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I think the followup at the end is wrong, unless I'm misreading. Originally all lines at coordinates k/2^n were drawn and the points not on those lines form an

I think the followup at the end is wrong, unless I'm misreading. Originally all lines at coordinates k/2^n were drawn and the points not on those lines form an uncountable set (also, between any two points a line we drew is separating them). Then at the end the intermediate representations are counted, which is a completely different thing. An infinite binary tree has countably many nodes and uncountably many paths, yes. The article was fine the way it was.

sukilot 12y ago
> points not on those lines form an uncountable set What do you mean? A countable set of lines cuts the screen into a nonsensical "set of regions", you can't do it. The infinite pixel scanline is the set of all finite paths (it's all *terminating binary strings), not the set of all infinite paths. Since the set of cutlines is dense, you can actually name all regions, so the only sensible way to assign pixels is to assign a pixel to every line, not to the space between lines (since there is no nonzero space between lines)
- SamReidHughes 12y ago
  >A countable set of lines cuts the screen into a nonsensical "set of regions", you can't do it. No, it (the set of lines in question) cuts the screen into a set of points (not regions) -- all the points without a coordinate of the form k/2^n.
pdonis 12y ago
> I think the followup at the end is wrong I think so too, but maybe not for the same reason you do. The original argument of the article is that there is a one-to-one mapping between the set produced as the end result of an infinite number of "pixel division" operations, and the set of infinite binary strings. Since the latter set is uncountable (by Cantor's diagonal argument), the former must be as well. This argument appears to me to be correct. The followup retracts the above argument, basically because of this counter-argument: "just as there’s no way to follow 0.333... until you reach exactly one-third, there’s no way to carve a pixel until you get to the exact point represented by a certain infinite binary string" However, this counter-argument is incorrect, because it leaves out a key qualifier. There is no way to "follow 0.333... until you reach exactly one-third" in a finite number of steps. If you take an infinite number of steps, you do reach exactly one-third: that is what the statement "one-third is the limit of the repeating decimal 0.333..." means. True, you can't explicitly take an infinite number of steps, but you can calculate, mathematically, what the result would be if you did; that is how we know what the limit points are of sequences like 0.333... Similarly, "there's no way to carve a pixel until you get to the exact point represented by a certain infinite binary string" in a finite number of steps. But you can do it in an infinite number of steps, and the article explicitly postulates exactly that--an infinite number of pixel divisions. And even though we can't explicitly do an infinite number of pixel divisions, we can still calculate, mathematically, what the result would be if we did. That result is the set of infinite binary strings; that set is the limit point of the sequence of pixel divisions, as the number of divisions goes to infinity, just as one-third is the limit point of the sequence 0.333... as the number of 3's goes to infinity.
- Dylan16807 12y ago
  Yes, that counter was confusing and kind of wrong. But the followup is mostly correct. It is possible to carve pixels into uncountably more pixels, but you need uncountably many lines to do it. Division by countably many lines, aka rational division, does not increase cardinality. The cardinality of rational numbers is the same as the cardinality of integers.
  - SamReidHughes 12y ago
    In your post what is your definition of pixel?
  - pdonis 12y ago
    > Division by countably many lines, aka rational division, does not increase cardinality. I understand that this sounds plausible, but I'm not sure it's true. The real numbers, which are uncountable, can be defined as the set of all limit points of countably infinite series of rational numbers. So the set of countably infinite sums is uncountable. Similarly, I think the set of countably infinite divisions is uncountable. Here's another way of looking at it: the number of pixels obtained by division by countably many lines is equal to the number of subsets of the rationals--which is uncountable.
    - Dylan16807 12y ago
      But the pixels aren't arbitrary subsets. Look at it in a single dimension for a moment. p | p | p | p | p | p There is a dead-obvious 1:1 correspondence between pixels and lines. You could even label each pixel by its bordering line, and then you're literally counting rationals. 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.
      - pdonis 12y ago
        > the pixels aren't arbitrary subsets. 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. > There is a dead-obvious 1:1 correspondence between pixels and lines Yes, but that in itself doesn't show the cardinality of the set of pixels/lines. Nor does it show that the cardinality of the set of pixels/lines when an infinite number of divisions have been made, is the same as the cardinality of the set of pixels/lines when only a finite number of divisions have been made. Assuming that they must be the same is the equivalent of assuming that the cardinality of the reals is the same as the cardinality of the rationals, which is obviously false. > it's also possible to construct them as a single rational in a countable way How? Your construction above does not show that.
Dylan16807 12y ago
The pixels are not points. The pixels are rational subdivisions of the screen. Rational numbers have the same cardinality as integers. The article clearly shows how you can give a unique number to every pixel, no matter how small. You can't just ignore that because you have another argument, you have to show how one of two incompatible arguments is wrong. The article shows how its original argument was wrong. Each specific pixel has a finite string describing it, not an infinite string. Therefore an argument based on infinite strings is wrong.
- cjhopman 12y ago
  If there are countably many rounds of division, then there is a vertical (and horizontal) "division" for each path (from the root) in an infinite complete binary tree. The number of paths in an infinite complete binary tree is uncountable, just like the infinitely divided screen is (each "pixel" corresponds to a path in an infinite complete 4-ary tree).
  - Dylan16807 12y ago
    Reified pixels can be described in finite terms, and can be counted. You can't just say you're at an infinite level of division, and ask for the name of the pixel you're at. There is a simple bijection between pixels and a subset of rational numbers. You wouldn't say rational numbers are uncountable, would you? I can make the same sort of tree argument about rational numbers. My set of rational numbers are constructed by multiplying numerator and denominator by ten and then adding 0-9 to the numerator. At each level they can represent any finite string of digits. Surely this is the same cardinality as the pixels. There is no way to jump to an "infinitely deep" number by this construction. You have to go a step at a time, and then they're clearly countable. In the way that real numbers aren't, because there's no way to generate all real numbers in any order. TL;DR Edit: Counting finite strings for an infinite time does not actually produce infinite strings. Each pixel is finite.
    - SamReidHughes 12y ago
      The word "reified pixel" is only being used by you so if you want to talk about it you need to define what that means.

