:
Y → Z
are continuous, then so is
g ◦ f
:
X → Z
.
Proof.
If
U ⊆ Z
is open,
g
is continuous, then
g
−1
(
U
) is open in
Y
. Since
f
is
also continuous, f
−1
(g
−1
(U)) = (g ◦ f)
−1
(U) is open in X.
In group theory, we had the notion of isomorphisms between groups. Isomor
phic groups are equaluptorenamingofelements, and are considered to be the
same for most purposes.
Similarly, we will define homeomorphisms between topological spaces, and
homeomorphic topological spaces will be considered to be the same (notice the
“e” in homeomorphism).
Definition (Homeomorphism). f : X → Y is a homeomorphism if
(i) f is a bijection
(ii) Both f and f
−1
are continuous
Equivalently, f is a bijection and U ⊆ X is open iff f (U) ⊆ Y is open.
Two spaces are homeomorphic if there exists a homeomorphism between
them, and we write X ' Y .
Note that we specifically require
f
and
f
−1
to be both continuous. In
group theory, if
φ
is a bijective homomorphism, then
φ
−1
is automatically a
homomorphism as well. However, this is not true for topological spaces.
f
being continuous does not imply
f
−1
is continuous, as illustrated by the example
below.
Example. Let
X
=
C
[0
,
1] with the topology induced by
k · k
1
and
Y
=
C
[0
,
1]
with the topology induced by
k · k
∞
. Then
F
:
Y → X
by
f 7→ f
is continuous
but F
−1
is not.
Example. Let
X
= [0
,
2
π
) and
Y
=
S
1
=
{z ∈ C
:
z
= 1
}
. Then
f
:
X → Y
given by f(x) = e
ix
is continuous but its inverse is not.
Similar to isomorphisms, we can show that homeomorphism is an equivalence
relation.
Lemma. Homeomorphism is an equivalence relation.
Proof.
(i) The identity map I
X
: X → X is always a homeomorphism. So X ' X.
(ii)
If
f
:
X → Y
is a homeomorphism, then so is
f
−1
:
Y → X
. So
X ' Y ⇒ Y ' X.
(iii) If f : X → Y and g : Y → Z are homeomorphisms, then g ◦ f : X → Z is
a homeomorphism. So X ' Y and Y ' Z implies X ' Z.
Example.
(i)
Under the usual topology, the open intervals (0
,
1)
'
(
a, b
) for all
a, b ∈ R
,
using the homeomorphism x 7→ a + (b − a)x.
Similarly, [0, 1] ' [a, b]
(ii) (−1, 1) ' R by x 7→ tan(
π
2
x).
(iii) R ' (0, ∞) by x 7→ e
x
.
(iv) (a, ∞) ' (b, ∞) by x 7→ x + (b − a).
The fact that
'
is an equivalence relation implies that any 2 open intervals in
R
are homeomorphic.
It is relatively easy to show that two spaces are homeomorphic. We just have
to write down a homeomorphism. However, it is rather difficult to prove that
two spaces are not homeomorphic.
For example, is (0
,
1) homeomorphic to [0
,
1]? No, but we aren’t able
to immediately prove it now. How about
R
and
R
2
? Again they are not
homeomorphic, but to show this we will need some tools that we’ll develop in
the next few lectures.
How about
R
m
and
R
n
in general? They are not homeomorphic, but we
won’t be able to prove this rigorously in the course. To properly prove this, we
will need tools from algebraic topology.
So how can we prove that two spaces are not homeomorphic? In group theory,
we could prove that two groups are not isomorphic by, say, showing that they
have different orders. Similarly, to distinguish between topological spaces, we
have to define certain topological properties. Then if two spaces have different
topological properties, we can show that they are not homeomorphic.
But before that, we will first define many many useful definitions in topological
spaces, including sequences, subspaces, products and quotients. The remainder
of the chapter will be mostly definitions that we will use later.
2.2 Sequences
To define the convergence of a sequence using open sets, we again need the
concept of open neighbourhoods.
