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I'm reading a book on axiomatic set theory, classic Set Theory: For Guided Independent Study, and at the beginning of chapter 4 it says:

So far in this book we have given the impression that sets are needed to help explain the important number systems on which so much of mathematics (and the science that exploits mathematics) is based. Dedekind's construction of the real numbers, along with the associated axioms for the reals, completes the process of putting the calculus (and much more) on a rigorous footing.

and then it says:

It is important to realize that there are schools of mathematics that would reject 'standard' real analysis and, along with it, Dedekind's work.

How is it possible that "schools of mathematics" reject standard real analysis and Dedekind's work? I don't know if I'm misinterpreting things but, how can people reject a whole branch of mathematics if everything has to be proved to be called a theorem and cannot be disproved unless a logical mistake is found?

I've even watched this video in the past: https://www.youtube.com/watch?reload=9&v=jlnBo3APRlU and this guy, who's supposed to be a teacher, says that real numbers don't exist and that they are only rational numbers. I don't know if this is a related problem but how is this possible?

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    $\begingroup$ I'm guessing the possible objections could come from ultrafinitism (which denies the existence of infinity) and intuitionistic logic (which denies the existence of the law of the excluded middle). $\endgroup$ – twnly Jul 23 at 14:13
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    $\begingroup$ Mathematics does not give you absolute truth. In fact, does absolute truth exist at all? Every truth is relative to what initial rules/assumptions you are playing by. I explain more in my answer below $\endgroup$ – Riemann'sPointyNose Jul 23 at 14:55
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    $\begingroup$ The video is one of many by Norman J. Wildberger. He is very much an outlier. Read the Wikipedia article Rational trigonometry for some of his more rational ideas. He has been mentioned on MSE several times. For example MSE question 962449 "Do mathematicians, in the end, always agree?". $\endgroup$ – Somos Jul 23 at 17:05
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    $\begingroup$ That video is by Norman Wildberger, whose views are, to say the least, a bit eccentric. $\endgroup$ – Brian M. Scott Jul 23 at 17:06
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    $\begingroup$ @Brian: Even for an understatement, "a bit eccentric" is an understatement. It's such an understatement that if $n$ is the largest number NJ Wildberger believes to exist, one can still chain $n+1$ times "understatement" and it would still not be enough to describe him. $\endgroup$ – Asaf Karagila Jul 23 at 18:48
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Although the possibility of different axioms is a concern, I think the major objection the author is speaking of is largely about constructivism (i.e. intuitionistic logic). There really is a big gap between rational numbers and real ones: with enough memory and time, a computer can represent any rational number and can do arithmetic on these numbers and compare them. This is not true for real numbers.

To be specific, but not too technical: let's start by agreeing that the rational numbers $\mathbb Q$ are a sensible concept - the only controversial bit of that involving infinite sets. A Dedekind cut is really just a function $f:\mathbb Q\rightarrow \{0,1\}$ such that (a) $f$ is surjective, (b) if $x<y$ and $f(y)=0$ then $f(x)=0$, and (c) for all $x$ such that $f(x)=0$ there exists a $y$ such that $x<y$ and $f(y)=0$.

Immediately we're into trouble with this definition - it is common that constructivists view a function $f:\mathbb Q\rightarrow\{0,1\}$ as some object or oracle that, given a rational number, yields either $0$ or $1$. So, I can ask about $f(0)$ or $f(1)$ or $f(1/2)$ and get answers - and maybe from these queries I could conclude $f$ was not a Dedekind cut (for instance, if $f(0)=1$ and $f(1)=0$). However, no matter how long I spend inquiring about $f$, I'm never going to even be able to verify that $f$ is a Dedekind cut. Even if I had two $f$ and $g$ that I knew to be Dedekind cuts, it would not be possible for me to, by asking for finitely many values, determine whether $f=g$ or not - and, in constructivism, there is no recourse to the law of the excluded middle, so we cannot say "either $f=g$ or it doesn't" and then have no path to discussing equality in the terms of "given two values, are they equal?"*.

