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I've increasingly come to doubt the utility of all these things. Real numbers and Lesbesgue integration that stem out of 19th c. analysis came before Computabi

I've increasingly come to doubt the utility of all these things. Real numbers and Lesbesgue integration that stem out of 19th c. analysis came before Computability was a thing... and it turns out that all but a measure 0 set of real-numbers are actually computable. So all these things that we worry about don't even "exist" in a computational sense. Mathematicians haven't come to terms with this (barring non-standard analysis etc.), and I suspect this is one of the reasons it's losing mind-share.

tsimionescu 2y ago
Note that "computable" just means "computable by a Turing machine or equivalent system", at this time. We don't know if we are missing some other form of computation that goes beyond the abilities of a Turing machine, that could upend the whole concept. And given that we haven't proven that general human thinking is a computable process, there is at least one significant candidate for possible computation beyond Turing machines. Note that I don't personally think it's likely that our thinking goes beyond the capabilities of a Turing machine, not at all. But I also think it would be premature to throw out all mathematics that doesn't conform to Turing machine computability before we are more sure that this is the best possible model. And even if it were, there is also the question of whether physical processes are computable or not. Right now there are plenty of physical processes where our best and only reliable models for predicting their behavior require assumptions from calculus, like the existence of all real numbers. For example, there is no successful formulation of quantum mechanics where the distance between two particles moving relative to each other can be constrained to any subset of the [minDistance, maxDistance] interval of the real number line. Which means that, as the particles move away from each other, at some times the distance between them will have to be an uncomputable number (and given their density in the real number line, this will be approximately all the time).
magneticnorth 2y ago
>> and it turns out that all but a measure 0 set of real-numbers are actually computable. This is the wrong way around. The computable numbers have measure 0. In fact, they are countable. The non-computable real numbers have full measure. (proof: by definition, computable numbers are generated by a program, which can be represented by a finite number of bits. The set of finite numbers is countable).
- xscott 2y ago
  What could it mean to have full measure but nobody can provide any method of generating the digits to a single element of the set? We can define things like Chaitin's constants, but we don't know very many digits, and Chaitin's constants are countable. We claim there are these bountiful numbers, more numerous than the countables, but can't even specify the value of any of them. I suspect some future generation is going to decide that diagonalizaation is based on questionable axioms. It leads to invisible pink elephants.
  - zmgsabst 2y ago
    They’re not finitely describable — for any member of that set, no matter how many digits you articulate, you’ve named a rational number. The same is true even for numbers that have finite predicates defining them — because there’s only measure zero finite predicates. Most real numbers are indescribable sequences (ie, not finitely describable), so it’s not surprising we can’t describe them. That’s what happens when you construct something by including all possible infinite sequences — almost all of them are total gibberish. Diagonalization is a consequence of that: we have a schema of descriptions forming our predicate and prove no matter how we index reals by the integers, there’s more stuff out there. To get rid of that, we need to stop completing sequences. Most people aren’t in favor of that finitism/constructive approach now — so I’m not sure why that would change.
    - xscott 2y ago
      Even Cantor's counterexample, where he traverses the diagonal of an enumerated list of numbers, he has a countable number of choices to pick his digits. So we've built a countable set, and he claims it's incomplete by picking one number from a countable set of counterexamples. So now lets union the original set with Cantor's countable set of counterexamples, and get a combined set which is also countable, right?
      - zmgsabst 2y ago
        But that schema can be applied to your new set, as well — to create yet more counter examples. Your method didn’t exhaust the real numbers outside the original list. You merely added one of them. For any listable subset of the real numbers, we can provide a counter example. And since we’re proving by contradiction in Cantor’s argument, we assume the original list is exhaustive, ie there’s a bijection between naturals and reals. We show that assumption leads to a contradiction, because there’s at least one real number for which no natural maps to it. Therefore, there can’t be such a bijection. - - - Assume f is a bijection between N and [0,1]. Let an be the nth digit of f(n). Then x = sum(1,inf)[10^-n * (an + 1 % 10)] is a real number in [0, 1] for which there is no f(k) = x. So f is not onto and therefore f is not a bijection. Therefore, no such bijection can exist — by contradiction. - - - You can’t patch that schema up just by adding counter examples to your first choice of bijection — because a counter example is constructable for any bijection attempt. The fundamental problem is that reals include limits — which is what allows the sum to be a real number.
paulfharrison 2y ago
Brownian motion mentioned at the end of the article and more generally Langevin Dynamics are incredibly useful. They're this weird interface between normal physics and statistical mechanics. When a big complex molecule is constantly jostled by smaller molecules, these nowhere-continuous motions are a good way to approximate what happens. Plus, rather bizarrely, it helps to understand this area to do actual statistics, in software such as Stan.
bjourne 2y ago
True. You could partition math with the question "Could a computer do that?" If the answer is yes, "that" is practical. If the answer is no, "that" is probably useless.
- nejsjsjsbsb 2y ago
  You could also partition roads to drive on base on "can I see a city with my naked eye". Seas to explore by "can I see land on the horizon" and so on.
deleted 2y ago
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roserevealed 2y ago
[dead]
wizzwizz4 2y ago
That measure 0 set of computable real numbers includes (nearly) every real number you'd ever describe – including all the reals on this page so far. The only exceptions are things like Chaitin's constant, which isn't really interesting for real analysis.
constantcrying 2y ago
>So all these things that we worry about don't even "exist" in a computational sense. They directly lead to weak derivatives which have certainly saved 10s of thousand of lives.
rssoconnor 2y ago
As a constructivist, I somewhat agree with your viewpoint, though my views are a bit more nuanced. (e.g. Dedekind's construction of the continuum as a collection of points is perhaps the primary source of corrupted thinking, which muddles even well-meaning thinkers into thinking that it is even possible to separate computable points from non-computable points. The real numbers form a /continuum/, and a continuum cannot be described as simply a collection of points.) That said I believe your comments are off the mark in regards to the topic of the Weierstrass function. The Weierstrass function is completely well behaved from a constructive point of view[0]; it is uniformly continuous and everything. In particular we can (constructively) map (constructive) real numbers to (constructive) real numbers via this function. [0]https://mathoverflow.net/a/43904 https://mathoverflow.net/a/43904
ginko 2y ago
This post reminds me of: https://old.reddit.com/r/mathmemes/comments/g3kolt/stop_doing_math/ https://old.reddit.com/r/mathmemes/comments/g3kolt/stop_doin...
nejsjsjsbsb 2y ago
No because Mathematics still has power. It would be awful if we couldn't dream beyond floating points or Mathematics were ties to the financial and physical forces that shape processor architectures.

