Lets take some un-constructible real number and call it r.

Now think of the function f(x)=0 if x<r and 1 if x>r.

Edit: I originally said this is a function over the un-constructible reals. That was wrong!

This is a function over the constructible reals. What I mean with that is that it only has a value for constructible real numbers, because r isn't constructible, f doesn't have a value at r. That's why I haven't defined what f(r) is.

While this function exists (or something like it does, I can prove this fact if you like), this is not a construction because of course r is un-constructible

This function is continuous.

Let's prove this with the definition of continuity. We say that a function is continuous at a point x if lim a->x f(x) = f(x). We say that a function is continuous if it is continuous at all points.

The only point where this function could hypothetically be discontinuous is at the boundary point r, but actually this boundary point doesn't exist for this function (because f(r) isn't defined), so this function is continuous at every point, and is thus a continuous function.

Edit: forgot to add, but based on this concept you can then show that almost all continuous functions over the constructible reals have these jumps at infinitely many places, and that means that you can't hope to define a way to integrate continuous functions.

I appreciate that this is a pretty complex topic, so that I probably haven't been that clear (or made a mistake), so I value any and all comments.

This argument is somewhat adapted from an argument about why we do calculus with the reals and not the rationals. See also this video: https://www.youtube.com/watch?v=vV7ZuouUSfs

▲ Nevermark 4 days ago | parent [-]

Thanks for that very clear argument.

But it doesn't hold up. (He says with an overconfident flourish!)

  So we have: f(x) = 0, over constructible reals

(I love the dead simple example.)

And we can prove it is continuous over constructible reals, because in that case,

  The limit of f(x), as x —> c, trends to f(c) = 0

We could then postulate un-constructible numbers are something that:

(1) Exist. This requires postulating some kind of infinite information oracle for each independent un-constructible number.

I have questions, but ... for now ok.

(2) That these un-constructible numbers are somehow "in" the constructible real line, even though they cannot be "on" the constructible real line, in a way that is coherent. Not all numbers are, i.e. imaginary numbers are not.

I have questions, but .. ok for now.

(3) That by defining f(r) = 1, for unconstructible r, we can create a case where f(x), as x —> r, does not trend to f(r)

I will concede this third postulate whole heartedly!

But that can't and doesn't invalidate our original constructible continuity proof.

We had to not only generalize "number", but redefine "f".

And I don't think it says much or anything about un-constructible numbers either.

For instance: Let's call regular numbers "blue reals", and define "red reals" such that for every blue real x, there is a red real red(x), that is exactly to the right (i.e. positive) of x. In such as way that they are ordered, but no number can come between them.

So (red(x)-x) = red(0), for all x, and there can never be a "blue-red" number smaller than red(0) (other than zero).

Then we can take our original "blue" proof, define f(red(1)) = 1, and declare we have broken continuity over blue-red reals.

So this breaking of continuity is a trivial trick, and it has no dependence on un-constructibility.

What we have really done, is define a new class of number and redefine "f" to get a motivated result.

We could just as easily have simply redefined f, to be the same but include f(1) = 1. If we get to redefine f, we get to redefine f.

So un-constructibility isn't needed to prove continuity (our original practical proof holds).

It can't "break" continuity either. Given redefining "f" was both a necessary and sufficient condition to do that. Nor does doing so shed any special light on unconstructibility or continuity.

Thoughts? :)

▲ NoahZuniga 4 days ago | parent [-]

> Thoughts? :)

You make some very good points! I have to commend your mathematical reasoning.

Because I've seen some more formal math, I have some pretty good answers to your points. But I want to emphasize again that you bring up some excellent concerns.

> We could then postulate un-constructible numbers are something that:

> (1) Exist

Very unfortunately, there isn't really a rigorous way to define a function that tells you if a real number is constructible, you can get pretty close! However, we don't need something like this.

There are only countably many constructible numbers. This is because that for every constructible number, there is at least one finite description. However, (because of Cantor's diagonal argument) we know there infinitely more real numbers than constructible real numbers. So there must be a large amount of un-constructible real numbers.

> (2) That these un-constructible numbers are somehow "in" the constructible number line

There's a pretty rigorous way to assert this. Lets say that r is an un-constructible number. Like all real numbers, it has a decimal expansion. Lets say that it starts:

0.1637289458946...

