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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 binar

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.
- cjhopman 12y ago
  The number of nodes in your tree is countable. The number of paths in your tree is uncountable. In the infinitely subdivided pixel case, a node in the tree corresponds to an x,y position at a particular depth (or, even better, the set of pixels within the square at x, y at that depth). That node, though, does not correspond to a pixel because that area is subdivided into many, many more pixels.
  - Dylan16807 12y ago
    Interesting way of putting it. So if we look at the problem as an infinite-pixel screen with backwards compatibility, then each pixel is a finite node with subnodes for increased resolution. If we look at the problem as an infinite-pixel screen that has a more pure layout without overlapping addressing, then each pixel can only be reached with an infinite path. I for one prefer the backwards compatibility, and think it makes more sense in the context of HD->4k->8k. But that's pretty hilarious if the API for the pixels changes how many you have.

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.
cjhopman 12y ago
The number of nodes in your tree is countable. The number of paths in your tree is uncountable. In the infinitely subdivided pixel case, a node in the tree corresponds to an x,y position at a particular depth (or, even better, the set of pixels within the square at x, y at that depth). That node, though, does not correspond to a pixel because that area is subdivided into many, many more pixels.
- Dylan16807 12y ago
  Interesting way of putting it. So if we look at the problem as an infinite-pixel screen with backwards compatibility, then each pixel is a finite node with subnodes for increased resolution. If we look at the problem as an infinite-pixel screen that has a more pure layout without overlapping addressing, then each pixel can only be reached with an infinite path. I for one prefer the backwards compatibility, and think it makes more sense in the context of HD->4k->8k. But that's pretty hilarious if the API for the pixels changes how many you have.