Tychonoff's theorem.pdf

Topology
Tychonoff’s Theorem
K. Christian Chávez
Yachay Tech University
February 24, 2023
Christian Chávez Topology February 24, 2023 1 / 23

Contents
1 Filters and ultrafilters
2 Compactness
3 Tychonoff’s theorem

Preliminaries
Definition 1
Let X be a set. A collection (Aα)α∈I of subsets of X has the finite
intersection property (FIP) if

α∈If
Aα ̸= ∅
for every finite set If ⊆ I.

Filters and ultrafilters
Definition 2
Let X be a set. A nonempty collection F ⊆ P(X) is a filter on X if
(i)
∅ /
∈ F,
(ii)
if A ∈ F and A ⊆ B, then B ∈ F, and
(iii)
if A, B ∈ F, then A ∩ B ∈ F.

Definition 3
An ultrafilter on a set X is a filter that on X that is not properly
contained in any other filter on X.
Remark 1
Every filter has the FIP. This follows from item iii of Definition 2.
Remark 2
If (X, T ) is a topological space and x ∈ X, the collection
NX (x) = {A ⊆ X : ∃U ∈ T such that x ∈ U ⊆ A}
is a filter on X, called the neighborhood filter of x.

Lemma 4
Let X be a set and E ⊆ P(X). If E has the FIP, then there is a
unique (and smallest) filter F on X such that E ⊆ F.

Proof.
Suppose E ⊆ P(X) has the FIP.
Denote by E′
the set of all the finite intersections of elements of E.
Let’s see that
E′′
= {A ⊆ X : A ⊇ U for some U ∈ E′
}
is a filter on X.
(i)
As E has the FIP, ∅ /
∈ E′′
(ii)
Suppose A ∈ E′′
and let B ⊇ A. Then there is U ∈ E′
such
that A ⊇ U. Clearly B ⊇ U, so B ∈ E′′
.
(iii)
Let A, B ∈ E′′
. There are U, V ∈ E′
such that A ⊇ U and
B ⊇ V. Because A ∩ B ⊇ U ∩ V and U ∩ V ∈ E′
, we have
A ∩ B ∈ E′′
.

Proof.
As a result, E′′
is a filter on X. In order to see that E′′
is the
smallest one that contains E, suppose there is another filter F on
X such that F ⊇ E. Let A ∈ E′′
. By definition, there is U ∈ E′
such
that A ⊇ U. We have U ∈ F because F ⊇ E has the FIP, but then
A ∈ F by item ii of Definition 2. Thus F ⊇ E′′
, whence E′′
cannot
contain any other filter containing E.

Lemma 5
Let U be a filter on a set X. Then U is an ultrafilter on X if and only
if either A ∈ U or XA ∈ U, for any A ⊆ X.

Lemma 6
Let X and Y be sets, let F be a filter on X, and let f : X → Y be a
function. Then
f(F) :=

A ⊆ Y : f−1
(A) ∈ F
is a filter on Y.
Remark 3
The notation f(F) must not be confused with the direct image of a subset
of X under f. Nevertheless we will refer to f(F) as the image of F under
f.

Proof.
First note that Y ∈ f(F) as f−1
(Y) = X ∈ F, so f(F) ̸= ∅.
Further:
(i)
Since f−1
(∅) = ∅ /
∈ F, we have ∅ /
∈ f(F).
(ii)
Let A ∈ f(F) and suppose Y ⊇ B ⊇ A. Since f−1
(A) ∈ F
and f−1
(A) ⊆ f−1
(B), it follows f−1
(B) ∈ F. Thus B ∈ f(F).
(iii)
Let A, B ∈ f(F). Because f−1
(A), f−1
(B) ∈ F,
f−1
(A ∩ B) = f−1
(A) ∩ f−1
(A) ∈ F,
whence A ∩ B ∈ f(F).
This proves f(F) is a filter on Y.

Corollary 7
If F is an ultrafilter on X, then f(F) is an ultrafilter on Y.
Proof.
Suppose F is an ultrafilter on X. Let A ⊆ Y and assume that
A /
∈ f(F). Notice that f−1
(YA) = Xf−1
(A). Since f−1
(A) /
∈ F,
we obtain Xf−1
(A) ∈ F. Thus f−1
(YA) ∈ F, whence, by
definition, YA ∈ f(F). This shows that either A ∈ f(F) or
YA ∈ f(F). We conclude that f(F) is an ultrafilter on Y by
Lemma 5.

Definition 8
Let (X, T ) be a topological space, F ⊆ P(X) a filter on X, and
x ∈ X. Then F converges to x if U ∈ F for every open set
U ∈ NX (x). In this case we write F → x.

Lemma 9
Let (X, T ) and (Y, U) be topological spaces, and consider a
continuous function f : X → Y. If F is a filter on X that converges
to a point x ∈ X, then f(F) → f(x).
Proof.
Suppose F → x and let U ∈ NY (f(x)) be open. Since f is
continuous, f−1
(U) ∈ NX (x) is open. Thus f−1
(U) ∈ F, whence
U ∈ f(F). Because U was arbitrarily chosen, we conclude that
f(F) → f(x).

