Download notes on the proof Tychonoff`s theorem

4 Tychonoff’s theorem
Theorem 4.1 (Tychonoff). If {(Xi, Oi ) : i 2 I} be a collection of compact topological
Q
spaces, then X := i 2I Xi with the product topology OQ is compact.
Definition 4.2. A subbase S of a topological space (X, O) is a subset S ⇢ O such that if
U is another topology on X with S ⇢ U , then O ⇢ U ; that is, O is the coarsest topology
containing S.
S
Proposition 4.3. Let X be a set and S ⇢ P (X ) a collection of subsets of X with S = X.
The collection B S of subsets of X defined by
8
9
>
>
> \
>
=
BS = <
S
:
k
2
N
and
S
,
.
.
.
,
S
2
S
i
1
k
>
>
> i=1,...,k
>
:
;
is a base of a topology O and S is a subbase of this topology. We call O the topology
generated by S.
Recall, that the product topology on X is the topology on X generated by
S := {⇡i 1 (U) : i 2 I, U 2 Oi }.
Proposition 4.4. Assume the situation of Theorem 4.1 If U is an open cover of X and
U ⇢ S, then U has a finite subcover.
Theorem 4.5 (James Waddell Alexander II). If (X, O) is a topological space and S is a
subbase of O such that any open cover U ⇢ S has a finite subcover, then X is compact.
Theorem 4.1 now follows by combining both of these results.
Proof of Proposition 4.4. Let U ⇢ S be a cover. For i 2 I set
Ui := {U 2 Oi : ⇡i 1 (U) 2 U }.
By definition and since U is a cover, we have
[
⇣[ ⌘ [
⇡i 1
Ui =
U = X.
i 2I
There must be an i 2 I such that Ui is a cover of Xi , for otherwise for each i 2 I
S
we can pick an x i 2 Xi such that x i < Ui ; that is, for all i 2 I, x = (x i )i 2I satisfies
S
⇡i (x) < Ui or, equivalently,
[
⇣[ ⌘
x<
⇡i 1
Ui = X .
i 2I
By compactness of Xi , Ui has a finite subcover Vi and then V = {V 2 Vi : ⇡i 1 (V )}
is a finite subcover of U.
⇤
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For the proof of Theorem 4.5 we will use Zorn’s Lemma.
Lemma 4.6 (Zorn). Let ( A, ) be a non-empty partically ordered set such that each
ordered subset B ⇢ A has an upper bound. Then A has an upper bound.
Recall, that at the very beginning of this class we declared this to be a foundational
axiom. Zorn’s Lemma can be proved if one assumes the axiom of choice; in fact, it is
equivalent to it. In any case, making use of Zorn’s Lemma means that one works within a
certain set-theoretic framework (say, ZFC). Most mathematicians today (secretely) assume
that they are working in ZFC or completely ignore this sort of question.
As a warm up for the proof of Theorem 4.5 we will recall the proof that any vector
space has a basis.
Definition 4.7. Let V be a k vector space. A subset B ⇢ V is called linearly independent
if for any n 2 N⇤ , 1, . . . , n 2 k and b1, . . . , bn 2 B we have
n
X
i bi
= 0 =)
1
i=1
= ··· =
n
= 0.
A subset B ⇢ V is say to span V if for any v 2 V there are n 2 N⇤ ,
b1, . . . , bn 2 B such that
n
X
v=
i bi .
1, . . . ,
n
2 k and
i=1
We say that B ⇢ V is a basis if it is linearly independent and spans V .
Proposition 4.8. If V is a vector space, then it has a basis B.
Proof. Let L denote the set of linearly independent subsets. This set is partially ordered
by the inclusion ⇢. Let K be an ordered subset; that is, if B1, B2 2 K , then either B1 ⇢ B2
or B2 ⇢ B1 .
S
B = K is an upper bound of K . To see this, we need to verify that B is linearly
independent. Suppose n 2 N⇤ , 1, . . . , n 2 k and b1, . . . , bn 2 B, such that
n
X
i bi
= 0.
i=1
Since there are only finitely many bi and K is ordered, there is a B 0 2 K such that
b1, . . . , bn 2 B 0. Because B 0 is linearly independent, it follows that 1 = · · · = n = 0.
The above shows that Zorn’s Lemma applies. Let B be an upper bound of L, that is, a
linearly independent set such that if B 0 is any other linearly independent set with B ⇢ B 0,
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then B = B 0. B spans V and thus is a basis. To see this, if v 2 V , then there exists n 2 N⇤ ,
1, . . . , n 2 k and b1, . . . , bn 2 B such that
v=
n
X
i bi
i=1
if and only if B [ {v} is not linearly dependent.2 Thus, if B would not span V , there would
be a v 2 V such that B [ {v} is linearly independent; contradicting the maximality of B. ⇤
Proof of Theorem 4.5. Let
C = {U ⇢ O : U covers X but does not admit a finite subcover }.
If X is not compact, then C is non-empty. We will derive a contradiction in this case.
First note that if C is partially ordered by inclusion ⇢ and if D is an ordered subset
S
of C, then U := D is an upper bound for D. To see this, we need to show that U
has is a cover with no finite subcover. It clearly is a cover. If it had a finite subcover
V = {V1, . . . , Vn } ⇢ U, then since there are only finitely many Vi they must be all contained
in a fixed U ⇢ D which would then have a V as a finite subcover.
The above shows that Zorn’s Lemma applies to C. Let U be a upper bound for C, that
is, a cover of X which contains no finite subcover and such that whenever U 0 has the same
property and U ⇢ U 0, then we have U = U 0.
If we can show that
V := U \ S
is a cover of X, then the assumption will show that V has a finite subcover; but then U
also has a finite subcover, leading to a contradiction with the definition of U and we are
done.
S
If V is not a cover, then there is some x 2 X \ V. Since S is a cover, there is a
V 2 S with x 2 V . Since U covers X, there is a U 2 U with x 2 U. Since S is a subbase,
there are V1, . . . , Vn 2 S such that x 2 V1 \ · · · \ Vn ⇢ U. By assumption all of the Vi
satisfy Vi < U, this U [ {Vi } is strictly larger than U and thus must have a finite subcover
Sn
of the form Ui [ {Vi }. But then i=1
Ui [ U is a finite subcover of U .
⇤
2This is the key point where invertibility in k is used.
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