The same trouble comes up when I try to add two cuts - if I had the Dedekind cut for $\sqrt{2}$ and the cut for $2-\sqrt{2}$ and wanted $g$ to be the Dedekind cut of the sum, I would never, by querying the given cuts, be able to determine $g(2)$ - I would never find two elements of the lower cut of the summands that added to at least $2$ nor two elements of the upper cut of the summands that added to no more than $2$.

There are some constructive ways around this obstacle - you can certainly say "real numbers are these functions alongside proofs that they are Dedekind cuts" and then you can define what a proof that $x<y$ or $x=y$ or $x=y+z$ looks like - and even then prove some theorems, but you never get to the typical axiomatizations where you get to say "an ordered ring is a set $S$ alongside functions $+,\times :S\times S \rightarrow S$ and $<:S\times S \rightarrow \{0,1\}$ such that..." because you can't define these functions constructively on $\mathbb R$.

(*To be more concrete - type theory discusses equality in the sense of "a proof that two functions $f,g$ are equal is a function that, for each input $x$, gives a proof that $f(x)=g(x)$" - and the fact that we can't figure this out by querying doesn't mean that we can't show specific functions to be equal by other means. However, it's a huge leap to go from "I can compare two rational numbers" - which is to say, I can always produce, from two rational numbers, a proof of equality or inequality - to "a proof that two real numbers is equal consists of..." understanding that the latter definition does not let us always produce a proof of equality or inequality for any pair of real numbers)

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    $\begingroup$ I'm too lazy to post an answer, but the asker might be interested in the fact that determining equality between two computable sets of rationals is as hard as solving the halting problem, which is why it is impossible. Similarly, equality between two computable reals is not computable. $\endgroup$ – user21820 Jul 24 at 13:37
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    $\begingroup$ From my POV as someone in computer science, where we basically do things constructively, nonconstructive systems like classical logic and the standard presentation of the reals are certainly useful mathematical tools or abstractions, I just don’t believe they correspond directly to physical phenomena, so I “reject” them in that sense. Philosophically, I find constructive (and linear) logic more compelling, because it seems like a more “honest” accounting of how things work that I care about, such as computational resources and effective algorithms. $\endgroup$ – Jon Purdy Jul 24 at 20:40
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    $\begingroup$ @JonPurdy: I don't know whether you're responding to me or not, but do you think the halting problem is well-defined? That is, given a program and the input it is run on, is it the case that either it halts or it does not halt? $\endgroup$ – user21820 Jul 25 at 13:20
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    $\begingroup$ @user21820: I was just offering some additional perspective on the answer, not necessarily responding to anyone specifically. I think the halting problem is well defined, but I prefer presentations of it like “Determine whether a machine will reach a state in a fixed number of steps”, i.e., about time bounds rather than halting. Alternatively, I think of a TC language program as a function A → ¬¬B, which, given an A, will not return anything that’s not a B. The halting function tries to decide ¬¬B = B—classically, this is decidable by definition; constructively it isn’t. $\endgroup$ – Jon Purdy Jul 25 at 21:37
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    $\begingroup$ @JonPurdy: I'm not sure what you're getting at. Just because the halting oracle has no real-world embedding does not mean that it has no real-world implication, such as the impossibility of implementing it via a program. But never mind. I know intuitionistic logic but it doesn't change the fact that the real world is fully classical. $\endgroup$ – user21820 Jul 26 at 4:36
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This is somewhat surprising if you're not used to this. But of course you're free to reject whatever mathematical statements you dislike. The real question is what else you are forced to reject with it, and what would remain of the mathematics that you know and love otherwise.

The onus is on you, as someone who decided that "everyone else is wrong", to convince people that your idea is better, and to get people to take interest in what and how to transfer mathematics from "the realm of error" into "the world of truth". That is, until someone will come in and reject your ideas, etc.


For example, Lebesgue is well-known as someone who rejected the axiom of choice. For him, the existence of non-measurable sets was unthinkable, so he was forced to reject the axiom of choice, and many other theorems that would contradict that.

Another example is in Kronecker who rejected the idea that infinite sets exist, this means that for Kronecker the axiom of infinity would be false. That implies that we want to work, in some sense, with some a second-order theory over the natural numbers, we can get some analysis done, and everything beyond that would be "a fiction".

