2Topological spaces
IB Metric and Topological Spaces
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.