Definition (Open neighbourhood). An open neighbourhood of
x ∈ X
is an open
set U ⊆ X with x ∈ U .
Now we can use this to define convergence of sequences.
Definition (Convergent sequence). A sequence
x
n
→ x
if for every open
neighbourhood U of x, ∃N such that x
n
∈ U for all n > N.
Example.
(i)
If
X
has the coarse topology, then any sequence
x
n
converges to every
x ∈ X, since there is only one open neighbourhood of x.
(ii)
If
X
has the cofinite topology, no two
x
n
s are the same, then
x
n
→ x
for every
x ∈ X
, since every open set can only have finitely many
x
n
not
inside it.
This looks weird. This is definitely not what we used to think of sequences.
At least, we would want to have unique limits.
Fortunately, there is a particular class of spaces where sequences are well
behaved and have at most one limit.
Definition (Hausdorff space). A topological space
X
is Hausdorff if for all
x
1
, x
2
∈ X
with
x
1
6
=
x
2
, there exist open neighbourhoods
U
1
of
x
1
,
U
2
of
x
2
such that U
1
∩ U
2
= ∅.
Lemma. If
X
is Hausdorff,
x
n
is a sequence in
X
with
x
n
→ x
and
x
n
→ x
0
,
then x = x
0
, i.e. limits are unique.
Proof.
Suppose the contrary that
x 6
=
x
0
. Then by definition of Hausdorff, there
exist open neighbourhoods U, U
0
of x, x
0
respectively with U ∩ U
0
= ∅.
Since
x
n
→ x
and
U
is a neighbourhood of
x
, by definition, there is some
N
such that whenever
n > N
, we have
x
n
∈ U
. Similarly, since
x
n
→ x
0
, there is
some N
0
such that whenever n > N
0
, we have x
n
∈ U
0
.
This means that whenever
n > max
(
N, N
0
), we have
x
n
∈ U
and
x
n
∈ U
0
.
So x
n
∈ U ∩ U
0
. This contradicts the fact that U ∩ U
0
= ∅.
Hence we must have x = x
0
.
Example.
(i)
If
X
has more than 1 element, then the coarse topology on
X
is not
Hausdorff.
(ii)
If
X
has infinitely many elements, the cofinite topology on
X
is not
Hausdorff.
(iii) The discrete topology is always Hausdorff.
(iv)
If (
X, d
) is a metric space, the topology induced by
d
is Hausdorff: for
x
1
6
=
x
2
, let
r
=
d
(
x
1
, x
2
)
>
0. Then take
U
i
=
B
r/2
(
x
i
). Then
U
1
∩U
2
=
∅
.
2.3 Closed sets
We will define closed sets similarly to what we did for metric spaces.
Definition (Closed sets). C ⊆ X is closed if X \ C is an open subset of X.
Lemma.
(i) If C
α
is a closed subset of X for all α ∈ A, then
T
α∈A
C
α
is closed in X.
(ii) If C
1
, ··· , C
n
are closed in X, then so is
S
n
i=1
C
i
.
Proof.
(i)
Since
C
α
is closed in
X
,
X \ C
α
is open in
X
. So
S
α∈A
(
X \ C
α
) =
X \
T
α∈A
C
α
is open. So
T
α∈A
C
α
is closed.
(ii)
If
C
i
is closed in
X
, then
X \ C
i
is open. So
T
n
i=1
(
X \ C
i
) =
X \
S
n
i=1
C
i
is open. So
S
n
i=1
C
i
is closed.
This time we can take infinite intersections and finite unions, which is the
opposite of what we have for open sets.
Note that it is entirely possible to define the topology to be the collection of
all closed sets instead of open sets, but people seem to like open sets more.
Corollary. If X is Hausdorff and x ∈ X, then {x} is closed in X.
Proof.
For all
y ∈ X
, there exist open subsets
U
y
, V
y
with
y ∈ U
y
, x ∈ V
y
,
U
y
∩ V
y
= ∅.
