
Mathematicians Bridge Finite-Infinite Divide - okket
https://www.quantamagazine.org/20160524-mathematicians-bridge-finite-infinite-divide/
======
p4bl0
Here is a little anecdote: I met Ludovic Patey at the École normale supérieure
that we both entered in 2009. Among our classmates, Ludovic was easily the
best programmer and mastered a lot of different technologies (from advanced
web stuff down to assembly languages), so we were all quite surprised when he
chose to turn to pure mathematics… But then during his PhD he has been
immensely successful (invited as keynote speaker to conferences, praised by
peers, and an impressive list of journal publications). A few weeks ago, he
has been ranked first in the CNRS 2016 recruitment competition, which is
internationally competitive. When that competition started he hadn't even
graduated from his PhD yet! Which means he (deservedly) bypassed the
cumbersome post-doc phase by which virtually all young researchers must go
through these days. This time, none of us were surprised :) !

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auggierose
Great article, very well written and actually understandable without hunting
down mathematical definitions.

I think it is quite cool that people are still hunting down the various
different incarnations of infinity and are able to prove deep results like
this.

On the other hand, I don't think that these questions are as essential as they
are made out to be. How do we know that everything finite is on unshakable
foundations? Doesn't even thinking about what that statement means involve
reasoning about infinities? Let's just accept that nothing ever is really
unshakable and let's use the tools that will get the job at hand done.

~~~
Smaug123
But we can be more sure of things that are less shakeable. Additionally, these
techniques sometimes give us more effective proofs: for example, if we know
that A is true, that is less useful than knowing that A is true _and that we
can find a finite certificate of its truth_. We can make computers verify
finite certificates, for instance.

A useful distinction to have in mind is: I know that the number 91 must have
prime factors, because I can prove that all numbers have prime factors. But
that's much less useful than knowing that its prime factors are 7 and 13.
Similarly, if I could prove A using infinitistic methods, that's in some sense
"a bit less useful" than a proof which uses finitistic and/or constructive
methods.

~~~
auggierose
My point is that if you prove that 91 has as its prime factors 7 and 13, I
don't care if you did that proof using infinite methods or not.

In general, I am only proving theorems that are useful to me, so "a bit less
useful" doesn't apply to these situations. If I needed that "bit more
usefulness", I would go ahead and prove it (if I could).

Your reasoning mostly only applies if you prove theorems for their own sake,
and not because you already know what you want them for.

~~~
Smaug123
It's only an analogy. The analogy is meant to be "prove that 91 has prime
factors" using infinitistic methods, and "prove that 91's prime factors are 7
and 13" using finitistic ones.

~~~
auggierose
I am aware of that. That's why analogies are usually not very useful.

~~~
Smaug123
I have no idea how you learn, but analogies are basically the only way I use
of absorbing new concepts.

~~~
auggierose
Interesting. I learn by playing around with the concept until I gain some
intuition about it. No analogies required.

------
okket
Related (a bit of Cantor and Hilbert, no direct relation to Ramsey’s theorem):

"How To Count Past Infinity"
[https://www.youtube.com/watch?v=SrU9YDoXE88](https://www.youtube.com/watch?v=SrU9YDoXE88)

~~~
avodonosov
How do people produce videos like that? It must be very expensive: shooting,
pictures, music, all these animations, montage. It can't be done a hobbyist?

~~~
goldenkey
Vsauce is a hobbyist. As he got more popular and ad revenue came in, he could
afford better cameras, mics, and props. Hes actually the most successful of
the VsauceX (X=1,2,3,4) channels but still keeps them as a network because
they are all friends. Honestly, YouTubing like anything else, takes effort
more than anything. Some of the most successful channels used a smartphone
camera for well past a million subscribers. And part of the rub is that there
are a ton of good channels with very little success. Good art and
entertainment is borne out of desire to share, not desire to bear. At least in
my opinion. Because there will always be garbage to cater to garbage.

I should mention, most of the editing can be easily done in After Effects with
a couple weeks of training. Its really amazing how accessible having your own
syndication is these days. I am not interesting enough or entertaining enough
so I leave it to those who are. Sticking to HN threads so I can hide my ugly
mug ;-)

------
Smaug123
I wrote a short note [1] to try and clear up some of the misconceptions in
this thread. Feedback welcome; Hacker News thread for the short note at [2].