Now I can compare constructible numbers and see if they are larger or smaller. Lets consider x = Pi-3 = 0.1415...

We'll look at it digit by digit. 0 = 0, so we don't know yet which one is larger. 1 = 1, so we still don't know. 4 < 6, so now we know that x=Pi-3 actually has to be smaller than r. This process finishes in finite time for any constructible number, because if there is no decimal place where they differ, they are the same number (which is impossible because x is constructible and r isn't).

Because I can compare the size of un-constructible numbers, the un-constructible numbers are "in between" the constructible numbers on the constructible number line. This is similar to how the irrational numbers (or if you prefer, the irrational constructible numbers) are "in between" the rational numbers on the rational number line.

> (3) That by defining f(1) = 1 (for instance)

I'm not sure exactly what your concern is, but I get the feeling that it will (hopefully, at least somewhat) be addressed by the next bit about "red" and "blue" numbers.

> Then I can take our original "blue" proof, define f(red(1)) = 1, and declare I have broken continuity, but again, I am just making up an arbitrary new class of numbers, and using it break continuity over the new class.

For a very subtle reason, this construction doesn't work. To see why, we'll have to take a look at the definition of the limit of a function.

When we say: lim a->x f(a) = L, we mean "if for every number ε > 0 , there exists a number δ > 0 such that whenever 0 < |a-x| < ε, we have that L - ε < a < L + ε.

(I'm happy to elaborate on this definition)

Now we can see why piecewise function I gave is continuous. let's consider x and f(x). (This is the f from my previous comment.) We know that x!=r, because x is constructible and r isn't), so we know that |x-r| is some positive real number. Now we can take δ=0.5|x-r| (or if you prefer, a constructible real number smaller than 0.5|x-r|).

Diagram of f:

    y=1                         ___________________________
                           
    y=0  _______________________

             ^     ^           ^   
             |     |           |   
             x    x+δ          r

Now because f is constant on x-δ to x+δ, for any delta this epsilon works and we've shown the limit is 0=f(x) if x is left to r. (If x is right to r, you use the same argument and show the limit is 1=f(x)).

This works because for any x, we can "zoom in" close enough to x that r is out of view and f is just a horizontal line. This is also the key difference between this and your color construction (at least what I think you mean).

My understanding is you define g (lets use a different function) as: g(x) = 0 if x < red(1) and 1 if x > red(1) Because red(1) and blue(1) are right next to each other, the function is discontinuous at g(blue(1)). Because for any ε, there exist some a>red(1), because blue(1+ε)>red(1) (by definition). So you see g(x)=0 and g(x)=1 no matter how much you zoom into the function at x=1. Your function is discontinuous, and continuity hasn't been broken!

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Nevermark 4 days ago | parent [-]

Ah! I get the point.

Because the discontinuity occurs at an un-constructible r, the constructible limit "process" of x approaching r, never includes r. So f(x->r) encounters a limit of 0. And 1 from the other side.

That does leaves the validity of defining continuity of constructible functions over constructible reals intact.

It is only a problem if un-constructible numbers are "brought" in.

Previously you stated:

> Basically all of calculus needs all the reals because if you don't you get some really pathological results.

This is then what I don't understand.

Why would un-constructible reals be needed?

I would agree that all constructible reals are needed (after all, constructible functions as a class are by definition, sensitive to constructible reals as a class).

But why would un-constructible reals be necessary for anything?

At best I can see them as a kind of contrivance that shortens some proofs, that could be made without them with more careful reasoning.

But ... ?

> However, (because of Cantor's diagonal argument) we know there infinitely more real numbers than constructible real numbers.

Cantor uses sleight of hand out of the gate. From out of nowhere, he postulates that there can be infinite digit numbers, without finite description.

That is really interesting thing to suggest.

Because until then, numbers were built up in steps, from the concepts of increment, repetition, and orthogonal units (to post-formalize the actual progression of informal to formal).

Constructibility (in practice and theory) was inherent in what it meant to "have" a number.

A number is a relationship. A relationship isn't a relationship because I can pick a name "r" and say it is a relationship.

But suddenly Cantor talks about infinite digit numbers with no expression. No defined relation. And then an (attempted) list of all of them.