Compactness
Compactness
Lemma 10
A topological space (X, T ) is compact if and only if for every
collection (Aα)α∈I of closed subsets of X with the FIP,

α∈I
Aα ̸= ∅.

Compactness
Lemma 11
A topological space (X, T ) is compact if and only if every ultrafilter
on X converges.
Proof.
(⇒) Suppose (X, T ) is compact, and suppose for the sake of
contradiction that U is an ultrafilter on X that does not
converge to any point. This means that for every x ∈ X, there
is Ux ∈ NX (x) open such that Ux /
∈ U. Note that
{Ux : x ∈ X} is an open cover of X, and so since X is
compact it has a finite subcover {Ux1
, Ux2
, . . . , Uxn }. Thus,
X =
n
[
i=1
Uxi
.

Compactness
Proof.
It must be the case that there is k ∈ {1, . . . , n} such that Uxk
∈ U.
If this were not the case, then Uxk
/
∈ U for all k ∈ {1, . . . , n} and,
by Lemma 5, XUxk
∈ U for all k ∈ {1, . . . , n}, which leads us to
U ∋
n

k=1
(XUxk
) = X
n
[
k=1
Uxk
!
= XX = ∅,
a contradiction. Therefore, Uxk
∈ U for some k ∈ {1, . . . , n}.
However this is again a contradiction as we stated earlier that
Ux /
∈ U for every x ∈ X. We conclude that U converges.

Compactness
Proof.
(⇐) Suppose every ultrafilter U on X converges. We will use
Lemma 10. Suppose (Aα)α∈I is a collection of closed subsets
of X with the FIP. By Lemma 4, there is a filter F on X such
that {Aα : α ∈ I} ⊆ F. By Zorn’s Lemma, F is contained in
an ultrafilter U on X. By hypothesis, U converges to some
x ∈ X. Let’s now see that
T
α∈I Aα ̸= ∅. Let U ∈ NX (x) be
open. Note U ∈ U as U → x. Fix Aα0
, α0 ∈ I. Since
Aα0
, U ∈ U, we have Aα0
∩ U ̸= ∅. Because U is arbitrary, we
obtain x ∈ Aα0
. Thus, x ∈ Aα0
as Aα0
is closed. Nevertheless,
Aα0
is also arbitrary so x ∈ Aα for every α ∈ I. Hence,

α∈I
Aα ̸= ∅.

Tychonoff’s theorem
Lemma 12
Let (Xα, Tα)α∈I be a family of topological spaces. Let Tp be the
product topology on
X =
Y
α∈I
Xα.
Let F be a filter on X, and let x ∈ X. Then F → x if and only if
pα(F) → pα(x) for all projections pα : X → Xα, α ∈ I.
Proof.
(⇒) Suppose F → x. Since pα is continuous for every α ∈ I, it
follows, by Lemma 9, that pα(F) → pα(x) for every α ∈ I.

Proof.
(⇐) Suppose pα(F) → pα(x) for all α ∈ I. We want to show that
U ∈ F for every open U ∈ NX (x). It will be enough to show
this for the basic open sets of Tp. Recall that
E =
[
α∈I

p−1
α (V) : V ∈ Tα
is a subbasis por Tp. If E ′
denotes the set of all the finite
intersections of elements of E (i.e., E ′
is a basis for Tp), then
for every U ∈ E ′
we can write
U =
n

i=1
p−1
αi
(Vi ), with Vi ∈ Tαi
for i ∈ {1, . . . , n} .

Proof.
Let U ∈ E ′
be such that x ∈ U; then, for all i ∈ {1, . . . , n}, it
follows that x ∈ p−1
αi
(V), whence pαi
(x) ∈ Vi . By hypothesis
Vi ∈ pαi
(F), so p−1
αi
(Vi ) ∈ F for all i ∈ {1, . . . , n}. Therefore, as F
has the FIP,
U =
n

i=1
p−1
αi
(Vi ) ∈ F.
Since every open neighborhood of x is the union of elements of E ′
that contain x, it follows by item ii of Definition 2 that F contains all
the neighborhoods of x. As a result, F → x.

Theorem 13 (Tychonoff’s)
Let (Xα, Tα)α∈I be a collection of topological spaces. Let Tp be
the product topology on
X =
Y
α∈I
Xα.
Then (X, Tp) is compact if, and only if, (Xα, Tα) is compact for
every α ∈ I.

Proof.
(⇒) Suppose X is compact when endowed with Tp. Since every
projection pα : (X, Tp) → (Xα, Tα) is continuous and onto,
pα (X) = Xα is compact for all α ∈ I.
(⇐) Suppose (Xα, Tα) is compact for every α ∈ I. Let U be an
ultrafilter on X. Let’s see that U converges in X. By Lemma 6,
pα(U) is an ultrafilter on Xα for every α ∈ I. Further, as Xα is
compact for every α ∈ I, it follows by Lemma 11 that each
pα(U) converges to some xα ∈ Xα. Let x ∈ X be such that
pα(x) = xα for every α ∈ I. Therefore, pα(U) → pα(x) for all
α ∈ I. Finally, by Lemma 12, we conclude that U → x. This
shows that every ultrafilter on X converges. Thereby X is
compact by Lemma 11.

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