Many people would reject large cardinal axioms, those are easily misunderstood and mistrusted outside of set theory (although often ignored just as well). But without inaccessible cardinals, there are not Grothendieck universes; without measurable cardinals there are some accessible categories which are not well-copowered. Even some set theorists reject large cardinal axioms such as Reinhardt and Berkeley cardinals, since they imply the negation of the axiom of choice, which (unlike Lebesgue) most set theorists readily accept as "obvious truth".

What is true, is that there is an implicit theory underlying mathematics, which lets us develop "most of working mathematics" without having to worry about foundations. But this theory is not without its controversies. It includes infinite sets, the axiom of choice, the law of excluded middle, and more. Sometimes it is just interesting to see what part actually depends on these axioms, and sometimes people outright feel that something is wrong with the axioms.

If you are more inclined to use computer assistance in your work (e.g. proof verification software), you might be more inclined to take a different foundation which is easier to understand from your proof assistant's point of view. This may be something that rejects the LEM, for example, or otherwise is incongruous with what "most people" would call "every day mathematics".

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  • $\begingroup$ I think most people would agree with Lebesgue if the alternative was not almost as bad (as far as I understand this) $\endgroup$ – lalala Jul 24 at 8:32
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    $\begingroup$ @lalala: We have known from the 1920s that measurable linear maps are continuous. After Solovay proved his seminal result on the consistency of "all sets are measurable", Garnir wrote a paper where he concluded that in that setting a very large class of topological vector spaces satisfy "all linear maps are continuous". He excitedly suggested that functional analysis will be split into Zornian (ZFC), Solovayan, and Constructive (ZF). But life had other plans. If all linear maps are continuous, Hahn–Banach fails, Banach-Alaoglu theorem fails, and so much more. It just won't stick. $\endgroup$ – Asaf Karagila Jul 24 at 8:35
  • $\begingroup$ @lalala: But one can simply reject assuming AC or its negation, without getting any funny consequences. Over ZF, it just means that one can say less than ZFC. I'm not saying I think we should throw away AC, but there is no reason to assume AC just because we cannot prove "there exists a non-measurable set of reals" without it. Nor should we assume "some set of reals is non-measurable" just because its negation is horrible. $\endgroup$ – user21820 Jul 24 at 13:28
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Remember that the book you are reading is on axiomatic set theory. Any time you do pure mathematics, you have to start with axioms. You can't prove them, you just specify them. And then you use them to prove other things.

The famous example of this is the parallel postulate. People were surprised when it was realized that you could have a perfectly consistent geometry where there was an infinite number of lines through a point and parallel to another line (not on the point).

In set theory, the axiom of choice plays a similar role. You cannot prove it from the other axioms, but yet it feels more like a theorem than a lot of the other axioms. Most people find it intuitively true, but some do not.

The different "schools" are people with different opinions about which sets of axioms you should use. They are not mainstream, but unlike fringe groups in other fields, nobody doubts the validity of the math that they do. The question "If you reject the axiom of choice, what can you prove?" is perfectly legitimate.

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Different people use different axiom systems.

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    $\begingroup$ yeah but you can't reject real analysis just because you started off from different axioms, other results found starting from other axioms have to be valid anyway $\endgroup$ – Andrea Burgio Jul 23 at 14:12
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    $\begingroup$ Why not? Without any form of axiom of choice, many theorems in analysis no longer can be proven! $\endgroup$ – ε-δ Jul 23 at 14:13
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    $\begingroup$ Yes, I believe so. And this takes us to Gödel's incompleteness theorem and more advanced depressing stuff. $\endgroup$ – ε-δ Jul 23 at 14:16
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    $\begingroup$ @AndreaBurgio how do you define truth? To a Mathematician, truth is relative to which system you are working in. From an outsider perspective, maybe you can ask "are the axioms actually true?" and you would be asking a Physics or a Philosophical question. But this isn't a Mathematics question. We always work within a system / framework, and if it's true from our postulates - it's true within that system. We aren't constrained to one universe $\endgroup$ – Riemann'sPointyNose Jul 23 at 14:25
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    $\begingroup$ Oversimplified perhaps -- but I really like the concision of this answer. $\endgroup$ – John Coleman Jul 24 at 1:27
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You need both a statement and an axiomatic system you are working within to evaluate whether a statement is true. You cannot deduce whether a statement is true simply from itself. It wouldn't make any sense. It's like me asking you - "is that ball over there on the floor green?". If you didn't understand the words ball, green, or even floor - how could you answer this question? Or let's say we both disagree on what the color green is - it might be true to me the ball is green. But you might say "no, that ball is blue! I can see with my own eyes that ball is blue!".

