Indicating the present boundary between what we can, and can not do.
We may prove the existence of a non-Borel set without using the axiom of
choice [Dug, p.76, Remark 6.6(1)]. However, there is no known method of proving
the existence of a nonmeasurable set without using the axiom of choice (see
[Roy, p.64, l.16]).
Establishing a loose relationship in a big picture (see the diagram
in [Dug, p.311]).
We prove metric Þ paracompact using the axiom of
choice.
Disadvantages.
The use of the axiom of choice will make an argument ineffective.
Complication is not compactification.
[Dug, p.244, l.1] says that Stone-Čech compactification is fairly complicated. The complication mainly comes from
Tychonoff's Theorem [Dug, p.224,
Theorem 1.4(4) & p.243, l.5] whose proof uses the axiom of choice [Dug, p.219,
Theorem 7.3].
The axiom of choice destroys the connection between existence and
uniqueness [Hew, p.189, l.-7-l.-1].
How to avoid using the axiom of choice.
We can find the misuses of the axiom of choice almost everywhere in
mathematical literature. Some of them are slips. Some of them stem from the bad
habit of using the axiom of choice indiscriminately [Dun, pt. I].
Zorn's lemma should be used as
a prediction rather than a realization. Finding an effective method to
realize the prediction is exactly what mathematics is for.
Algebraic method: Finding the maximal ideals [Hun, p.128, Theorem
2.18] in Z using prime ideals [Hun, p.128, Theorem 2.19].
The shrinking Lemma.
The proof of [Dug, p.152, Theorem 6.1] immitates the proof of [Spi, vol.1,
p.67, Theorem 14]. Their essential difference is that the former uses
transfinite induction, while the latter uses mathematical induction. Thus the connection
in the former argument is greatly weakened. Furthermore, [Dug, p.152, l.!3
& p.153, l.1] uses the axiom of choice (see [Dug, p.31, Theorem 2.1]).
Partitions of unity.
[Dug, p.170, Theorem 4.2] ® [Spi, vol.1, p.68,
Theorem 15].
[Dug, p.170, l.!18-l.!16]
uses [Dug, p.152, Theorem 6.1] twice and [Dug, p.152, Theorem 6.1] uses the
axiom of choice twice.
Maximal filterbases-Characterizing the maximal in an effective way.
[Dug, p.219, Theorem 7.3] uses Zorn's
lemma. However, for the case in [Dug, p.219, l.15], we may effectively determine
whether a filterbase is maximal by using [Dug, p.219, Theorem 7.2].
(Bolzano-Weierstrass Property) We need the axiom of choice because we
use inadequate terms to describe the situation.
Transfinite induction gives a formal sequence-version [Po3, p.71, Theorem 4,
4)] of [Po3, p.71, Theorem 4, 2)]. Obviously, sequence is not an adequate term
to describe compactness. This is the reason why we need an extra axiom to
reformulate the statement. In fact, this new version produces no extra
significant applications as its benefit.
Correct attitude. The term "infinite sequence" belongs to the category of
countable compactness [Dug, p.229, Theorem 3.2]. Any exaggeration of its
function like infinite set or generalized sequence (indexed by a well-ordered
set) will cost us by requiring us to add the axiom of choice into our theory (see [Po3, p.71,
Theorem 4, 3)Þ 2); 4)Þ
2)]).
(Tychonoff's Theorem) Reduction
to the finite case is the only way to preserve a theorem=s
true value.
[Dug, p.133, Theorem 1.4(4)] ® [Per, p.133,
Theorem 8.2.2].
The Haar integral. Many mathematicians misuse the axiom of choice in
proving the existence of the Haar integral [Hew, p.214, l.17-l.18]. [Hew, p.189,
l.!7!1] finds an
effective method to avoid using the axiom of choice. The idea (mean [Po3, p.194,
(7)]) becomes more clear when the setting ([Hew, p.185, Theorem 5.15]'s:
locally compact group ® partition of unity [Hew,
p.190, l.11], outer measure [Hew, p.186, l.17]; [Po3, p.193, Theorem 24]=s:
compact group, measure) is simpler.
