
Was Cantor Wrong? Are the real numbers countable? - freefrancisco
https://medium.com/p/a30802cf3152
======
dmfdmf
Cantor is wrong, yes, but you won't prove that to contemporary mathematicians
who believe his ideas because this is an epistemological issue and can't be
resolve with mathematical arguments.

This is really an age-old philosophic dispute that the mystics won decades
ago. As David Hubert described it "No one shall expel us from the Paradise
that Cantor has created." It is the "paradise" of deuces wild for the
mathematicians and it rests on an equivocation of the meaning of "infinity".
Cantor's idea of a "completed infinity" is a self contradiction if you grasp
what the concept of infinity actually means and keep it tied to reality. It is
the error of treating infinity as a real thing and not an abstraction. A
similar error, for similar reasons, is made in the history of the philosophy
by the mystics of nihil who wanted to treat nothingness as on par with
existence via the Reification of Zero.

~~~
jfarmer
You out yourself by using the phrase "Reification of Zero," which is something
I've only ever heard from Objectivists. It's funny to hear you say this, since
Objectivists are usually so apoplectic about axioms and premises.

You can object to ZF(C) as the basis for set theory, but I don't see how one
can call it "wrong." If you want to build up a theory without the Axiom of
Infinity you're free to do so — many mathematicians have. Either Cantor's
arguments follow from ZF(C) or they don't (hint: they do).

It's not a philosophical problem at all. One can certainly explore what it
would mean to do mathematics without access to "infinity", as many
mathematicians in the first half of the 20th century did. IMO this
conversation is all air until you propose your own precise set of axioms.

~~~
dmfdmf
> It's not a philosophical problem at all....

Well, that was my main point. You are free to disagree.

> IMO this conversation is all air until you propose your own precise set of
> axioms.

Apoplectically, the three basic axioms are Existence, Identity and
Consciousness. Whatever the proper mathematical axioms (and methods) are they
have to be consistent with these three. Cantor's method treats infinity as a
real thing, which violates the axiom of identity. You could even call it the
Reification of Infinity but the error is easier to see in Rand's
identification of Reification of Zero by the mystics who attempt to treat
nothingness as real.

~~~
jfarmer
The three basic (metaphysical) axioms according to Objectivists are Existence,
Identity and Consciousness. Anyhow, as far as I can tell, all you're saying is
that either (1) Cantor's argument is invalid within ZF(C) or (2) you find
ZF(C) objectionable on the grounds that it conflicts with the basis of your
metaphysics.

That's fine — I don't care whether it does or not — but mathematics has no
opinion on the matter. You're free to choose a set of mathematical axioms
which you don't find objectionable and go about doing your math in that
universe.

If you want to claim Cantor's argument is invalid then you need to tell me
what axioms you're taking as the foundation for your mathematics. It is valid
in ZF(C) and I'm 99% sure it's valid in plain ZF. Which of the axioms of ZF do
you find objectionable and what would you replace them with?

Assuming you're trying to engage sincerely, I'd ask you to be precise. This
means listing whatever foundations of mathematics you don't find objectionable
and talking about how they differ from ZF(C).

If you're unfamiliar with it, I'd read up on proof theory:
[http://en.wikipedia.org/wiki/Proof_theory](http://en.wikipedia.org/wiki/Proof_theory)

------
pesenti
Your argument is indeed completely wrong.

 _Let me construct a new positive even number as follows. Take the first
positive even number, then add the second positive even number, then add the
third positive even number, etc._

The set of even number is infinite. And this is a divergent sum, it's
infinite, not an even number.

~~~
freefrancisco
Thanks for the response, a few of my friends in Facebook told me the same
thing, the constructed number is infinite, so it is not a member of the set
and my argument is invalid. Do you have any insight as to why my attempt to
show that the reals are countable by placing them in a binary tree is also
wrong? We know the rationals are countable by putting them in a grid and
mapping them to the natural numbers, how is this fundamentally different from
putting the reals in a binary tree?

~~~
colanderman
You sound like a computer scientist, so think of it this way. When would your
breadth-first search terminate, given an irrational number to locate? Can you
even bound it? The answer is no, because there are an infinite number of
digits in any irrational number. The breadth-first search would not terminate;
the result would diverge toward infinity.

With the rationals in a grid, even ignoring that I can trivially
arithmetically compute the corresponding natural (interleave the (finite!)
digits of the numerator and denominator), I can give a bound on a similar
search (say Z-order): O(n²).

I think the fundamental idea you are missing is that a natural number _must_
have a finite number of digits. (This stems from the construction of naturals
as finite objects.) Reals are not subject to this limitation, as evidenced by
the existence of the irrationals.

[Actually, that last paragraph begs another question: what are the theoretical
implications of the fact that any _specific_ real must have a finite
representation of some form? Does the reason we can reason about reals which
we can't specifically identify with a finite construction have anything to do
with the axiom of choice?]