You can think of a combination of both statement and system,

$${(p,S)}$$

where ${p}$ denotes some proposition and ${S}$ denotes some system you are working within, and you could think of some operator ${T(p,S)}$ which returns true or false depending on whether that proposition is true. It might be that within a system $S$, ${p}$ doesn't even make sense. That is - you cannot even evaluate

$${T(p,S)=?}$$

Or it's truth value could change depending on which $S$ you use:

$${T(p,S_1) = \text{true}}$$

$${T(p,S_2) = \text{false}}$$

So don't think of Mathematics as "This is definitely 100% how everything must work". It's more like "Given these set of axioms, and given they are consistent - this is definitely 100% how things must work". Notice we had to assume also the system was consistent - there are some systems which are inconsistent and hence evaluating a statement could be both false and true simultaneously - hence it's inconsistent. It doesn't agree with itself:

$${\text{A system $S$ is inconsistent iff both }T(p,S)=\text{true}, \text{ and }T(p,S)=\text{false}}$$

In your case - it's quite easy to see how it could be rejected. If someone uses a different ${S}$, as stated above - the results could be quite different. Some people may disagree with the choice of $S$. Ultimately what's true to a Mathematician is governed by the axioms they are using, which can very easily change. It's not so black and white. There's no such thing as "absolute truth" in Mathematics. You need to state what rules you are playing by. In fact - I'm not sure it even makes sense to ask what "absolute truth" is in any field - you always need some context, some background.

Edit: Just to restate - this operator "$T$" I have defined here is not to be taken too seriously. It's just a symbolic way of writing out this viewpoint on Mathematics, and how Mathematics works. As I said - there are a lot of statement + system combinations where $T$ isn't evaluate-able in any meaningful way. For example, take

$${S = \{false\}}$$

That is, the axiomatic system that tells us everything is false. Now take the statement

$${p=\text{this statement is false}}$$

Assuming $S$ is consistent, try to evaluate ${T(p,S)}$. You cannot do it. The statement can't be true, since the axioms tell us everything must be false. But if it's false, it's true, which again contradicts the axiom.

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  • $\begingroup$ I suspect that many of the people who would object to Dedekind's particular construction of real analysis would also object to the notion that there is a function that tells you whether something is true or false (not to mention that a system of axioms can have different models, in which different statements are true/false - i.e. some statements are independent of some axiom system) - and this is all leaving behind the questions of consistency and of defining $T$, which both raise serious and deep concerns that largely undermine this point of view. $\endgroup$ – Milo Brandt Jul 23 at 14:54
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    $\begingroup$ @MiloBrandt The "$T$" operator I talk about in this post shouldn't be taken too literally, and as I said - in some cases we might not even be able to properly define ${T}$ when applied to some statement and system. It was just a symbolic way of trying to get across the point that truth really does depend on where you start from and what assumptions you make. I'm not sure how it would undermine this argument. For example, ${T(2n\ |\ n \in \mathbb{N}\text{ is always divisible by }2,\text{Real Analysis})}$ can be very easily evaluated and shown to be true $\endgroup$ – Riemann'sPointyNose Jul 23 at 15:09
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    $\begingroup$ @MiloBrandt Now, a better counterargument to my point of view is "how do you know that's true? What assumptions are you starting from?" - and honestly I feel like this point of view I have written is more of an unspoken, Philosophical axiom - it's how any serious Mathematician treats Mathematics in the modern-day $\endgroup$ – Riemann'sPointyNose Jul 23 at 15:13
  • $\begingroup$ Axiomatic systems in math are more than definitions, axioms, and theorems. There are also "undefined concepts". I remember from projective geometry that addition of the undefined concept "sense" could be substituted for an axiom about the order of points on a line. $\endgroup$ – richard1941 Jul 29 at 14:14
  • $\begingroup$ @richard1941 I think if your theory contains undefined concepts - then you should really be working towards finding a suitable working axiom. That's the whole point of pure Mathematics - to cut out intuitivistic logic and replace it with a strong foundation. I think the moment you are working with intuitivistic logic - you are no longer doing things from a strict set of definitions, and so I'd argue you are no longer working within an axiomatic system $\endgroup$ – Riemann'sPointyNose Jul 29 at 15:02
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"Rejecting" standard real analysis is (usually) not rejecting that real analysis's theorems can be derived from the axioms used.