[1]: [https://www.patrickstevens.co.uk/finitistic-
reducibility/](https://www.patrickstevens.co.uk/finitistic-reducibility/)

[2]:
[https://news.ycombinator.com/item?id=11769425](https://news.ycombinator.com/item?id=11769425)

------
egjerlow
>Whereas almost all theorems can be shown to be equivalent to one of a handful
of major systems of logic — sets of starting assumptions that may or may not
include infinity, and which span the finite-infinite divide — RT22 falls
between these lines.

Could anyone explain this statement? I feel like it's key for understanding
the rest! How exactly does RT22 fall between 'these lines'?

~~~
thaumasiotes
From later in the article:

> Almost all of the thousands of theorems studied by Simpson and his followers
> over the past four decades have turned out (somewhat mysteriously) to be
> reducible to one of five systems of logic spanning both sides of the finite-
> infinite divide. For instance, Ramsey’s theorem for triples (and all ordered
> sets with more than three elements) was shown in 1972 to belong at the third
> level up in the hierarchy, which is infinitistic.

> A breakthrough came in 1995, when the British logician David Seetapun,
> working with Slaman at Berkeley, proved that RT^2_2 is logically weaker than
> RT^3_2 and thus below the third level in the hierarchy.

> “Since then, many seminal papers regarding RT^2_2 have been published,” said
> Weiermann — most importantly, a 2012 result by Jiayi Liu (paired with a
> result by Carl Jockusch from the 1960s) showed that RT^2_2 cannot prove, nor
> be proved by, the logical system located at the second level in the
> hierarchy, one rung below RT^3_2.

So: there are five nested axiomatic systems that have been in common use to
classify how much a theorem relies on infinite concepts. RT^2_2 is weaker than
the third of those, and hard-to-compare with the second of them. (The second
system can't prove it, but it also can't prove the second system.)

The new result says that RT^2_2 is reducible to primitive recursive
arithmetic, which should mean that PRA is capable of proving anything RT^2_2
can prove. The article mentions that Mysterious Classification Level 2 is also
reducible to primitive recursive arithmetic, so, as far as I understand
things, PRA was already a system that fell "between the lines" of the five
Mysterious Classification Levels (since Mysterious Level 3 is infinitistic and
PRA is not).

~~~
sxyuan
In case you're curious, the five systems in the article are mentioned by the
original paper, and correspond to the ones described here:
[https://en.wikipedia.org/wiki/Reverse_mathematics#The_big_fi...](https://en.wikipedia.org/wiki/Reverse_mathematics#The_big_five_subsystems_of_second_order_arithmetic)

------
scribu
As a noob, I'm having a hard time grasping why the Ramsey pairing is
interesting: If you pair up every member of an infinite set with every member
of that same infinite set, of course you'll get an infinite subset for almost
every predicate about a pairing, just by virtue of starting from an infinite
superset.

It seems like the theorem would be interesting if it said something about the
"magnitude" of the subset, not just that it's infinite.

~~~
n4r9
It's hard to know what you're finding uninteresting/confusing from your
description. Perhaps it would help to note that _every possible_ pairing is
given a colour, and in your infinite subset _every possible_ pairing within
that subset has the same colour. To me this is quite counter-intuitive.

There is a somewhat enlightening comparison to be made between Ramsey on pairs
and the result that every real sequence x_n has a monotonic subsequence. You
get this result as a corollary by colouring a pair of natural numbers a<b red
if x_a <= x_b and blue otherwise.

Interestingly, the Ramsey theorem can be extended to three, four or any number
of colours. I believe you can use this to show that any real sequence has
either a convex or concave subsequence, but I'll omit the details :p

There's also a delicate balance between the infinite and finite going on.
Clearly, with an infinite number of colours you could get a colouring without
a monochromatic subset. More surprisingly perhaps, if you colour the _infinite
subsets_ of the natural numbers red or blue, then there exist colourings for
which there is no monochromatic subset.

~~~
SilasX
I have the same confusion as the GP, so let me try to ask a clarifying
question:

Is the challenge to partition the pairs (of natural numbers) such that both
partitions are infinite and monochromatic? Is that the hard thing that was
proved here?

If so, how about "color red if a = b-1, blue otherwise"? Then the infinite
subset (0, 1), (1, 2), (2, 3), ... is monochromatic.