Let
C
y
=
X \ U
y
. Then
C
y
is closed,
y 6∈ C
y
,
x ∈ C
y
. So
{x}
=
T
y6=x
C
y
is
closed since it is an intersection of closed subsets.
2.4 Closure and interior
2.4.1 Closure
Given a subset
A ⊆ X
, if
A
is not closed, we would like to find the smallest
closed subset containing A. This is known as the closure of A.
Officially, we define the closure as follows:
Definition. Let X be a topological space and A ⊆ X. Define
C
A
= {C ⊆ X : A ⊆ C and C is closed in X}
Then the closure of A in X is
¯
A =
\
C∈C
A
C.
First we do a sanity check: since
¯
A
is defined as an intersection, we should
make sure we are not taking an intersection of no sets. This is easy: since
X
is closed in
X
(its complement
∅
is open),
C
A
6
=
∅
. So we can safely take the
intersection.
Since
¯
A
is an intersection of closed sets, it is closed in
X
. Also, if
C ∈ C
A
,
then A ⊆ C. So A ⊆
T
C∈C
A
C =
¯
A. In fact, we have
Proposition.
¯
A is the smallest closed subset of X which contains A.
Proof.
Let
K ⊆ X
be a closed set containing
A
. Then
K ∈ C
A
. So
¯
A
=
T
C∈C
A
C ⊆ K. So
¯
A ⊆ K.
We basically defined the closure such that it is the smallest closed subset of
X which contains A.
However, while this “clever” definition makes it easy to prove the above
property, it is rather difficult to directly use it to compute the closure.
To compute the closure, we define the limit point analogous to what we did
for metric spaces.
Definition (Limit point). A limit point of
A
is an
x ∈ X
such that there is a
sequence x
n
→ x with x
n
∈ A for all n.
In general, limit points are easier to compute, and can be a useful tool for
determining the closure of A.
Now let
L(A) = {x ∈ X : x is a limit point of A}.
We immediately get the following lemma.
Lemma. If C ⊆ X is closed, then L(C) = C.
Proof.
Exactly the same as that for metric spaces. We will also prove a more
general result very soon that implies this.
Recall that we proved the converse of this statement for metric spaces.
However, the converse is not true for topological spaces in general.
Example. Let
X
be an uncountable set (e.g.
R
), and define a topology on
X
by saying a set is open if it is empty or has countable complement. One can
check that this indeed defines a topology. We claim that the only sequences that
converge are those that are eventually constant.
Indeed, if x
n
is a sequence and x ∈ X, then consider the open set
U = (X \ {x
n
: n ∈ N}) ∪ {x}.
Then the only element in the sequence
x
n
that can possibly be contained in
U
is x itself. So if x
n
→ x, this implies that x
n
is eventually always x.
In particular, it follows that L(A) = A for all A ⊆ X.
However, we do have the following result:
Proposition. L(A) ⊆
¯
A.
Proof.
If
A ⊆ C
, then
L
(
A
)
⊆ L
(
C
). If
C
is closed, then
L
(
C
) =
C
. So
C ∈ C
A
⇒ L(A) ⊆ C. So L(A) ⊆
T
C∈C
A
C =
¯
A.
This in particular implies the previous lemma, since for any
A
, we have
A ⊆ L(A) ⊆
¯
A, and when A is closed, we have A =
¯
A.
Finally, we have the following corollary that can help us find the closure of
subsets:
Corollary. Given a subset
A ⊆ X
, if we can find some closed
C
such that
A ⊆ C ⊆ L(A), then we in fact have C =
¯
A.
Proof. C ⊆ L
(
A
)
⊆
¯
A ⊆ C
, where the last step is since
¯
A
is the smallest closed
set containing A. So C = L(A) =
¯
A.
Example.
– Let (a, b) ⊆ R. Then (a, b) = [a, b].
– Let Q ⊆ R. Then
¯
Q = R.
– R \ Q = R.
