
How to use Zorn's lemma - ColinWright
http://www.tricki.org/article/How_to_use_Zorns_lemma
======
ColinWright
There is a standard and well-known joke that goes:

    
    
        “The Axiom of Choice is obviously true,
         the well-ordering principle obviously
         false, and who can tell about Zorn’s
         lemma?” — Jerry Bona
    

It's a joke because all three statements are equivalent, and yet AC seems like
it should be true, and the well-ordering principle seems too much to ask for.

The submitted page is intended to help those who don't understand what Zorn's
lemma is about, or how or why one might use it.

Some math required ...

~~~
nerdponx
Why does WO seem like too much to ask for, and why are people so-so on Zorn's?
I find them both just as intuitive as AC if not moreso.

~~~
v64
I think the canonical example is that WO means there's a well ordering of the
reals. We have to accept it as true, but what the hell would that look like?
It's generally assumed that such a well ordering wouldn't be constructible, so
it does seem like a little much to ask for.

Edit: For those curious why this is problematic, a well ordering on a set
means that for any non-empty subset of the set, there's a consistent way to
define the least element of the set.

An easy example is the natural numbers starting from 0 going up. The well
ordering on the naturals is the usual <, and 0 is the lowest of the entire
set.

This doesn't work this way for the integers. Because the integers go to
negative infinity, we have to order them in a different way (but we know it's
possible because the well ordering principle says so). It turns out the way to
do it is to define the order on the integers this way: 0 1 -1 2 -2 3 -3 etc.
Now the integers are ordered in such a way that we have a well defined least
element, 0.

Now, WO says the real numbers have a well ordering as well. What would that
look like? For instance, < fails us because the open interval (0,1) has no
least element. The well ordering has to hold for any non-empty subset, and
there are ways to create very strange and pathological subsets of R. To think
that the entirety of R can be ordered in such a way like the integers is
asking for a lot. But the well ordering principle states that even if we can't
figure out the order, we can assume that one exists and use it in proofs. But
it's weird because we're pretty much resigned to never knowing how it'd
actually look.

~~~
danbruc
I guess the following won't work then.

0 / 0.1, 0.2, 0.3...0.9 / 0.01, 0.02, 0.03...0.09, 0.11, 0.12, 0.13...0.19,
0.21...0.99 / 0.001, 0.002, 0.003...0.009, 0.011, 0.012...

And when ever you list x, you also list -x, 1/x, and -1/x. Except for 0 of
course. And you have to throw in +1 and -1 somewhere. Somehow it feels like
you should cover all the reals eventually but that can't be true because that
would imply there are only countable infinite many reals. It actually looks
more like a convoluted version of the standard way to enumerate the rationals
by visiting diagonals in order of increasing distance from the origin on a two
dimensional integer lattice.

~~~
Retra
Yeah, that's obviously just an ordering of the rationals.

~~~
danbruc
I don't think it's that obvious, actually I think it's not actually an
ordering of the rationals. If this were an ordering of the rationals, then it
better included rationals with an infinite number of non-zero decimal digits,
for example one third. But if this would include one third, then I am unable
to see why it wouldn't also include square root two. Why would this be able to
include repeating decimal expansions but not non-repeating ones?

~~~
Retra
You're right, I misread what you wrote. You're actually only covering a small
portion of the rationals.

Either way, to order the reals, you have to include the non-computable ones,
which cannot be written down in any systematic way. This also means that
ordering and equality relations cannot be systematically defined either.

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unprovable
Doing Reverse Mathematics, AC is waaaay to strong for me :P (We stick to weak
subsystems of second order arithmetic, by and large) But this is a really nice
treatment of the Lemma :D Thanks!

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xamuel
My favorite application of the Axiom of Choice, which Computer Scientists will
appreciate:

Theorem ("Konig's Lemma"): If a tree has infinitely many nodes, and every node
has only finitely many children, then the tree must have an infinite branch.

Proof: The tree's root, X1, has infinitely many descendants. Since the root
has only finitely many children, some child, X2, of the root, must have
infinitely many descendants. Since X2 has only finitely many children, some
child, X3, of X2, must have infinitely many descendants. This process repeats
forever, producing a sequence X1, X2, X3, ... which is an infinite branch
through the tree.

~~~
cuspycode
Why is the AC necessary for this recursive tree traversal?

~~~
xamuel
The AC is necessary[1] because at each of the infinitely many steps, you have
to choose which of possibly several candidates to use.

[1] As mrmyers points out, you actually only need a weaker version of AC,
called Countable Choice.

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matthewrudy
Apologies for the appalling Mathmo joke, but...

* What's yellow and equivalent to the Axiom of Choice?

* Zorn's Lemon

~~~
giomasce
That was the Bananach space...

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fjsolwmv
> then Zorn's lemma may well be able to help you.

Sure , stuff is easier to prove if you add unrealistic axioms.

~~~
giomasce
Which other ZFC axiom would be "realistic"?