tsimionescu 2y ago

Note that "computable" just means "computable by a Turing machine or equivalent system", at this time. We don't know if we are missing some other form of computation that goes beyond the abilities of a Turing machine, that could upend the whole concept. And given that we haven't proven that general human thinking is a computable process, there is at least one significant candidate for possible computation beyond Turing machines. Note that I don't personally think it's likely that our thinking goes beyond the capabilities of a Turing machine, not at all. But I also think it would be premature to throw out all mathematics that doesn't conform to Turing machine computability before we are more sure that this is the best possible model. And even if it were, there is also the question of whether physical processes are computable or not. Right now there are plenty of physical processes where our best and only reliable models for predicting their behavior require assumptions from calculus, like the existence of all real numbers. For example, there is no successful formulation of quantum mechanics where the distance between two particles moving relative to each other can be constrained to any subset of the [minDistance, maxDistance] interval of the real number line. Which means that, as the particles move away from each other, at some times the distance between them will have to be an uncomputable number (and given their density in the real number line, this will be approximately all the time).

magneticnorth 2y ago

>> and it turns out that all but a measure 0 set of real-numbers are actually computable. This is the wrong way around. The computable numbers have measure 0. In fact, they are countable. The non-computable real numbers have full measure. (proof: by definition, computable numbers are generated by a program, which can be represented by a finite number of bits. The set of finite numbers is countable).