What criterion did my partition there fail to satisfy?

~~~
NotAPerson
The hard part is to show that for _any_ coloring, there's some infinite subset
without relying on "well, we can just pick one for each of these infinitely
many numbers".

I'm pretty sure that the challenge is to prove that you can for _any_ coloring
construct a finitely defined rule for picking the members of the subset which
is guaranteed to give you a monochromatic subset.

The question is more about if you can always find such a (finite) rule to
partition the set, rather than if you can in a few easily constructed
examples.

~~~
SilasX
But it's not true for any coloring, e.g. "red if a and b < 20, blue
otherwise".

~~~
NotAPerson
Wouldn't the subset {(a, b) | a,b > 20} be monochromatic?

Ed:

Perhaps I phrased it poorly, but I think the point was to show that you can
always construct a predicate, P over a and b, such that P(a, b) is finitely
defined (such as "a > 20 and b > 20"), but {(a, b) | P(a, b) is true} is
infinite and monochromatic.

Instead of having some cases of colorings where your only option is to
construct things of the form "(a = 5 and b = 17) or (a = 3 and b = 47) or ..."
where you just list out every pair that matches (in an infinite subset).

------
igravious
> The boundary does not pass between some huge finite number and the next,
> infinitely large one. Rather, it separates two kinds of mathematical
> statements: “finitistic” ones, which can be proved without invoking the
> concept of infinity, and “infinitistic” ones, which rest on the assumption —
> not evident in nature — that infinite objects exist.

Let Y = some ridiculously huge finite number. The notion that ∞ = Y + 1 has to
be one of the most insidious defects in reasoning. Even to acknowledge the
defect by dismissing it is almost like saying things could have turned out
that way when they _never_ could.

Finite means limited or bounded or quantifiable. Infinite means unlimited or
unbounded or unquantifiable. They are two wholly separate classes of things.
Another way to say this is discrete versus continuous (in nature).

Imagine I said that I could bridge the coloured/non-coloured divide. What
could I possibly mean by that. It would mean that I have found a way to talk
about both coloured things and non-coloured things that applies to both class
of things.

Put another way. Imagine I separate coloured things into green ones and not
green ones. Both subclasses belong to the class of coloured things. What class
then do the subclasses of finite things and non-finite things belong to?
Negation constructs the distinction, negation _is_ the bridge in a way and all
the could be said of both subclasses of things is that they are both things.
(Which, to be clear, is not saying very much at all, is it?) If those things
are numbers then all we are saying is that both things are numbers, if both
things are procedures then all we are saying is that both things are
procedures. And so on.

If I'm thinking about this properly then the paper is trying to more precisely
define the term "countably". That's as most as it can do. To say otherwise is
not just vastly overstating what the paper is about but outright misleading.

~~~
Smaug123
You're not thinking about this properly. There are some facts about finite
objects which simply cannot be proved without reasoning about infinite objects
- for example, the well-definedness of the TREE function. These facts are true
if you have access to infinite objects, but not necessarily if you don't.

This paper shows that a certain class of statements about finite objects,
which we knew were true by virtue of reasoning about a certain infinite
object, in fact remain true if you're restricted only to reasoning about
finite objects.

~~~
igravious
I tried to give my reasons by using only appeals to objects of our common
understanding. You have countered my claim by appealing to "the well-
definedness of the TREE function". Do you expect me (and everyone else) to
know what that is? In order for us to be able to follow your argument you
would need to spell that out in a bit more detail.

If I'm not thinking about this properly, which of the assertions I made was
incorrect? Where is the error in my reasoning? Is there an error in how I am
conceiving things?

> This paper shows that a certain class of statements about finite objects

From what I can gather, it's not that simple. For a start, the initial set is
the set of natural numbers. This is not finite. The procedure for
generating/enumerating them is. The paper deals with pairs of inequalities
based on the natural numbers and sub-sequences to be found therein. This set
of pairs is also non-finite (but I'm happy with the assertion that in some
sense it is a different order of infinity from the natural numbers). We are
now trying to reason about the nature of the sequencing of these sub-
sequences.

Are you saying that sometimes the sub-sequences are finite? If so, their
complement would be infinite. And it is the partitioning that makes this so.

~~~
Smaug123
Your understanding of the point of the theorem is very different to mine, and
I'm moderately sure my understanding is pretty close to correct.

