by NoahZuniga | 2025-08-29 21:55:21+00:00
> 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!