H.1
Tychonoff via well-ordering,
We present a proof of the Tychonoff theorem that uses the well-ordering
theorem rather than Zorn's lemma.
Lemma H.1.
A be a collection of basis elements for the topology
Let
of the product space
If
of
yA
X
It follows the outline of Exercise 5 of §37.
X X Y,
such that no finite subcollection of A
is compact, there is a point
x4X
covers
x Y.
such that no finite subcollection
Ex}x Y.
covers the slice
Proof.Suppose there is no such point
one can choose finitely many elements of
x.
A
Then, given a point
x
t}hat cover the slice
of
X,
x>x Y.
Then, as in the proof of the tube lemma, one can find a neighborhood
such that these elements of A
can cover
U of
x
is compact, we
U x Y. Because X
x
bl finitely many such neighborhoods U ; then all of
X
cover
Theorem
Proof.
H,2.
Let
~
*
Products of compact spaces are compact.
IXT(J
be
a family of compact spaces; let X
rTJ x
X=
and let
XXY
,
.
can be covered by finitely many elements of
x
: X-+XL
be their
product,
;
be the projection map.
Well-order J in such a way that
it has a largest element.
Step 1. Le!t
Xi
be an element of
has been specified for all
supspace of
Then for each
o(<A
,
Define
Note that as
Z3
Pi
of
to be the following
, then tA
predecessor.
for
i
li(
) =
Yd
Pi
for
=
X:
i <
shrinks, and that
Z
equals the
finite collection of basis elements for
actually covers the larger space
Y
}
to be the following subspace of
hats an immediate predecessor in
Tn-en
i
YL for all k</3.
,4is a
We show that if
If
=
Y.
, increases, the space
intersection of the spaces
(X) =
W2
define
Y
Z
.
ar;d suppose that a point
X:
9
covers
i<
J;
Z
J, let o
Y&,
for some
X
t.at
c
be that immediate
, and the result is trivial.
H.2.
A
,
of
then
Iri(A) /.Xi;
all
A
Then
.A,
in
and
A
A
We show that
y
Y
Then
of
covers
Z
Since
Given
T i( Y)
= F
i
( x)
y
we know that
for
i
TTi(x)
Assume that
i>
A
of
A..
x.
for all
Ti(A)
i
.
TTi(x) E r(A)
i.
and for
(A)
because in this case
X.
We show that
for
We also
And finally, for
<d
.
7(A) = Xi.
LA is a collection-dfbasis elements for
We shall choose points
A
element of
~ J,
J,
X
Pi
,
for all
can be finitely covered by
the space
finitely covered by %A,
i,
X
siuch
LA itself does not
such that none of the spaces
A.
When oC i: the largest
is a one-point space.
Yj
Since it cannot be
it is not contained in any element of
To begin, let o/ be the smallest element of
XTFi.
X
Define a
.
The theorem follows.
X.
Snce
i
ai;d
we know that TT(y)
that no finite subcollection covers
for
for
lies in some element
also contains
A.
cit
Yke ,
for
y
so that
LA
7f
i (x) = 7T()
dover
xY~ , we show that it lies
i <,
for
= Pi
know that
Step 2.
.
is a basis element, we need only to show that
A
i J.
Y
TTi()
,
We show this element of
all
<
by setting
belongs to
Since
< i
is an index such that
i
4. We know that 7i(x) = i
in an element of
for
be the largest element of this union.
whenever
X.
JA,
iA.
is an element of
point
let ol
is also finite;
for which
The union of the sets
is a finite set.
JA
and Ti.(A)
i<Cd3,
i <3
denote the set of those indices
JA
let
For each eement
has no immediate predecessor.
Now suppose that
and since
the preceding lemma implies that there is a point
space
cannot be finitely covered by
X
=
A.
We write
X
in the form
Xi
c~annot be finitely covered by
Y4
J.
A .
i/
Xi1p
X
p. e XC
is compact,
sch that the
H.3
Now suppose
the space
is defined for all
cannot be finitely
LA
, Step 1 implies that
zB
Because
such that
X
for
d </
P,
such that for each V4 ,
A4·
We seek to define the point
, can be finitely covered
cannot be finitely covered by
7
4A.
Let
in the form
Zp
us write
i<
overed by
Y
Since none of the spaces
pi.
by
Yk
Pi
= iC26tpi X X X i>3 i.
p
is compact , the lemma tells us there is a point
Xp
he space
cannot be finitely covered by
.
Tthis is just the space
By the general principle of recursive definition (see
defined for all
i.
Y .72),
Pi
is
Note of course that we have used the axiom of choice
repeatedly to choose the points
Pi.