Ascoli's Theorem. [Dug, p.267,
Theorem 6.4] uses the concept of maximal filterbase, hence Zorn's
lemma, while [Ru1, p.144, Theorem 7.23 (b)] and [Po3, p.189, Theorem 23] avoid
using Zorn's lemma.
A correct attitude towards the axiom of choice.
We would like to preserve the progress of the theory of well-ordered
sets even though there is no effective method of constructing a well-order on an
arbitrary set.
We can find a set with card (S)> À0.
Furthermore, we can effectively construct a well-order on the set S.
Once a fixed well-order is given, the problem of effective construction
is eliminated. Then the theory of a well-ordered set becomes effective.
A well-ordered set W is a generalization of the set N of positive
integers. Transfinite induction is a generalization of mathematical induction.
The method to establish the inductive hypothesis which extends from ]¬
,x[ to x should be divided into two cases:
x has an immediate predecessor. This case should be reduced to
ordinary inductive hypothesis if W=N.
x is a limit number. This case requires a different strategy from
that of mathematical induction.
In view of (B) & (C), Zermelo's
theorem [Dug, p.32, Theorem 2.1 (3)] is probably the best form (see the proof of
[Po3, p.269, Theorem 51]) of the axiom of choice if we want to directly retrieve
an effective argument when applying a theory involving the axiom of choice to the
case W=N. See [Po3, p.326, Theorem 68] also.
Using a more constructive argument than the axiom of choice may produce a
much stronger result.
Compare [Karl, p.21, Theorem 12] with [Ru3, p.70, Theorem 3.21]. In proving
[Karl, p.21, Theorem 12], Karloff uses the downward induction rather than the
axiom of choice. However, in proving [Ru3, p.70, Theorem 3.21], Rudin uses Hahn-Banach
theorem whose proof assumes the axiom of choice is true.
The foundation of a theory may deteriorate if certain dubious axioms are
used.
The root of a theory should be based on concrete examples. Through concrete
patterns we may make certain assumptions to generalize the theory to a
hypothetical level. Rudin does it the other way around: He starts the theory of
convex sets with the empty axiom of choice [Ru3, Chap.3, p.55, l.8
The Hahn-Banach Theorem]. Thus his foundation is false. How can anything that
follows [Ru3, p.60, Theorem 3.7] be true? Also we may easily lose many effective
algorithms which really support the theory [Baz, p.45, Theorem 2.4.4; p.50,
Theorem 2.4.8; p.51, Theorem 2.4.10].
Kadison uses Zorn's lemma [Kad,
vol.1, p.5, l.18] to prove [Kad, vol.1, p.4, Theorem 1.1.2; p.7, Theorem 1.1.4;
p.20, Theorem 1.2.9 & Theorem 1.2.10]. Quoting the axiom of choice
inappropriately would diminish the quality of his theory.
If one uses the axiom of choice [Ru3, p.71, l.7] in proving a theorem, one
should not trace back and determine what hypothesis is actually required [Ru3,
p.71, l.14] because his argument is basically supported by an unproven
assumption [Baz, p.58, Theorem 2.6.5].
How we repair a proof that uses the axiom of choice
(Peano curves) [Dug, p105, Theorem 4.4]: [Dug, p.105, l.3] uses [Dug, p.103, Proposition 2.7] to prove that h is
a homeomorphism [Dug, p.105, l.4]. In this case, the axiom of choice used in the proof of [Dug, p.103, Proposition 2.7] can be replaced by mathematical induction.
[Dug, p.344, Lemma 4.3(2)]: For the method of repair, see [Dug, p.345, Ex. 3].
[Mun00, p.514, Theorem 85.1]: If F is finitely generated, the method of
repair is given by [1].
(the existence of a basis in a vector space)
Only under the condition that
everything is ready except one axiom may one truly understand and appreciate a
theorem's proof better. If one directly reads the proof given in [Dug, p.36,
l.13], one cannot the importance of the axiom of choice. However, after one
reads [Ru2, p.90, Theorem 4.18], one will appreciate the role that the axiom of
choice plays in the proof given in [Dug, p.36, l.13].