–
In
R
n
with the Euclidean metric,
B
r
(x)
=
¯
B
r
(
x
). In general,
B
r
(x) ⊆
¯
B
r
(
x
), since
¯
B
r
(
x
) is closed and
B
r
(
x
)
⊆
¯
B
r
(
x
), but these need not be
equal.
For example, if
X
has the discrete metric, then
B
1
(
x
) =
{x}
. Then
B
1
(x) = {x}, but
¯
B
1
(x) = X.
In the above example, we had
¯
Q
=
R
. In some sense, all points of
R
are
“surrounded” by points in Q. We say that Q is dense in R.
Definition (Dense subset). A ⊆ X is dense in X if
¯
A = X.
Example. Q and R \ Q are both dense in R with the usual topology.
2.4.2 Interior
We defined the closure of
A
to be the smallest closed subset containing
A
. We
can similarly define the interior of
A
to be the largest open subset contained in
A.
Definition (Interior). Let A ⊆ X, and let
O
A
= {U ⊆ X : U ⊆ A, U is open in X}.
The interior of A is
Int(A) =
[
U∈O
A
U.
Proposition. Int(A) is the largest open subset of X contained in A.
The proof is similar to proof for closure.
To find the closure, we could use limit points. What trick do we have to find
the interior?
Proposition. X \ Int(A) = X \ A
Proof. U ⊆ A ⇔
(
X \U
)
⊇
(
X \A
). Also,
U open in X ⇔ X \U is closed in X
.
So the complement of the largest open subset of
X
contained in
A
will be
the smallest closed subset containing X \ A.
Example. Int(Q) = Int(R \ Q) = ∅.
2.5 New topologies from old
In group theory, we had the notions of subgroups, products and quotients. There
are exact analogies for topological spaces. In this section, we will study the
subspace topology, product topology and quotient topology.
2.5.1 Subspace topology
Definition (Subspace topology). Let
X
be a topological space and
Y ⊆ X
.
The subspace topology on
Y
is given by:
V
is an open subset of
Y
if there is
some U open in X such that V = Y ∩ U.
If we simply write
Y ⊆ X
and don’t specify a topology, the subspace topology
is assumed. For example, when we write Q ⊆ R, we are thinking of Q with the
subspace topology inherited from R.
Example. If (
X, d
) is a metric space and
Y ⊆ X
, then the metric topology on
(Y, d) is the subspace topology, since B
Y
r
(y) = Y ∩ B
X
r
(y).
To show that this is indeed a topology, we need the following set theory facts:
Y ∩
[
α∈A
V
α
!
=
[
α∈A
(Y ∩ V
α
)
Y ∩
\
α∈A
V
α
!
=
\
α∈A
(Y ∩ V
α
)
Proposition. The subspace topology is a topology.
Proof.
(i) Since ∅ is open in X, ∅ = Y ∩ ∅ is open in Y .
Since X is open in X, Y = Y ∩ X is open in Y .
(ii) If V
α
is open in Y , then V
α
= Y ∩ U
α
for some U
α
open in X. Then
[
α∈A
V
α
=
[
α∈A
(Y ∩ U
α
) = Y ∩
[
α∈U
U
α
!
.
Since
S
U
α
is open in X, so
S
V
α
is open in Y .
(iii) If V
i
is open in Y , then V
i
= Y ∩ U
i
for some open U
i
⊆ X. Then
n
\
i=1
V
i
=
n
\
i=1
(Y ∩ U
i
) = Y ∩
n
\
i=1
U
i
!
.
Since
T
U
i
is open,
T
V
i
is open.
Recall that if
Y ⊆ X
, there is an inclusion function
ι
:
Y → X
that sends
y 7→ y
. We can use this to obtain the following defining property of a subspace.
Proposition. If
Y
has the subspace topology,
f
:
Z → Y
is continuous iff
ι ◦ f : Z → X is continuous.
Proof.
(
⇒
) If
U ⊆ X
is open, then
ι
−1
(
U
) =
Y ∩ U
is open in
Y
. So
ι
is
continuous. So if f is continuous, so is ι ◦ f .
