4 ms·

> It's possible to construct the pixels with an infinite series of rationals, but it's also possible to construct them as a single rational in a countable way.

> It's possible to construct the pixels with an infinite series of rationals, but it's also possible to construct them as a single rational in a countable way. Only if you want overlapping pixels or pixels of varying size. Or pixels that are points, in which case you don't need a convoluted recursive mechanism to define where and what they are.

Dylan16807 12y ago
I don't think they have to overlap just because you defined them rationally. I think it just becomes painful to address pixels if you don't have them overlap. I'm not certain on this point, however. But it's painful to use an infinite string as an address too.
- SamReidHughes 12y ago
  > I don't think they have to overlap just because you defined them rationally. They don't have to overlap, if you just define them as points or sets of points with zero area, or let them have varying size, or define them as weird subsets of the screen that aren't measurable. If your pixels are rectangles, though, they have to overlap or have varying size. Edit: I'm taking countable to mean "countably infinite" since this conversation already assumes we're talking about infinite sets of pixels.
  - Dylan16807 12y ago
    Can they be points but only rational points? How can you tell the difference between dense rational points and dense real points? It seems like you can mark regions of zero size either through countable or uncountable methods...
    - SamReidHughes 12y ago
      > Can they be points but only rational points? They can be, but then the total area of your pixels would be zero. Which is fine, since this isn't a real monitor we're talking about, this is a math monitor. But: One property of our computer monitor that we'd want is that when displaying white, it should have uniform brightness -- the amount of light emitted from a rectangle R should be proportional to area(R). The total amount of light emitted from the monitor should be nonzero, too. Let's say the area of the whole monitor is 1 and so is the luminosity. Your pixels are what's emitting light -- they're point-sources? Suppose p is one of your pixels (or its location). Then let's consider a sequence smaller and smaller squares around p, with side lengths 1/20, 1/21, 1/22, 1/23, ... The luminosity of each of these squares of sidelength 1/n must be the square's area, 1/n^2, because the brightness is uniform. So the pixel p must have zero luminosity, because if it had nonzero luminosity, then some very small rectangle around it would be emitting more light than it should be, and the monitor would not have uniform brightness. From this argument it seems that each pixel's luminosity is zero. And then you can enumerate the pixels, since they form a countable set: p_1, p_2, p_3, .... The luminosity of the whole screen is a countable sum, which is sum[i = 1 to infinity](luminosity(p_i)) = sum[i = 1 to infinity](0) = 0. So I don't think the idea of dividing a math-monitor into countably many pixels, each with zero area, works. It's OK that the total area of the pixels is zero, but you have to accept non-uniform luminosity or say that the pixels aren't point sources, they "bleed". However, if you have uncountably many pixels, then it's okay if each "pixel" (whatever) has zero luminosity. You only get nonzero luminosity when talking about sets of pixels with nonzero area -- their luminosity on the uniformly bright screen is proportional to the set's area. That's perfectly compatible with intuition -- the same way it's intuitive that the area of a unit square is 1, even though it's an uncountable union of sets having one point each, each of whose area is zero.
      - Dylan16807 12y ago
        Well the other discussion of non-overlapping pixels is definitely teaching me a lot, but I think this settles what the chosen screen should do. It should use overlapped addressing so you can not only put a picture on it with a countably infinite computer, you can put a picture on it with a finite computer.
        SamReidHughes 12y ago
        Agreed.

Dylan16807 12y ago

I don't think they have to overlap just because you defined them rationally. I think it just becomes painful to address pixels if you don't have them overlap. I'm not certain on this point, however. But it's painful to use an infinite string as an address too.

SamReidHughes 12y ago
> I don't think they have to overlap just because you defined them rationally. They don't have to overlap, if you just define them as points or sets of points with zero area, or let them have varying size, or define them as weird subsets of the screen that aren't measurable. If your pixels are rectangles, though, they have to overlap or have varying size. Edit: I'm taking countable to mean "countably infinite" since this conversation already assumes we're talking about infinite sets of pixels.
- Dylan16807 12y ago
  Can they be points but only rational points? How can you tell the difference between dense rational points and dense real points? It seems like you can mark regions of zero size either through countable or uncountable methods...
  - SamReidHughes 12y ago
    > Can they be points but only rational points? They can be, but then the total area of your pixels would be zero. Which is fine, since this isn't a real monitor we're talking about, this is a math monitor. But: One property of our computer monitor that we'd want is that when displaying white, it should have uniform brightness -- the amount of light emitted from a rectangle R should be proportional to area(R). The total amount of light emitted from the monitor should be nonzero, too. Let's say the area of the whole monitor is 1 and so is the luminosity. Your pixels are what's emitting light -- they're point-sources? Suppose p is one of your pixels (or its location). Then let's consider a sequence smaller and smaller squares around p, with side lengths 1/20, 1/21, 1/22, 1/23, ... The luminosity of each of these squares of sidelength 1/n must be the square's area, 1/n^2, because the brightness is uniform. So the pixel p must have zero luminosity, because if it had nonzero luminosity, then some very small rectangle around it would be emitting more light than it should be, and the monitor would not have uniform brightness. From this argument it seems that each pixel's luminosity is zero. And then you can enumerate the pixels, since they form a countable set: p_1, p_2, p_3, .... The luminosity of the whole screen is a countable sum, which is sum[i = 1 to infinity](luminosity(p_i)) = sum[i = 1 to infinity](0) = 0. So I don't think the idea of dividing a math-monitor into countably many pixels, each with zero area, works. It's OK that the total area of the pixels is zero, but you have to accept non-uniform luminosity or say that the pixels aren't point sources, they "bleed". However, if you have uncountably many pixels, then it's okay if each "pixel" (whatever) has zero luminosity. You only get nonzero luminosity when talking about sets of pixels with nonzero area -- their luminosity on the uniformly bright screen is proportional to the set's area. That's perfectly compatible with intuition -- the same way it's intuitive that the area of a unit square is 1, even though it's an uncountable union of sets having one point each, each of whose area is zero.
    - Dylan16807 12y ago
      Well the other discussion of non-overlapping pixels is definitely teaching me a lot, but I think this settles what the chosen screen should do. It should use overlapped addressing so you can not only put a picture on it with a countably infinite computer, you can put a picture on it with a finite computer.
      - SamReidHughes 12y ago
        Agreed.