Rather, it is rejecting it as a reasonable model of numbers.

When you take axiomatic mathematics and you run with it, you end up with strange results. Things like Gödel numbering lets you show that your axiomatic system for even something as simple as addition, subtraction, multiplication and division isn't able to exclusively model what you'd intuitively think was numbers.

To evaluate an axiom system, we have to look at what it is used for. Sure, this is icky applied mathematics, but if our axiom system of counting numbers (for example) doesn't model what you do when you go "one chicken, two chicken, three chickens, ..., 5,272,992 chickens, ...", maybe you should reconsider your axiom system for counting numbers.

People who "challenge" or "don't accept" standard real analysis think that alternative sets of assumptions -- axioms, rules of proof, etc -- produce better or maybe just otherwise useful results.

One example is constructive real analysis. Here, we start with pretty much the same assumptions, but we remove the law of excluded middle -- that a statement can be assumed to be either true or false. We still claim that no statement can be both. There just isn't a rule that goes "!!X implies X". There is still a rule that "!!!X implies !X", which can be derived from the other axioms of logic.

There are some other subtle changes as well.

That, plus being a bit careful about what other axioms we use, and changes to some of the definitions of terms in analysis (these "redefinitions" can be shown to be equivalent in standard set theory to the standard definitions usually), gives us an interesting property; that you can take any proof of existence of a object and mechanically turn the proof into an algorithm that produces the object.

So if you have a proof that says "there exists an X with property P(X)", you can always write out the digits of X (well, the algorithm to do so might be expensive).

There are other non-standard real analysis as well. Some permit infinitesimals -- values that are bigger than 0, but smaller than any number you can write down -- and do calculus where dx/dy can be calculated by doing infinitesimal mathematics.

We call all of these a form of "real analysis", because at the scale of doing calculus in the service of physics, these all end up agreeing. They can all agree and derive that a car with an acceleration of a over time t travels 1/2 a t^2 distance, they call can produce their equivalent of the fundamental theorem of calculus, etc.

Sometimes there will be slight differences. For example, the intermediate value theorem states that any continuous function of one variable that starts above a line and ends below a line crosses the line. The constructive version instead concludes that it gets within any arbitrary distance of the line.

Because there is no effective procedure that permits you to take an arbitrary continuous function, a proof that it is above the line at one point, and below at the other, and produce a decimal expansion (or equivalent) of the location where it crosses the line ... constructive analysis doesn't give it to you.

Constructive analysis does give you a sequence of points $p_i$ such that $|f(p_i) -k|$ converges to 0, and the points $p_i$ all lie within a closed interval; in classical analysis, this guarantees a convergent subsequence. In constructive analysis, this doesn't guarantee a convergent subsequence, because there is no way to find that convergent subsequence!

No physical experiment could distinguish between these two claims, because they disagree in the limit. So both model reality. One just models a reality with additional, untestable claims (and that is the classical analysis version).

One can find nonstandard analysis useful without "rejecting" standard analysis. As an example, when you are doing geometry on a computer, being aware of constructive analysis theorems and their difference with classical analysis can help illuminate some things you shouldn't assume.

And this isn't just games. A recent paper - popular article - uses intuitionalist/constructive reals and logic to describe a non-time-symmetrical general relativity universe. Because a fully time-symmetrical universe requires infinitely dense information at the big bang; sort of like an infinitely precise real number.

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    $\begingroup$ I'd be careful in your (very) informal description of Gödel's theorem. Real-closed fields are systems of numbers with addition, multiplication, and inverses to both, but it is a recursive theory which is both consistent and complete (the reason being that you cannot define the natural numbers in this first order theory). $\endgroup$ – Asaf Karagila Jul 26 at 17:31
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Real analysis based on Dedekind cuts (or Cauchy sequences) has proven extremely useful in modelling many aspects of physical reality in science and engineering. Proponents of alternative systems have not found any internal inconsistencies in it, but they seem to enjoy the challenge of doing math with less powerful tools and have met with limited success in some areas of application.

