>Doesn't matter. They are the set of all subsets of a countably infinite set (the set of rationals of the form k / 2^n in the 2d example), and any such set of subsets is uncountable. There is a theorem in set theory to this effect.
They are not the set of all subsets. Not even close. Each pixel is a very particular kind of contiguous subset. You can describe it with one or two rationals (Edit: two or four for 2D).
"Set of all subsets" requires the ability to either include or not include an infinite number of elements independently.
It's obvious that the sets {1,2,3}, {5,6,7}, {9,10,11}, etc. are countable, right? It's a list of subsets of integers but not all subsets.
>How? Your construction above does not show that.
The pixels are defined by their borders, the algorithm for outputting all borders looks like this:
print 0
print 1
d = 2
loop:
n = 1
while n < d:
print (n / d)
n += 2
d *= 2
This constructs all the pixels nice and rationally. Rational numbers are dense, and there are no irrational pixels.
You can define "pixel" however you want, but that has nothing to do with your claim that you can't divide an interval into uncountably many disjoint sets with countably many lines (points).
pdonis and I are not arguing about which way you'd prefer to define the word pixel. But I think pdonis lost the train of what he/she was precisely speaking about in one sentence of the last post.
Edit:
> You can describe it with one or two rationals (Edit: two or four for 2D).
Considering the one-dimensional case for a second: if you define a pixel with two "rationals" (I think you mean two numbers of the form k/2^n here, you're describing (I presume) its left and right boundary, creating an interval (a, b) where a < b. Fine.
How do you describe a pixel with one rational (I think you mean one number of the form k/2^n)? For example, which pixel is described by the rational number 3/8?
Edit: And by the way, the immediate direction I'm walking you into with that edit is that, if for some a < b, (a,b) is one of your pixels, then there are no pixel boundaries in that interval (a,b), unless your pixels overlap.
Well it's the non-overlapping pixels that are the interesting case, so let's stick with those. So no a and b, just a. Then I suppose my first attempt to define a pixel would be the zero-width area stating at a, non-inclusive. So the zero width area starting at 3/8, non-inclusive.
Is that fundamentally flawed? (Probably. Edit 2: I mean seriously I have no idea how to handle ranges that are infinitely small. But they clearly exist in some form, I can look at the set of irrational numbers...)
> one rational (I think you mean one number of the form k/2^n)
Yes, in this context. It doesn't really matter if it's k/2^n or any rational, but for consistency's sake let's stick with k/2^n dividers.
Edit: How about this. Is it meaningful to talk about rounding reals to the nearest rational? If not, why not?
> So the zero width area starting at 3/8, non-inclusive.
So yeah, is that just the interval (3/8, 3/8), i.e. the empty set, or what? If it were inclusive, you'd have [3/8, 3/8], which means the same thing as {3/8} -- it's the set that contains just the value 3/8.
> Is it meaningful to talk about rounding reals to the nearest rational? If not, why not?
It is not. The reason is, there is no nearest rational (if it's not already rational). Suppose x is an irrational number, and p is some rational number. Then p != x. That means |x-p| > 0. That means there's some integer n such that 1/n < |x-p|. If p > x, then x < (p - 1/n) < p. If p < x, then p < (p + 1/n) < x. In each case we've constructed a rational number (p-1/n or p+1/n) that is closer to x than p was.
In general, between any two irrational numbers, there's infinitely many rational numbers, and between any two rational numbers, there's (uncountably) infinitely many irrational numbers.
Pretty much. It was not the most accurate first attempt. I was hoping to find a way to fix it.
> no rounding
Well that argument is the obvious answer yes, but it requires a certain framework.
Kind of like arguing that there is no biggest number, but there is ω...
Anyway I accept that with normal semantics you can't define non-point non-overlapping ranges between dense rationals. I was wrong. Thank you for your time.
They are not the set of all subsets. Not even close. Each pixel is a very particular kind of contiguous subset. You can describe it with one or two rationals (Edit: two or four for 2D).
"Set of all subsets" requires the ability to either include or not include an infinite number of elements independently.
It's obvious that the sets {1,2,3}, {5,6,7}, {9,10,11}, etc. are countable, right? It's a list of subsets of integers but not all subsets.
>How? Your construction above does not show that.
The pixels are defined by their borders, the algorithm for outputting all borders looks like this:
This constructs all the pixels nice and rationally. Rational numbers are dense, and there are no irrational pixels.