xscott 2y ago
What could it mean to have full measure but nobody can provide any method of generating the digits to a single element of the set? We can define things like Chaitin's constants, but we don't know very many digits, and Chaitin's constants are countable. We claim there are these bountiful numbers, more numerous than the countables, but can't even specify the value of any of them. I suspect some future generation is going to decide that diagonalizaation is based on questionable axioms. It leads to invisible pink elephants.
- zmgsabst 2y ago
  They’re not finitely describable — for any member of that set, no matter how many digits you articulate, you’ve named a rational number. The same is true even for numbers that have finite predicates defining them — because there’s only measure zero finite predicates. Most real numbers are indescribable sequences (ie, not finitely describable), so it’s not surprising we can’t describe them. That’s what happens when you construct something by including all possible infinite sequences — almost all of them are total gibberish. Diagonalization is a consequence of that: we have a schema of descriptions forming our predicate and prove no matter how we index reals by the integers, there’s more stuff out there. To get rid of that, we need to stop completing sequences. Most people aren’t in favor of that finitism/constructive approach now — so I’m not sure why that would change.
  - xscott 2y ago
    Even Cantor's counterexample, where he traverses the diagonal of an enumerated list of numbers, he has a countable number of choices to pick his digits. So we've built a countable set, and he claims it's incomplete by picking one number from a countable set of counterexamples. So now lets union the original set with Cantor's countable set of counterexamples, and get a combined set which is also countable, right?
    - zmgsabst 2y ago
      But that schema can be applied to your new set, as well — to create yet more counter examples. Your method didn’t exhaust the real numbers outside the original list. You merely added one of them. For any listable subset of the real numbers, we can provide a counter example. And since we’re proving by contradiction in Cantor’s argument, we assume the original list is exhaustive, ie there’s a bijection between naturals and reals. We show that assumption leads to a contradiction, because there’s at least one real number for which no natural maps to it. Therefore, there can’t be such a bijection. - - - Assume f is a bijection between N and [0,1]. Let an be the nth digit of f(n). Then x = sum(1,inf)[10^-n * (an + 1 % 10)] is a real number in [0, 1] for which there is no f(k) = x. So f is not onto and therefore f is not a bijection. Therefore, no such bijection can exist — by contradiction. - - - You can’t patch that schema up just by adding counter examples to your first choice of bijection — because a counter example is constructable for any bijection attempt. The fundamental problem is that reals include limits — which is what allows the sum to be a real number.

paulfharrison 2y ago

Brownian motion mentioned at the end of the article and more generally Langevin Dynamics are incredibly useful. They're this weird interface between normal physics and statistical mechanics. When a big complex molecule is constantly jostled by smaller molecules, these nowhere-continuous motions are a good way to approximate what happens. Plus, rather bizarrely, it helps to understand this area to do actual statistics, in software such as Stan.

bjourne 2y ago

True. You could partition math with the question "Could a computer do that?" If the answer is yes, "that" is practical. If the answer is no, "that" is probably useless.

nejsjsjsbsb 2y ago
You could also partition roads to drive on base on "can I see a city with my naked eye". Seas to explore by "can I see land on the horizon" and so on.

deleted 2y ago

[deleted]

roserevealed 2y ago

[dead]

wizzwizz4 2y ago

That measure 0 set of computable real numbers includes (nearly) every real number you'd ever describe – including all the reals on this page so far. The only exceptions are things like Chaitin's constant, which isn't really interesting for real analysis.

constantcrying 2y ago

>So all these things that we worry about don't even "exist" in a computational sense. They directly lead to weak derivatives which have certainly saved 10s of thousand of lives.

rssoconnor 2y ago

As a constructivist, I somewhat agree with your viewpoint, though my views are a bit more nuanced. (e.g. Dedekind's construction of the continuum as a collection of points is perhaps the primary source of corrupted thinking, which muddles even well-meaning thinkers into thinking that it is even possible to separate computable points from non-computable points. The real numbers form a /continuum/, and a continuum cannot be described as simply a collection of points.) That said I believe your comments are off the mark in regards to the topic of the Weierstrass function. The Weierstrass function is completely well behaved from a constructive point of view[0]; it is uniformly continuous and everything. In particular we can (constructively) map (constructive) real numbers to (constructive) real numbers via this function. [0]https://mathoverflow.net/a/43904 https://mathoverflow.net/a/43904

ginko 2y ago

This post reminds me of: https://old.reddit.com/r/mathmemes/comments/g3kolt/stop_doing_math/ https://old.reddit.com/r/mathmemes/comments/g3kolt/stop_doin...

nejsjsjsbsb 2y ago

No because Mathematics still has power. It would be awful if we couldn't dream beyond floating points or Mathematics were ties to the financial and physical forces that shape processor architectures.