
Choose Your Paradox – the downside of the Axiom of Choice - rutenspitz
https://billwadge.wordpress.com/2017/06/12/thanks-axiom-of-choice-the-banachtarski-paradox/
======
danharaj
Constructive/Computable insights into variants of choice:

[https://mathoverflow.net/questions/22990/choice-vs-
countable...](https://mathoverflow.net/questions/22990/choice-vs-countable-
choice)

The way mathematics is taught it's easy to think that there's only one notion
of truth and logical justification. For mathematics of finite objects that can
be a very compelling story. Once you let infinity in, and you will because
infinity is where all the fun's at, there are a multiplicity of concepts of
truth.

The axiom of choice is an incredibly powerful statement about infinite objects
in a system, namely logic, which is inherently finitary[1]. Any strong
statements about how infinity works is going to imply bizarre things to us
finite mortals.

For example, if you assume the _negation_ of AC then you can prove that
there's a collection of nonempty sets whose cartesian product is empty.
Equally bizarre!

AC and not AC are both independent of ZF, which makes sense because ZF is an
axiom system for reasoning about monstrously infinite objects. Stronger
infinities are harder and harder to say anything about.

In a constructive mathematical universe (yet another notion of truth that's
perfectly good), as usual, you have to be careful in how you state the axiom
of choice. One formulation is a theorem and another implies law of excluded
middle. The latter one ought to be considered the right importation of
classical AC into constructive universes [2].

[1] There are infinitary logics but they don't magically allow you to perform
infinite amounts of reasoning. All physically plausible logical reasoning is
inherently a finite process.

[2]
[https://pdfs.semanticscholar.org/68d6/790cdbfda26cf71311e137...](https://pdfs.semanticscholar.org/68d6/790cdbfda26cf71311e13756d87f164838da.pdf)

~~~
delhanty
>For example, if you assume the negation of AC then you can prove that there's
a collection of nonempty sets whose cartesian product is empty. Equally
bizarre!

Intuitively, that's not "obviously" false to me.

Similarly, when I first encountered AC it wasn't obviously true to me.

~~~
yaks_hairbrush
> Intuitively, that's not "obviously" false to me.

That's fair -- intuitions are pretty personal things. It is "obviously" false
to me, and Bertrand Russell fairly well articulated the reason why by calling
AC the "multiplicative axiom".

If you have sets A, B with 3 and 5 elements, respectively, the Cartesian
product has 15 elements.

So, if we have an infinite series of sets, each with at least one element, I'd
expect there to be something in the infinite Cartesian project just by
multiplication -- the number of elements should (and I say "should" here to
clearly denote my intuition) be in some sense the product of each of the
individual numbers of elements. (I avoided the word "cardinality" here
explicitly because I'm talking intuitively).

The idea that there's a collection of non-empty sets whose Cartesian product
is empty is therefore akin to saying that you can multiply numbers bigger than
zero and end up with a zero result.

------
DonbunEf7
Quote: "One possibility is to treat AC as a powerful drug and take it only
when necessary. Theorems should come with consumer labels saying what went
into them. So if you see a box on the shelf of 'Banach and Tarski’s Miracle
Duplicator! Feed Multitudes!', it will say on the back of the box 'Contains
AC'."

Indeed, many mathematicians are sensitive to this and try to point out when
they invoke the Axiom of Choice, and there are other mathematicians who
deliberately seek non-AC-powered variants of theorems in order to put them on
less nebulous and more constructive ground.

Personally, I view AC as another reason to consider more generalized
foundations of mathematics. For example, if you know anything about topoi, it
turns out that we can view set theory as a topos on a point, and any topos on
a complete Boolean algebra comes pre-equipped with an inherent axiom of
choice! [0]

[0]
[https://ncatlab.org/nlab/show/axiom+of+choice](https://ncatlab.org/nlab/show/axiom+of+choice)

------
ikeboy
>But it also implies that ZF is consistent. This sounds nice, too, but is
actually a disaster. It means that we can’t prove the consistency of AD with
ZF (assuming the consistency of ZF).

I don't think this is strictly speaking correct. For instance, Con(ZF)
_itself_ trivially implies that ZF is consistent (which is literally what it
says). But we can prove that ZF+Con(ZF) is consistent, assuming that ZF itself
is consistent. (For that matter, we can prove that ZF+not(Con(ZF)) is
consistent, again assuming that ZF itself is consistent. Don't ask.) So just
because something implies Con(ZF) doesn't mean we can't prove that it plus ZF
is consistent, assuming ZF is consistent.

If ZF is allowed access to Con(ZF) it can prove a great many things that ZF
alone cannot, without generating any paradoxes, as far as I know.

I may be missing something, this always blows my brain.

~~~
kirrent
> But we can prove that ZF+Con(ZF) is consistent

Perhaps I'm naive, but doesn't that violate the incompleteness theorem? Surely
you could only prove such a statement in ZF + Con(ZF) + Con(ZF + Con(ZF)) or
something equivalent?

In regards to what you quoted, the incompleteness theorem says that because ZF
is provable in AD+ZF then AD is not provable in ZF otherwise ZF would prove
its own consistency.

