Sometimes all functions are continuous (2006)

http://math.andrej.com/2006/03/27/sometimes-all-functions-are-continuous/

by lelf • 7 years ago

60 42 7 years ago

math.andrej.com

The article is a slow burn with a hilariously clickbait title but damn, those are some fascinating and significant consequences.

An interesting question comes to mind: which programming features, apart from the ones we mentioned above, allow us to program [computably continuous integer streams]? In terms of Haskell, the same question is which monads allow us to program [these streams]?
Can we do it without any extra features and use just “pure” functional programming?
If by pure functional programming we understand a functional programming language with natural numbers, booleans and recusive definitions, also known as PCF, the answer is no. The proof of this uses denotational semantics and domain theory, and I will not go into it now. You may entertain yourself by trying and failing to define [computably continuous integer streams] in pure Haskell or ML, i.e., no side effects, no exceptions, no mutable store, no parallelism, no monads.
... The lesson is for those “experts” who “know” that all reasonable models of computation are equivalent to Turing machines. This is true if one looks just at functions from N to N. However, at higher types, such as the type of our function m, questions of representation become important, and it does matter which model of computation is used.

It takes a hit on the trend that was in vogue in early 2010s to do AI and metaphysics on top of Kolmogorov Complexity, Solmonoff Induction, information physics... people took the Turing Equivalence as a carte blanche to say whatever willy-nilly thing you want about the (potentially) computable nature of reality without being informed by more naturalistic methods.

This is harshed even further given that these information-theoretic arguments often used non-computable notions.

mturmon• 7 years ago

Note that the comments thread below the article addresses some of the natural objections to the novel results exhibited in the post.

quotemstr• 7 years ago

In particular,

the real number is not given by an infinite sequence that is actually written down anywhere, but rather as an algorithm, or a black box, which accepts a number n and outputs the n-th digit. At no point do I manipulate an infinite amount of information, yet I can look at any digit I want. Of course, I cannot look at all digits in a finite time, which is sort of the point of the post.

edflsafoiewq5• 7 years ago

A related fact: all functions are defined on closed sets. You cannot "tell" whether a number is in a set or is a limit point of a set. Since all functions are continuous, they work at limit points too.

As a "practical" consequence: the common design of a random() function returning values in [0, 1) is wrong. Any function you write cannot "tell" whether it got a 1 or not.

These are still very good guidelines even for floating-point code, as long as you, like most everyone else, treat floats as a blackbox abstraction for "real" real numbers.

poizan42• 7 years ago

Well, random() functions that are normally used are defined on the [0, 1) interval of 32-bit or 64-bit IEEE 754 floating point numbers, for which it is very much computable to determine whether they are exactly 1 (there's nothing "real number" about them though, they all belong to a small finite subset of the rations). Of course that isn't really an open interval in the topological sense either.

edflsafoiewq5• 7 years ago

But that's why it's often a bad idea to test exact equality of floats: the equality test isn't continuous.

In fact, for float computations it's usually even more important to use continuous (ie. error tolerant) functions, since floating-point operations can introduce errors.

tel2• 7 years ago

The punchline at the end is worth calling out: Turing Equivalence applies at first-order computations like `Nat -> Nat`, but fails at higher-order computation like `(Nat -> Nat) -> {0, 1}`.

alain940402• 7 years ago

I'm not making much sense of that article. It's rejecting both the mathematical definition of continuity, and also wants to compute with infinite real numbers, which is just not realistic to computers and languages. So the author ends up with a lot of absurd results.

throwawaymath2• 7 years ago

That's because the author is using constructive mathematics (i.e. computable analysis instead of standard real analysis). It is true that all computable functions are continuous. Likewise discontinuous real functions aren't computable.

The continuum hypothesis is tangentially related to this topic and makes for fun reading.

alain940402• 7 years ago

But this is begging the question. The author uses the definition that numbers have infinite decimals, and all you are allowed to do is ask for decimal n to find out a number. You can never grasp the full number, since you are given a finite amount of time (compute) to discover an infinite sequence of decimals.

From that definition, it's quite obvious that everything you compute has to be continuous, because you are never sure of what other decimals may be coming up, so whatever you compute has to be close enough.

That sounds more like an argument that representing numbers that way is not particularly useful since you can't do much with them (you can't even provide an equality operator).

jayd16• 7 years ago

In fact, no finite amount of computation will guarantee that we will be able to tell whether x=0 or x>0

Hmm, I'm still stuck at this assertion. Why can we not assume the number is finite?

If we assume an infinitely long number takes an infinite amount of time to read, can we not also assume it must take an infinite amount of time to write? If we only have finite time, can we not assume all numbers given to the function in that time are finite?

edflsafoiewq5• 7 years ago

You may like the whole list of constructive gems and stones then: http://math.andrej.com/category/gems-and-stones/ :)

As a sampler, the following are all constructively false (ie. imply the law of excluded middle)

* Every finite set can be enumerated without repetitions

* A subset of a finite set is finite

* The intersection of finite sets is finite

lonelappde3• 7 years ago

Note that these results rely on a technical definition of set that is rather abstract mathematically, in a way programers may find amusing. It defines a set as a potentially infinite list with potential duplicates, from which we can observe that some lists cannot be deduplicated in finite time, making certain operations that are easy on computer science sets not always possible on computer science lists.

