The diagonal proof is arguing that there is no bijection from N to {0, 1}∞ , despite the fact that both are infinite. The sense in which the latter is ”bigger” is that there are always elements left over that are covered by N.
Neither. That's my point. Literally any definition of something infinite can always be reduced to a procedure that recursively transforms or observes some prior state. To say that one of these functions can produce more distinct states than another is pointless, because the procedure that produces the most states will always be the one that you ran the most times.
There is nothing observably infinite, since it would take infinite time to observe that any given thing was infinite. The only possible proof of infinity would be a machine that runs infinitely quickly. e.g. https://qntm.org/responsibility
This combination of words seems strange.
Like: proof of ‘zero’ or proof of ‘left’.
All of them have definitions, not proofs.
(Qualitative distinction of different infinity types has a proof though)
Could you come up with or point to such a procedure for R (the reals)?
As I understand the diagonalization argument you can do that for N, but not for R.
This will never reach 2, so it will not generate all real numbers. (Which was what parent was asking for, to recusively generate all R)
This is how natural numbers work in the first place. You're just adding a decimal point to all of the possible places it could go.
When will this reach PI or e or sqrt(2)?
(There are infinitely many numbers that will not be reached by this procedure)
How is this? This is because PI is not actually a number. It's a procedure that generates digits for approximating things about circles.
This is the same with sqrt(2). The sqrt procedure emits digits just as the procedure to find all real numbers does.
You can't "reach" PI for the same reason that a natural number can't reach "f(x) => x + 1". That is, natural numbers aren't procedures or functions.
What about 1/3, 1/7, …? Previously outlines recursive procedure doesn’t generate those.
But yeah, if you deny existance of irrational numbers, and redefine Real:=Rational, then you can generate these “real” numbers recursively and it does follow that all infinities have same cardinality here.
Btw. what is the diagonal of a unit square formed by 4 objects at the corners? I assume it is a rational number. Btw2. If you take that answer and multiply by itself, what do you get?
Important to note: When ggp asked for a recursive procedure to generate real numbers, they wanted that exactly same proceedure would generate all reals (not special procedure for each number)
If we have special procedure for each number, then procedure to generate 1/3 is just 1/3. …of course naively assuming notation of 1/3 is as valid as 0.33333…, and that base 10 is not the only possible base.
1/3 isn't a real, it's a fraction. Fractions can be used to generate reals, and they can be used in algebra along with reals.
0.(3) is also not a real. It's also just representative of a procedure that can generate reals.
Both 1/3 and 0.(3) can still be used in algebra in the same way as before. You don't lose any capability because you can't practically expand 0.(3) to infinite decimal places in the first place.
What about same number expressed in base 3? (I think in base 3 that would be written as 1/31)
And what about number 0.1 in base 3? (Which is equivalent to 1/3 in base 10)
Does it really have an infinite representation? I can't imagine an infinite representation fitting on a page. I'm pretty sure you're representing it as 1/3 or 0.(3). Neither of those representations are infinite. They're only a few characters really.
Are these numbers the same: “0.5 in base10” and “0.1 in base2”?
1 => first step in the procedure 2 => second step ...
That in turn would mean that N and R have the same cardinality. This would be news.
But in all probability we are discussing the wrong thing here. Our difference it's likely at a deeper conceptual level than this.
You can’t generate R this way. This is a consequence of Cantor’s proof.