▲ | Nevermark 4 days ago | |||||||||||||||||||||||||
Thanks for that very clear argument. But it doesn't hold up. (He says with an overconfident flourish!)
(I love the dead simple example.)And we can prove it is continuous over constructible reals, because in that case,
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:
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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