For example, is A a subset of B={0}, if all you know about A is that it is a Java iteratable that returns 0 for the first 10^3000 elements you examine?

blablabla123• 7 years ago

It's a bit unclear, also the conclusion. However most definitions of continuity are just covering real functions. I remember my math prof stating that there are also continuous functions on the natural numbers (N->N), IIRC n -> n^2 would be an example. Given that numerical calculations are not on R - at best on Q - this doesn't sound so far-fetched

lonelappde3• 7 years ago

The results are only absurd if you hold absurd axioms like believing that uncomputable numbers can be computed, or that the real number set is real.

svat• 7 years ago

More precisely, all well-defined computable functions are continuous.

For another exposition of the same (or similar) results, see also Dan Piponi's post from 2008 "What does topology have to do with computability?" at http://blog.sigfpe.com/2008/01/what-does-topology-have-to-do...

Also, as a follow-up, you may be interested in Martin Escardo's 2007 post on the same blog: http://math.andrej.com/2007/09/28/seemingly-impossible-funct...

carapace• 7 years ago

Greg Buchholz in the site's comments linked to Gregory Chaitin's "How real are real numbers?"[1] and asks,

Isn't it a little bit crazy to talk about computations on real numbers, when most of them are uncomputable/unnameable?

[1] https://arxiv.org/pdf/math/0411418.pdf

Experimental physicists know how difficult accurate measurements are. No physical quantity has ever been measured with more than 15 or so digits of accuracy. Mathematicians, however, freely fantasize with infinite-precision real numbers. Nevertheless within pure math the notion of a real number is extremely problematic. We’ll compare and contrast two parallel historical episodes:
1. the diagonal and probabilistic proofs that reals are uncountable, and
2. the diagonal and probabilistic proofs that there are uncomputable reals.
Both case histories open chasms beneath the feet of mathematicians.

skybrian2• 7 years ago

I'm not a mathematician, but it sounds like what this is saying is that as soon as you allow a number to contain an infinite number of digits, you can get an infinite loop when you try to do a simple comparison. (For example, try to compare pi to an number that's not-quite pi. It might be a little more or a little less, but you can't know this without looking at all the digits, which is impossible.)

As a consequence, there are only two boolean-valued total functions: always return true, or always return false.

I guess that's why we don't use infinite streams of numbers for computation.

lonelappde3• 7 years ago

You don't need to look at all the digits, but there is no finite stopping point you can choose in advance.

It's similar to the problem of comparing two natural numbers given in least significant bit order.

skybrian2• 7 years ago

Given a comparison function, an adversary can construct streams of digits that will make it run forever. This would be true either way. The problem seems to be with allowing algorithmically generated streams as input?

Maybe our domain should be limited to numbers generated by algorithms that are known to terminate? But despite eliminating irrational numbers, this guarantee doesn't seem like enough for practical computing. Even arbitrary precision algorithms require their inputs not only to be computable, but actually computed in advance, showing at least that they fit in memory.

Nevermark2• 7 years ago

I don't understand the point of arguments about constructible functions on unconstructable reals. I.e. reals with infinite digits and no finite symbolic representation.

For anything actually computable (in finite time), on computable (in finite time) values, you will always get discrete (not continuous) functions on discrete (not continuous) values.

vilhelm_s• 7 years ago

The idea of computable reals go back to at least Turing. The idea is that a real number (or an operation on real numbers) is computable if we can compute the decimal digits, one by one. This way, if the user wants to know it to any desired precision, they just have to wait until the computer has printed that many decimals. You can equivalently treat them as terminating computable functions of type (nat -> bool), which computes the n:th binary decimal.

This is actually a quite constructive notion! At least, it works well with intuinistic mathematics, which is why Andrej Bauer is writing about it.

Nevermark2• 7 years ago

Ok, that makes sense. Even when restricting numbers to reals with finite symbolic/program definition, comparing two such reals requires an unknown and unbounded number of their digits to be computed. If two symbolic representations define the same number, but it isn't obvious, a complete digit comparison will never end. Got it.

Thank you!

mianos• 7 years ago

This reminds me of the coastline paradox: https://en.wikipedia.org/wiki/Coastline_paradox

Specifically, due to the fractal nature of a coastline the length diverges to infinity. The theoretical length at the infinite scale is infinity.

improv32• 7 years ago

For a very nice into to constructive math, see this[0] link as well as some good HN discussion here[1].

[0] https://home.sandiego.edu/~shulman/papers/rabbithole.pdf [1]: https://news.ycombinator.com/item?id=18411935

Gehinnn• 7 years ago

How do you implement algorithms for real numbers if there uncountable many of them? There is always a real number your algorithm cannot read/process.

loicd• 7 years ago

The result presented in TFA is that all computable functions over the real numbers are continuous. To generalize that to 'all functions are continuous', you not only need to believe that all functions should be computable, which sounds reasonable for a constructivist, but also that all numbers should be real, which is not. Back when I was in academia, I remember a talk by a mathematician who thought that we should all be doing analysis with rational numbers rather than real numbers. With such a perspective, the sign function defined over rational numbers (e.g. presented as a pair numerator, denominator) is both discontinuous at 0 and computable.

Sometimes all functions are continuous (2006)

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