4Compactness

IB Metric and Topological Spaces

4.2 Products and quotients

4.2.1 Products

Recall the product topology on

X × Y

.

U ⊆ X × Y

is open if it is a union of

sets of the form V × W such that V ⊆ X, W ⊆ Y are open.

The major takeaway of this section is the following theorem:

Theorem. If X and Y are compact, then so is X × Y .

Proof.

First consider the special type of open cover

V

of

X × Y

such that every

U ∈ V has the form U = V × W, where V ⊆ X and W ⊆ Y are open.

For every (x, y) ∈ X × Y , there is U

xy

∈ V with (x, y) ∈ U

xy

. Write

U

xy

= V

xy

× W

xy

,

where V

xy

⊆ X, W

xy

⊆ Y are open, x ∈ V

xy

, y ∈ W

xy

.

Fix

x ∈ X

. Then

W

x

=

{W

xy

:

y ∈ Y }

is an open cover of

Y

. Since

Y

is

compact, there is a finite subcover {W

xy

1

, ··· , W

xy

n

}.

Then

V

x

=

T

n

i=1

V

xy

i

is a finite intersection of open sets. So

V

x

is open in

X

.

Moreover, V

x

= {U

xy

1

, ··· , U

xy

n

} covers V

x

× Y .

x

U

xy

i

V

x

× Y

X

Y

Now

O

=

{V

x

:

x ∈ X}

is an open cover of

X

. Since

X

is compact, there is

a finite subcover

{V

x

1

, ··· , V

x

m

}

. Then

V

0

=

S

m

i=1

V

x

i

is a finite subset of

V

,

which covers all of X × Y .

In the general, case, suppose

V

is an open cover of

X × Y

. For each (

x, y

)

∈

X × Y

,

∃U

xy

∈ V

with (

x, y

)

∈ U

xy

. Since

U

xy

is open,

∃V

xy

⊆ X, W

xy

⊆ Y

open with V

xy

× W

xy

⊆ U

xy

and x ∈ V

xy

, y ∈ W

xy

.

Then

Q

=

{V

xy

× W

xy

: (

x, y

)

∈

(

X, Y

)

}

is an open cover of

X × Y

of

the type we already considered above. So it has a finite subcover

{V

x

1

y

1

×

W

x

1

y

1

, ··· , V

x

n

y

n

×W

x

n

y

n

}

. Now

V

x

i

y

i

×W

x

i

y

i

⊆ U

x

i

y

i

. So

{U

x

1

y

1

, ··· , U

x

n

y

n

}

is a finite subcover of X × Y .

Example. The unit cube [0, 1]

n

= [0, 1] × [0, 1] × ··· × [0, 1] is compact.

Corollary (Heine-Borel in

R

n

).

C ⊆ R

n

is compact iff

C

is closed and bounded.

Proof.

If

C

is bounded,

C ⊆

[

−N, N

]

n

for some

N ∈ R

, which is compact. The

rest of the proof is exactly the same as for n = 1.

4.2.2 Quotients

It is easy to show that the quotient of a compact space is compact, since every

open subset in the quotient space can be projected back to an open subset the

original space. Hence we can project an open cover from the quotient space to

the original space, and get a finite subcover. The details are easy to fill in.

Instead of proving the above, in this section, we will prove that compact

quotients have some nice properties. We start with a handy proposition.

Proposition. Suppose

f

:

X → Y

is a continuous bijection. If

X

is compact

and Y is Hausdorff, then f is a homeomorphism.

Proof.

We show that

f

−1

is continuous. To do this, it suffices to show (

f

−1

)

−1

(

C

)

is closed in

Y

whenever

C

is closed in

X

. By hypothesis,

f

is a bijection . So

(f

−1

)

−1

(C) = f(C).

Supposed

C

is closed in

X

. Since

X

is compact,

C

is compact. Since

f

is

continuous,

f

(

C

) = (

im f|

C

) is compact. Since

Y

is Hausdorff and

f

(

C

)

⊆ Y

is

compact, f (C) is closed.

We will apply this to quotients.

Recall that if

∼

is an equivalence relation on

X

,

π

:

X → X/∼

is continuous

iff f ◦ π : X → Y is continuous.

Corollary. Suppose

f

:

X/∼ → Y

is a bijection,

X

is compact,

Y

is Hausdorff,

and f ◦ π is continuous, then f is a homeomorphism.

Proof.

Since

X

is compact and

π

:

X 7→ X/∼

is continuous,

im π ⊆ X/∼

is

compact. Since

f ◦ π

is continuous,

f

is continuous. So we can apply the

proposition.

Example. Let

X

=

D

2

and

A

=

S

1

⊆ X

. Then

f

:

X/A 7→ S

2

by (

r, θ

)

7→

(1, πr, θ) in spherical coordinates is a homeomorphism.

We can check that

f

is a continuous bijection and

D

2

is compact. So

X/A ' S

2

.