IMHO it seems to me that Wildberger is just being provocative by saying that he "rejects" real analysis. If he wants to be taken seriously in this, he must either demonstrate an internal inconsistency or formulate a workable alternative. As it stands, the current system works remarkably well and cannot so easily be dismissed.

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    $\begingroup$ What aspects of modelling physical reality do you lose if you reject the axiom of choice? All of the physics I'm aware can be done without it (or at least with weaker versions of it). $\endgroup$ – Alexis Olson Jul 24 at 20:40
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    $\begingroup$ @AlexisOlson Do you mean that none of the mathematics used in physics required AC in its development? $\endgroup$ – Dan Christensen Jul 25 at 2:02
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    $\begingroup$ @AlexisOlson: Nothing at all. Indeed all applications of mathematics to the real world do not seem to need anything beyond higher-order arithmetic, which is not even anywhere near Z, much less ZF. $\endgroup$ – user21820 Jul 25 at 13:24
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    $\begingroup$ @AlexisOlson I would be surprised if any paper on theoretical physics cited AC, if that's what you mean, but I find it hard to believe a formal development of real (and complex) analysis can be done without using AC at some point starting from say, Peano's Axioms, constructing the reals, etc. $\endgroup$ – Dan Christensen Jul 26 at 3:09
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    $\begingroup$ Here is the simplest application of AC that I can think of: Given a surjective function $f:X \to Y$, formally prove the existence of a function $g: Y\to X$ such that $g(a)=b \implies f(b)=a$. (We say the $g$ is the half-inverse of $f$.) Can this be done without AC? $\endgroup$ – Dan Christensen Jul 26 at 4:50
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I believe the important point to be that real analysis as most mathematics is like a house. It is built on a foundation of axioms that create the "floorplan" of the house. The foundation is then built upon and decorated.

The current foundation is not the only possibility. Indeed over time it has changed. And it is quite possible, almost inevitable, that it will change in the future. What happens when the foundation changes is that you tear down the whole house and start over again. It has happened over time.

Now, you might believe that tearing down the foundation is destructive, and it is. But you may reach a situation where the current foundation simply does not manage to handle all the requirements, and then it is time to rebuild from bottom up.

This, tearing up the foundations, is where the different schools are. You are, as a student, being "brain washed" to accept the current main stream thinking. This is in a way good - you need to know the current state in order to build on it and change it in the future, and could be very bad as it might hinder you from thinking the really new thoughts needed for the field to go forward.

To make an analogy, look at physics (simplified). There has been a change from Newton's mechanics to Einstein's relativity to quantum mechanics and the standard model. Each of these has demolished the foundation, but sort of kept the old one as a special case. The current Physics cannot fully describe the basis of some of the real world phenomenons, as example gravity. You might believe gravity to be well known by now, but do a little search for "quantum gravity" and you end up in an a place where the foundations are being currently under review.

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Counterexample: A cellular automaton can in principle end up simulating a virtual world where virtual mathematicians appear who invent real numbers. All the theorems they prove are therefore also theorems within the context of the theory of cellular automata, which only involves discrete mathematics.

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    $\begingroup$ Counterexample to what? $\endgroup$ – Asaf Karagila Jul 24 at 11:38
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    $\begingroup$ @AsafKaragila The whole premise of the question and also to most answers given here. Whatever we do in mathematics can always be reinterpreted in finitistic terms. Infinite sets can be introduced, but the formalism that is then used involving these infinite sets can be reinterpreted in a way that that does not invoke infinite quantities at all. $\endgroup$ – Count Iblis Jul 26 at 0:10
  • $\begingroup$ Why do you need cellular automata for this? Proofs are already finite... $\endgroup$ – Asaf Karagila Jul 26 at 1:07
  • $\begingroup$ @AsafKaragila Yes, I could also have invoked the formalist argument, but then I'm more into physics than abstract math, using a thought experiment to illustrate something is something I'm used to do. $\endgroup$ – Count Iblis Jul 26 at 5:55

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