sukilot 12y ago

> points not on those lines form an uncountable set What do you mean? A countable set of lines cuts the screen into a nonsensical "set of regions", you can't do it. The infinite pixel scanline is the set of all finite paths (it's all *terminating binary strings), not the set of all infinite paths. Since the set of cutlines is dense, you can actually name all regions, so the only sensible way to assign pixels is to assign a pixel to every line, not to the space between lines (since there is no nonzero space between lines)

SamReidHughes 12y ago
>A countable set of lines cuts the screen into a nonsensical "set of regions", you can't do it. No, it (the set of lines in question) cuts the screen into a set of points (not regions) -- all the points without a coordinate of the form k/2^n.

pdonis 12y ago

> I think the followup at the end is wrong I think so too, but maybe not for the same reason you do. The original argument of the article is that there is a one-to-one mapping between the set produced as the end result of an infinite number of "pixel division" operations, and the set of infinite binary strings. Since the latter set is uncountable (by Cantor's diagonal argument), the former must be as well. This argument appears to me to be correct. The followup retracts the above argument, basically because of this counter-argument: "just as there’s no way to follow 0.333... until you reach exactly one-third, there’s no way to carve a pixel until you get to the exact point represented by a certain infinite binary string" However, this counter-argument is incorrect, because it leaves out a key qualifier. There is no way to "follow 0.333... until you reach exactly one-third" in a finite number of steps. If you take an infinite number of steps, you do reach exactly one-third: that is what the statement "one-third is the limit of the repeating decimal 0.333..." means. True, you can't explicitly take an infinite number of steps, but you can calculate, mathematically, what the result would be if you did; that is how we know what the limit points are of sequences like 0.333... Similarly, "there's no way to carve a pixel until you get to the exact point represented by a certain infinite binary string" in a finite number of steps. But you can do it in an infinite number of steps, and the article explicitly postulates exactly that--an infinite number of pixel divisions. And even though we can't explicitly do an infinite number of pixel divisions, we can still calculate, mathematically, what the result would be if we did. That result is the set of infinite binary strings; that set is the limit point of the sequence of pixel divisions, as the number of divisions goes to infinity, just as one-third is the limit point of the sequence 0.333... as the number of 3's goes to infinity.