~~~
ikeboy
You're not proving it formally, you're proving that if ZF is consistent, then
that system is consistent as well.

> AD is not provable in ZF

That's not what it said. It said you can't prove the consistency of AD
(presumably AD+ZF). But under the same conditions (assuming Con(ZF) that
doesn't rule it out and definitely not for the reason it states.

Edit: a meta proof that ZF+Con(ZF) is consistent if ZF is consistent:

Suppose it wasn't consistent. Then ZF+Con(ZF) proves anything. But then we can
prove not(Con(ZF) from ZF as well:

1\. Assume Con(ZF) 2\. Prove contradiction using ZF. 3\. Therefore,
not(Con(ZF)), proof by contradiction.

But if ZF proves that it isn't consistent, clearly it can't be consistent. But
by assumption ZF is consistent.

Therefore, if ZF is consistent, so is ZF+Con(ZF).

I think this proof works but again, mind blowing material so I can never be
100%.

~~~
kirrent
> Suppose it wasn't consistent.

That means the model you're investigating isn't ZF+Con(ZF). It's
ZF+Con(ZF)+Not(Con(ZF)) because you've added the extra axiom as part of your
proof by contradiction. Obviously you trivially get not(Con(ZF)) from this
model, let alone anything else via the principle of explosion.

ZF+Con(ZF)+Not(Con(ZF)) is not consistent and it shouldn't surprise you that
you can use it to trivially derive contradictory results.

Ultimately, Godel's second incompleteness theorem says that you can't prove
the consistency of a formal system within that system. The Wikipedia article
is dense and awful to read but has a reasonably succinct statement of the
theorem.
[https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_...](https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#Second_incompleteness_theorem)

EDIT: "mind blowing material" \- on that we can both agree. This is something
I really wish I got to study more back in uni.

Further editing to clean up some minor errors.

~~~
ikeboy
>It's ZF+Con(ZF)+Not(Con(ZF)) because you've added the extra axiom as part of
your proof by contradiction.

No. It's ZF+Con(ZF)+not(con(ZF+con(ZF)). Which is, in fact consistent
(assuming ZF is).

~~~
kirrent
Of course! Thankyou, that makes me much more comfortable.
ZF+Con(ZF)+not(con(ZF+con(ZF)) is an example of what's sometimes called the
self hating theory, which is known to be consistent if ZF+Con(ZF) is
consistent (you said much the same thing in your original comment). As it's
known to be consistent, you can't apply the principle of explosion to produce
Not(Con(ZF)).

EDIT: Obviously I'm not really the sort of person to talk about this stuff
with you because of how much of a neophyte I am. Perhaps this lecture would
make it clearer?
[http://www.scottaaronson.com/democritus/lec3.html](http://www.scottaaronson.com/democritus/lec3.html)

In particular, the excerpts:

The Second Incompleteness Theorem establishes what we maybe should have
expected all along: that the only mathematical theories pompous enough to
prove their own consistency, are the ones that don't have any consistency to
brag about! If we want to prove that a theory F is consistent, then we can
only do it within a more powerful theory -- a trivial example being F+Con(F)
(F plus the axiom that F is consistent). But then how do we know that F+Con(F)
is itself consistent? Well, we can only prove that in a still stronger theory:
F+Con(F)+Con(F+Con(F)) (F+Con(F) plus the axiom that F+Con(F) is consistent).
And so on infinitely. (Indeed, even beyond infinitely, into the countable
ordinals.)

On the other hand, again by the Second Incompleteness Theorem, ZF can't prove
its own consistency. If we want to prove Con(ZF), the simplest way to do it is
to posit the existence of infinities bigger than anything that can be defined
in ZF. Such infinities are called "large cardinals." (When set theorists say
large, they mean large.) Once again, we can prove the consistency of ZF in
ZF+LC (where LC is the axiom that large cardinals exist). But if we want to
prove that ZF+LC is itself consistent, then we need a still more powerful
theory, such as one with even bigger infinities.

------
soVeryTired
>the union of two minorities is a minority, and the intersection of two
majorities is a majority

Is that a typo? Shouldn't it be the other way around?

~~~
taejo
The other way around is already implied by the previous property.

Maybe it helps to consider finite sets. A finite set should clearly a
minority, and the union of two finite sets, though bigger, is still finite;
similarly, the intersection of two co-finite sets is smaller, but still co-
finite. Minority and majority are generalizations of finite and co-finite,
respectively. (A set is co-finite if the set of elements in some universe that
it _doesn 't_ contain is finite).

~~~
soVeryTired
Yep - I see it now, you're right. Thanks for the clarification.

------
ocfnash
There are famously-many other axioms that can be added to ZF, and which turn
out to be equivalent to Choice.

Of the following three equivalent extensions to ZF:

    
    
      * Axiom of Choice
    
      * Zorn's Lemma
    
      * Well-ordering Principle
    
    

Jerry Bona quipped:

"The Axiom of Choice is obviously true, the Well-ordering Principle obviously
false, and who can tell about Zorn's Lemma?"

