2Topological spaces

IB Metric and 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 equal-up-to-renaming-of-elements, 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.