Dylan16807 12y ago
Yes, that counter was confusing and kind of wrong. But the followup is mostly correct. It is possible to carve pixels into uncountably more pixels, but you need uncountably many lines to do it. Division by countably many lines, aka rational division, does not increase cardinality. The cardinality of rational numbers is the same as the cardinality of integers.
- SamReidHughes 12y ago
  In your post what is your definition of pixel?
- pdonis 12y ago
  > Division by countably many lines, aka rational division, does not increase cardinality. I understand that this sounds plausible, but I'm not sure it's true. The real numbers, which are uncountable, can be defined as the set of all limit points of countably infinite series of rational numbers. So the set of countably infinite sums is uncountable. Similarly, I think the set of countably infinite divisions is uncountable. Here's another way of looking at it: the number of pixels obtained by division by countably many lines is equal to the number of subsets of the rationals--which is uncountable.
  - Dylan16807 12y ago
    But the pixels aren't arbitrary subsets. Look at it in a single dimension for a moment. p | p | p | p | p | p There is a dead-obvious 1:1 correspondence between pixels and lines. You could even label each pixel by its bordering line, and then you're literally counting rationals. 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.
    - pdonis 12y ago
      > the pixels aren't arbitrary subsets. 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. > There is a dead-obvious 1:1 correspondence between pixels and lines Yes, but that in itself doesn't show the cardinality of the set of pixels/lines. Nor does it show that the cardinality of the set of pixels/lines when an infinite number of divisions have been made, is the same as the cardinality of the set of pixels/lines when only a finite number of divisions have been made. Assuming that they must be the same is the equivalent of assuming that the cardinality of the reals is the same as the cardinality of the rationals, which is obviously false. > it's also possible to construct them as a single rational in a countable way How? Your construction above does not show that.

Dylan16807 12y ago

The pixels are not points. The pixels are rational subdivisions of the screen. Rational numbers have the same cardinality as integers. The article clearly shows how you can give a unique number to every pixel, no matter how small. You can't just ignore that because you have another argument, you have to show how one of two incompatible arguments is wrong. The article shows how its original argument was wrong. Each specific pixel has a finite string describing it, not an infinite string. Therefore an argument based on infinite strings is wrong.

cjhopman 12y ago
If there are countably many rounds of division, then there is a vertical (and horizontal) "division" for each path (from the root) in an infinite complete binary tree. The number of paths in an infinite complete binary tree is uncountable, just like the infinitely divided screen is (each "pixel" corresponds to a path in an infinite complete 4-ary tree).
- Dylan16807 12y ago
  Reified pixels can be described in finite terms, and can be counted. You can't just say you're at an infinite level of division, and ask for the name of the pixel you're at. There is a simple bijection between pixels and a subset of rational numbers. You wouldn't say rational numbers are uncountable, would you? I can make the same sort of tree argument about rational numbers. My set of rational numbers are constructed by multiplying numerator and denominator by ten and then adding 0-9 to the numerator. At each level they can represent any finite string of digits. Surely this is the same cardinality as the pixels. There is no way to jump to an "infinitely deep" number by this construction. You have to go a step at a time, and then they're clearly countable. In the way that real numbers aren't, because there's no way to generate all real numbers in any order. TL;DR Edit: Counting finite strings for an infinite time does not actually produce infinite strings. Each pixel is finite.
  - SamReidHughes 12y ago
    The word "reified pixel" is only being used by you so if you want to talk about it you need to define what that means.