~~~
freefrancisco
This actually helps a lot with the intuition, thanks! So in a sense every
irrational number is an infinite number because there is no way to actually
represent it in a finite way.

~~~
colanderman
Almost! See pesenti's link about definable numbers: _most_ irrationals have no
finite representation, although almost all the ones we are familiar with (e.g.
sqrt(2), π, e) do: they can be represented as mathematical formulæ. (And
indeed, note that such "definable" reals _are_ countable!)

So in my example above, π would not work if, say, your tree was a parse tree
of mathematical formulæ: one _could_ locate the formula for π in finite time.
Unfortunately I cannot give an example of a single number which does _not_
have such a finite parse tree, as, by its very nature, I cannot write down a
finite description of it!

------
thenerdfiles
The new string has to be placed in the assumed list of Reals. If you place it
anywhere, you have to recast the Reals.

It's not just that it's new by algorithm, but it does not have a deterministic
place within the assumed list of Reals.

It's a new "Cantor's Paradise" every time you place that new number. And it
does not come into existence by the same procedure used to START generating
the list. It's a metanumber -- exception to rule. This is why one has to have
a metaphysics of number before one accepts the proof. Why Hilbert et al accept
the proof on aesthetic grounds, and a whole class of mathematics on aesthetic
grounds. The appropriate response is: That's just another real number.

Naturals, etc START somewhere but you are not forced to reindex with each new
number.

~~~
thenerdfiles
It's not necessary to believe in all those classes of infinities to see the
sui generis nature of the _mathematical gesture_ used to generate the
Metanumber.

Between any two mathematicians, likely the two will pluck very different Rules
from their imaginations or backgrounds to produce _some other real number._

What is _interesting_ is that we have a new real but also a new list. The
lists are contiguous games, not a continuous "paradise."

Number is the market wherein we SHARE rules. If we give up rule-sharing, the
market itself has no value nor do the numbers (and then all order is
defeated).

------
mdxn
In your set of positive even numbers example, you are only showing that a
specific enumeration scheme does not generate the set of even numbers. A
diagonalization argument requires that you show that no such enumeration
scheme can exist (not just a particular chosen one). In the typical proof of
diagonalization, it does not presuppose any particular scheme of listing the
numbers. It just assumes that some arbitrary one exists, then derives a
contradiction.

For your binary tree construction, a trancedental real number can be
represented by a path of infinite length down the tree. Your argument is that
since a breadth first traversal will eventualy exhaust that whole path, that
the real number described by it will have been encountered. There are two ways
to interpret what you are doing wrong:

\- If we are indexing/pairing these nodes by time steps (an index), your
construction is using a countably infinite time step to express the numbers
described by an entire path (which defeats the point of being countable).

\- For countable sets, you have to give me an index of finite size. If I give
you a real number, you need to return a natural number (or equivalent) that
indicates where it is. To test this, give me the index of pi in your claimed
"countable" enumeration. The reason you wont be able to somewhat follows from
Cantor's diagonalization scheme.

You might want to do some Googling before spending the time to write up an
entire blog post about it (god forbid posting it to HN). The mistake you made
is very common and has been discussed to death. You would have caught it.

~~~
freefrancisco
Thanks for your reply, the index of finite size is the part where I was
confused. I have actually done some googling about this before, and never
caught it, because I wasn't exactly sure where I was uneasy. This has bothered
me for a while, but reading that post in HN prompted me to think about it
more, and I started writing my thoughts in a communicable manner in order to
organize my own thinking. I figured publishing it and submitting it to HN
would be a very effective way to get immediate feedback. In HN if you are
wrong people will gladly let you know quickly. Also putting it in HN helps
prompt people to post and argue about it and other ideas that are related, so
it's an interesting and fun discussion.

------
banku_brougham
Why do programmers assume they are mathematicians? There is a huge difference
between those disciplines.

Trying to uphold the HN principles here, with a civil but pointed question.
Computer science is deep, so is physics, and mathematics as well but they are
distinct and quite different. Steven Hawking fumbled the proof of the
infinitude of primes in his excellent survey of mathematical history "God
Created the Integers," as usual for physicists he saw his way to the proof
without traversing each step.

Infinity is a difficult but knowable concept. Calculus is a good teacher of
this, and Euclid is a good teacher of proof. That's what's missing I say: the
standard of proof. For physicists and computer scientists that standard
depends on the outcome of physical events, but mathematics requires a purity
that doesn't exist in the physical world.