(
⇐
) Suppose we know that
ι ◦ f
is continuous. Given
V ⊆ Y
is open, we
know that
V
=
Y ∩ U
=
ι
−1
(
U
). So
f
−1
(
V
) =
f
−1
(
ι
−1
(
U
))) = (
ι ◦ f
)
−1
(
U
) is
open since ι ◦ f is continuous. So f is continuous.
This property is “defining” in the sense that it can be used to define a
subspace:
Y
is a subspace of
X
if there exists some function
ι
:
Y → X
such
that for any f , f is continuous iff ι ◦ f is continuous.
Example.
D
n
=
{
v
∈ R
n
:

v
 ≤
1
}
is the
n
dimensional closed unit disk.
S
n−1
= {v ∈ R
n
: v = 1} is the n − 1dimensional sphere.
We have
Int(D
n
) = {v ∈ R
n
: v < 1} = B
1
(0).
This is, in fact, homeomorphic to
R
n
. To show this, we can first pick our favorite
homeomorphism
f
: [0
,
1)
7→
[1
, ∞
). Then v
7→ f
(

v

)v is a homeomorphism
Int(D
n
) → R
n
.
2.5.2 Product topology
If X and Y are sets, the product is defined as
X × Y = {(x, y) : x ∈ X, y ∈ Y }
We have the projection functions π
1
: X × Y → X, π
2
: X × Y → Y given by
π
1
(x, y) = x, π
2
(x, y) = y.
If A ⊆ X, B ⊆ Y , then we have A × B ⊆ X × Y .
Given topological spaces
X
,
Y
, we can define a topology on
X ×Y
as follows:
Definition (Product topology). Let
X
and
Y
be topological spaces. The product
topology on X × Y is given by:
U ⊆ X × Y
is open if: for every (
x, y
)
∈ U
, there exist
V
x
⊆ X, W
y
⊆ Y
open neighbourhoods of x and y such that V
x
× W
y
⊆ U .
Example.
–
If
V ⊆ X
and
W ⊆ Y
are open, then
V × W ⊆ X × Y
is open (take
V
x
= V , W
y
= W ).
–
The product topology on
R × R
is same as the topology induced by the
k · k
∞
, hence is also the same as the topology induced by
k · k
2
or
k · k
1
.
Similarly, the product topology on
R
n
=
R
n−1
×R
is also the same as that
induced by k · k
∞
.
– (0, 1) × (0, 1) × ··· × (0, 1) ⊆ R
n
is the open ndimensional cube in R
n
.
Since (0, 1) ' R, we have (0, 1)
n
' R
n
' Int(D
n
).
–
[0
,
1]
× S
n
'
[1
,
2]
× S
n
' {
v
∈ R
n+1
: 1
≤ 
v
 ≤
2
}
, where the last
homeomorphism is given by (
t,
w)
7→ t
w with inverse v
7→
(

v
,
ˆ
v
). This is
a thickened sphere.
–
Let
A ⊆ {
(
r, z
) :
r >
0
} ⊆ R
2
,
R
(
A
) be the set obtained by rotating
A
around the z axis. Then R(A) ' S × A by
(x, y, z) = (v, z) 7→ (
ˆ
v, (v, z)).
In particular, if
A
is a circle, then
R
(
A
)
' S
1
× S
1
=
T
2
is the two
dimensional torus.
r
z
The defining property is that
f
:
Z → X × Y
is continuous iff
π
1
◦ f
and
π
2
◦ f are continuous.
Note that our definition of the product topology is rather similar to the
definition of open sets for metrics. We have a special class of subsets of the
form
V × W
, and a subset
U
is open iff every point
x ∈ U
is contained in some
V × W ⊆ U . In some sense, these subsets “generate” the open sets.
Alternatively, if U ⊆ X × Y is open, then
U =
[
(x,y)∈U
V
x
× W
y
.
So
U ⊆ X ×Y
is open if and only if it is a union of members of our special class
of subsets.
We call this special class the basis.