So constructibility of numbers is abandoned as a precondition for his arguments. Before any point of diagonal incompleteness is made.

It gets worse. Even if we had access to magic oracles that will generate any number of un-constructible reals we desire, with any number of digits supplied, so we can add, subtract and compare them: The process would remain indistinguishable from the same process in which we are actually being given constructible numbers.

My conviction (very open to being shot down of course!) is that un-constructible reals are an interesting concept, and interesting to reason about as mathematical puzzles, perhaps good exercises for inspiring new proof tactics, but in direct relation to anything we do with "actual" real numbers, they are unnecessary.

Also, any actual reasoning about infinite information structures and higher order infinities is going to itself be isomorphic to a finite (or countably infinite) system with 1-to-1 behaviors not interpreted as about un-constructible things. Because anything we do, even reasoning about the un-constructible, remains a constructible domain.

If I am wrong, I would be very interested to understand why.

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NoahZuniga 3 days ago | parent [-]

> That does leaves the validity of defining continuity of constructible functions over constructible reals intact.

What I think you're getting at is that even though my function f breaks what we think of as continuity of functions over the constructible reals, f is clearly un-constructible. So if we only do analysis using constructible functions, all is well.

I was thinking about how this works and trying to think of an example proof that doesn't work with only constructible reals, but actually the same proof basically works, so I'll just share that instead:

Intermediate value theorem: if f is continuous, f(a) is negative and f(b) is positive, then there is some c such that f(c)=0.

Proof for real functions: Define S = { x ∈ [a,b] : f(x) ≤ 0 }.

S is nonempty because a ∈ S (f(a) < 0).

S is bounded above by b, so S has a least upper bound c = sup S with c ∈ [a,b].

We claim f(c) = 0.

Suppose first that f(c) > 0. By continuity at c there is δ > 0 such that for all x with |x−c| < δ we have |f(x)−f(c)| < f(c). In particular for such x we get f(x) > 0 (since f(c) − |f(x)−f(c)| > 0). But then every x in (c−δ, c+δ)∩[a,b] is not in S, so there is no point of S greater than or equal to c−δ/2. That contradicts c being the least upper bound of S because then c−δ/2 would be an upper bound smaller than c. Hence f(c) ≤ 0.

Now suppose f(c) < 0. By continuity at c there is ε > 0 such that for all x with |x−c| < ε we have |f(x)−f(c)| < −f(c) (note −f(c) > 0). Then for such x we get f(x) < 0, so every x in (c, c+ε)∩[a,b] also satisfies f(x) ≤ 0 and hence belongs to S. But that gives points of S strictly greater than c, contradicting that c is an upper bound of S. Thus f(c) < 0 is impossible.

If we instead talk about constructible functions, note that f is constructible, so S is constructible, so c = sup S is constructible. We know that c is in the domain of f, and using the proof above we can show f(c)=0.

So maybe if we limit ourselves to constructible functions analysis works out. There are still two reasons why you might not want to do this. Adding a line at the end of every proof explaining why all the numbers you're talking about are constructible feels unnecessary when you can just talk about the reals. Secondly, (as far as I understand) its impossible to actually formalize our idea of constructible.

	▲	Nevermark 3 days ago \| parent [-]
		I think we can formalize constructible structures as any mathematical structure which can be uniquely defined by at least one finite sequence of symbols. As far as not wanting to pedantically refer to "constructible reals", I agree that doesn't sound fun. The better solution would be having a clear common pithy term for "unconstructible reals", for: 1. Teaching math related to numbers, until unconstructible reals or other structures had any relevance. I.e. most people never, ever. 2. For talking about algorithms, physics and other constructible structures, where the term reals is pervasivably used to mean constructible reals. 3. Most students get introduced to the fact that the cardinality of reals is greater than the cardinality of integers. But would be surprised, and get more use, out of knowing that the cardinality of instantiatable numbers (the ones they could define, calculate, measure or apply in virtually every situation but highly abstract math games), is EXACTLY the same as the integers. Un-constructible structures are an interesting but exotic concept that shouldn't be riding around sereptitiously in common vocabulary. A fair number of responses here involved confusion about what "reals" covers. And this is HN.