Tychonoff ’s theorem implies AC∗
CWoo†
2013-03-22 2:29:52
In this entry, we prove that Tychonoff’s theorem implies that product of
non-empty set of non-empty sets is non-empty, which is equivalent to the axiom
of choice (AC). This fact, together with the fact that AC implies Tychonoff’s
theorem, shows that Tychonoff’s theorem is equivalent to AC (under ZF). The
proof was first discovered by John Kelley in 1950, and is now an exercise in
axiomatic set theory.
Proof. Let C be a non-empty collection of non-empty sets. Let Y be the generalized cartesian product of all the elements in C. Our objective is show that
Y is non-empty.
First, some notations: for each A ∈ C, set XA := A ∪ {A}, D := {XA | A ∈
C}, X the generalized cartesian product of all the XA ’s, and pA the projection
from X onto XA .
We break down the proof into several steps:
T
1. Y is equipollent to Z := {p−1
A (A) | A ∈ C}.
S
An element of X is a function f : D → D, such that f (XA ) ∈ XA
for each A ∈ C. In other words, either
S f (XA ) ∈ A, or f (XA ) = A. An
element of Y is a function g : C → C such that g(A) ∈ A for each
A ∈ C. Finally, h ∈ p−1
A (A) iff h(XA ) ∈ A.
Given g ∈ Y , define g ∗ ∈ X by g ∗ (XA ) := g(A) ∈ A. Since A is arbitrary,
g ∗ ∈ Z. Conversely, given h ∈ Z, define h0 ∈ Y by h0 (A) := h(XA ), which
is well-defined, since h(XA ) ∈ A. Now, it is easy to see that the function
φ : Y → Z given by φ(g) = g ∗ is a bijection, whose inverse φ−1 : Z → Y
is given by φ−1 (h) = h0 . This shows that Y and Z are equipollent.
2. Next, we topologize each XA in such a way that XA is compact.
Let TA be the coarsest topology containing the cofinite topology on XA
and the singleton {A}. A typical open set of XA is either the empty set,
or has the form S ∪ {A}, where S is cofinite in A.
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1
To show that XA is compact under TA , let D be an open cover for XA . We
want to show that there is a finite subset of D covering XA . If XA ∈ D,
then we are done. Otherwise, pick a non-empty element S ∪ {A} in D,
so that A − S 6= ∅, and is finite. By assumption, each element in A − S
belongs to some open set in D. So to cover A − S, only a finite number of
open sets in D are needed. These open sets, together with S ∪ {A}, cover
XA . Hence XA is compact.
3. Finally, we prove that Z, and therefore Y , is non-empty.
Apply Tychonoff’s theorem, X is compact under the product topology.
Furthermore, πA is continuous for each A ∈ C. Since {A} is open in XA ,
and A = XA − {A}, A is closed in XA , and thus so is p−1
A (A) closed in X.
To show that Z is non-empty, we employ a characterization of compact
space: X is compact iff every collection of closed sets in X having FIP
has non-empty intersection. Let us look at the collection S := {p−1
A (A) |
A ∈ C}. Given A1 , . . . , An ∈ C, pick an element ai ∈ Ai , since Ai 6= ∅
by assumption. Note that this S
is possible, since there are only a finite
number of sets. Define f : D → D as follows:
ai if A = Ai for some i = 1, . . . , n,
f (XA ) :=
A otherwise.
Since f (XAi ) = ai ∈ Ai , f ∈ p−1
Ai (Ai ) for each i = 1, . . . , n. Therefore,
−1
f ∈ p−1
A1 (A1 ) ∩ · · · ∩ pAn (An ).
−1
Since p−1
collection
A1 (A1 ), . . . , pAn (An ) are arbitrarily picked from S, the T
S has finite intersection property, and since X is compact, Z = S must
be non-empty.
This completes the proof.
Remark. In the proof, we see that the trick is to adjoin the set {A} to
each set A ∈ C. Instead of {A}, we could have picked some arbitrary, but
fixed singleton {B}, as long as B ∈
/ A for each A ∈ C, and the proof follows
essentially the same way.
References
[1] T. J. Jech, The Axiom of Choice. North-Holland Pub. Co., Amsterdam,
1973.
[2] J. L. Kelley, The Tychonoff ’s product theorem implies the axiom of choice.
Fund. Math. 37, pp. 75-76, 1950.
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