Definition (Basis). Let
U
be a topology on
X
. A subset
B ⊆ U
is a basis if
“U ∈ U iff U is a union of sets in B”.
Example.
– {V ×W
:
V ⊆ X, W ⊆ Y are open}
is a basis for the product topology for
X × Y .
– If (X, d) is a metric space, then
{B
1/n
(x) : n ∈ N
+
, x ∈ X}
is a basis for the topology induced by d.
2.5.3 Quotient topology
If
X
is a set and
∼
is an equivalence relation on
X
, then the quotient
X/∼
is the
set of equivalence classes. The projection
π
:
X → X/∼
is defined as
π
(
x
) = [
x
],
the equivalence class containing x.
Definition (Quotient topology). If
X
is a topological space, the quotient topology
on X/∼ is given by: U is open in X/∼ if π
−1
(U) is open in X.
We can think of the quotient as “gluing” the points identified by
∼
together.
The defining property is
f
:
X/∼ → Y
is continuous iff
f ◦ π
:
X → Y
is
continuous.
Example.
–
Let
X
=
R
,
x ∼ y
iff
x − y ∈ Z
. Then
X/∼
=
R/Z ' S
1
, given by
[x] 7→ (cos 2πx, sin 2πx).
–
Let
X
=
R
2
. Then v
∼
w iff v
−
w
∈ Z
2
. Then
X/∼
=
R
2
/Z
2
=
(
R/Z
)
×
(
R/Z
)
' S
1
×S
1
=
T
2
. Similarly,
R
n
/Z
n
=
T
n
=
S
1
×S
1
×···×S
1
.
–
If
A ⊆ X
, define
∼
by
x ∼ y
iff
x
=
y
or
x, y ∈ A
. This glues everything
in A together and leaves everything else alone.
We often write this as
X/A
. Note that this is not consistent with the
notation we just used above!
◦
Let
X
= [0
,
1] and
A
=
{
0
,
1
}
, then
X/A ' S
1
by, say,
t 7→
(
cos
2
πt, sin
2
πt
). Intuitively, the equivalence relation says that the
two end points of [0
,
1] are “the same”. So we join the ends together
to get a circle.
◦
Let
X
=
D
n
and
A
=
S
n−1
. Then
X/A ' S
n
. This can be pictured
as pulling the boundary of the disk together to a point to create a
closed surface
–
Let
X
= [0
,
1]
×
[0
,
1] with
∼
given by (0
, y
)
∼
(1
, y
) and (
x,
0)
∼
(
x,
1),
then X/∼ ' S
1
× S
1
= T
2
, by, say
(x, y) 7→
(cos 2πx, sin 2πx), (cos 2πy, sin 2πy)
Similarly, T
3
= [0, 1]
3
/∼, where the equivalence is analogous to above.
Note that even if
X
is Hausdorff,
X/∼
may not be! For example,
R/Q
is not
Hausdorff.
3 Connectivity
Finally we can get to something more interesting. In this chapter, we will study
the connectivity of spaces. Intuitively, we would want to say that a space is
“connected” if it is onepiece. For example,
R
is connected, while
R \ {
0
}
is
not. We will come up with two different definitions of connectivity  normal
connectivity and path connectivity, where the latter implies the former, but not
the other way round.
3.1 Connectivity
We will first look at normal connectivity.
Definition (Connected space). A topological space
X
is disconnected if
X
can
be written as
A ∪ B
, where
A
and
B
are disjoint, nonempty open subsets of
X
.
We say A and B disconnect X.
A space is connected if it is not disconnected.
Note that being connected is a property of a space, not a subset. When we
say “
A
is a connected subset of
X
”, it means
A
is connected with the subspace
topology inherited from X.
Being (dis)connected is a topological property, i.e. if
X
is (dis)connected,
and
X ' Y
, then
Y
is (dis)connected. To show this, let
f
:
X → Y
be the
homeomorphism. By definition,
A
is open in
X
iff
f
(
A
) is open in
Y
. So
A
and
B disconnect X iff f(A) and f (B) disconnect Y .
E