It is a fact of mathematics that there are some statements which are solely
about finite objects, but to prove them requires reasoning about an infinite
object. For a more accessible example than TREE, I think the Ackermann
function falls into this category. The Ackermann function A(n+1, m+1) = A(n,
A(n+1, m)) is well-defined for all n and m (we prove this by induction over
NxN), but the proof relies on considering the lexicographic order on NxN which
is inherently infinite. (I'm not totally certain that all proofs of
Ackermann's well-definedness rely on an infinite object, but the only proof
known to me does.) Ackermann's function itself is in some sense a "finite"
object, but the proof of its well-definedness is in some sense "infinite".
Whatever the status of my conjecture that "you can't prove that Ackermann's
function is well-defined without considering an infinite object", it is
certainly a fact that Ackermann is not primitive-recursive, and "primitive-
recursive functions" corresponds to the lowest level of the five "mysterious
levels" the article talks about.

So the analogy is as follows. Imagine that we knew of this "infinitary" proof
that Ackermann is well-defined, but we hadn't proved that no "finitary" proof
exists. (So finitists are not happy to use Ackermann, because it might not
actually be well-defined according to them: any known proof requires dealing
with an infinite object.) Now, this paper comes along and proves that actually
a finitary proof exists. Suddenly the finitists are happy to use the Ackermann
function.

The actual definition of TREE is a bit too long for me to explain here, but it
is an example of a function like Ackermann, which is well-defined, but in fact
if you're not allowed to consider infinite objects during the proof then it is
provably impossible to prove that TREE is well-defined. So the statement "TREE
is well-defined" is, in some sense, "less constructive" or "more infinitary"
than R_2^2.

~~~
igravious
I really appreciate your reasonable and measured tone but I'll need some time
to digest your comment. It's a brain-full :)

------
wutf
The book Where Mathematics Comes From demonstrates that the concept of
infinity itself is finitistically reducible via conceptual metaphor. This
really isn't that surprising though, is it? Our experience is grounded in the
interaction of concrete, finite objects.

------
arenaninja
So if I get the gist of this correctly, this makes proofs that are applicable
to finite sets/structures/objects easier to translate to a proof that's
applicable to infinite sets?

------
jsprogrammer
You may be interested in Wildberger's recent Math Foundations series lectures.
180 [* * _] may be a decent starting point.

[_ * *] [https://youtu.be/LJR24_Povzw](https://youtu.be/LJR24_Povzw)

------
openasocket
I'm having trouble imagining what finitism actually is, based on what I
remember from set theory. Is it equivalent to Zermelo-Frankel set theory
without the axiom of infinity?

~~~
pohl
I found this helpful:

"The main idea of finitistic mathematics is not accepting the existence of
infinite objects such as infinite sets. While all natural numbers are accepted
as existing, the set of all natural numbers is not considered to exist as a
mathematical object. Therefore quantification over infinite domains is not
considered meaningful."

[https://en.wikipedia.org/wiki/Finitism](https://en.wikipedia.org/wiki/Finitism)

------
sewercake
does this 'heirarchy of logic' have any direct relationship to computational
models, or do all (non-hyper) models fall under a single category?

------
robinhouston
I love the fact there's a long popular article devoted to the proof “that
RT^2_2 is Π^0_3 conservative over IΣ^0_1”, as the abstract of the paper
summarises it. Reverse mathematics is a fairly niche area even within
mathematical logic. Kudos to Natalie Wolchover for taking it on.

~~~
xviia
It makes me curious how Quanta is funded. The journal is very high quality and
in a niche market: pure science journalism. I can't imagine they make enough
subscription revenue; maybe it's funded mostly through the Simons Foundation?

Either way, glad it exists!

~~~
defen
I believe it's funded by James Simons

~~~
xviia
Ah, now I understand. For others:

Simons Foundations --> James Simons, who is a hedge fund mananger worth ~
$14billion.

Excellent interview with Numberphile here:

[https://www.youtube.com/watch?v=QNznD9hMEh0](https://www.youtube.com/watch?v=QNznD9hMEh0)

~~~
antognini
In addition, James Simons was a mathematician prior to his work in the finance
industry, hence his partiality toward funding a magazine that often writes
about pure mathematics.

~~~
bradleyjg
I believe he is now back at Stony Brook in the mathematics department. Not
full time, he has a lot going on, but some kind of affiliation.

~~~
xviia
I bet being an academic is a lot better when you don't have to worry about
being a post-doc searching for a faculty position :/

~~~
tikhonj
To be fair, he was a successful researcher who became professor and later
chair of the Mathematics department at Stony Brook _before_ starting
Renaissance, so he got past all that the normal way.

