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What... ouch. This is not math, this is... I don't even. Read the post closely. The author introduces a set S of squarey curves approximating a circle. This se

What... ouch. This is not math, this is... I don't even. Read the post closely. The author introduces a set S of squarey curves approximating a circle. This set has an obvious correspondence with the natural numbers: there is curve #1, curve #2, etc. Then the author defines a function f: S->S that takes curve #n to curve #n+1. Wait, that doesn't sound right! The function f has no interesting structure whatsoever, it's exactly equivalent to defining f(n)=n+1 on the naturals. Of course, taking the "limit" of a function f: S->S makes no sense at all. What would make sense is taking the limit of a certain function N->C, where N is the naturals and C is the set of all curves on the plane. That is, the limit of a sequence of curves (not of a function from curves to curves as the OP tried to say). To talk about such limits, you need to define what it means for a sequence of curves to converge - a "topology" on C. There are many ways to do that, some more outlandish than others. One way is pointwise convergence: assume a parameterization t->C_i(t) on each curves in the sequence, and require that C_i(t) converges co C_lim(t) for each t separately. Now, pretty much any reasonable notion of "convergence" on the space of all curves has to imply pointwise convergence. That is, pointwise convergence is a very "weak" notion of convergence: if we have a sequence of curves that has a limit in some reasonable sense, then it had better coincide with the pointwise limit, dontcha think? And here we come to the second facepalm moment in the post. Under pointwise convergence, the sequence of squarey curves under discussion does not converge to some "right angled fractal beast". It converges to the circle. As n grows, every point on S_n comes closer and closer to some point on the circle. Ain't nothing more to it. Now the correct explanation for the original puzzle. Pointwise convergence of curves doesn't imply that their lengths converge to the limit's length. Hell, we don't even need 2-dimensional space to show that! A simple straight line will do. Imagine a human traveling a straight road of 1km length in this fashion: he takes two steps forward, then one step back, then repeats. In the end he will have traveled about 3km instead of 1km. As we make the human and his steps tinier and tinier, his movement looks smoother and smoother to an external observer, but he still travels 3km in total instead of 1. Or maybe (going back to the 2D space) the human could take a step left, then forward, then right; this would make his path look like a fine comb that approximates the straight line more and more closely, but it's always 3x longer. Something like this is happening in the original puzzle. Finishing touch: there are notions of convergence where it's true that the length of the curves in the sequence always converges to the length of the limit. One such notion says that the direction of travel (velocity vector) must also pointwise converge to the velocity vector of the limit curve. Under this definition of convergence, the original sequence of curves does not converge, because it makes too many sharp turns.

neilk 16y ago
On a less trollish note, I had always been puzzled by the related question you discuss: why Manhattan Distance does not converge to the Euclidean distance. I'm pretty uneducated about math, so please let me know what trap I'm falling into. Your answer isn't yet helping me because you just note that Manhattan Distance remains the same no matter how many twists and turns there are. Yes, this is the premise of the question! But turn up the twistiness to infinity. Now the Manhattan Distance line is identical to the hypotenuse. There is no point on the Manhattan distance line isn't also on the hypotenuse and vice versa. They both have an infinite number of points, of the same aleph-number, I think. You mention a hypothetical path that doubled back on itself -- it is easy to see why that would result in a different answer. But the Manhattan Distance line is not doubling back on itself. Every point is a step towards the goal, and it doesn't cross itself. All of its deviations from the simple hypotenuse are infinitely small. So why are the distances still different?
- simcop2387 16y ago
  The reason it won't converge is that a very large number of the points will ALWAYS lie off of the line. While you might be able to make something that looks like a line on a computer screen when you make it smaller and smaller segments of manhattan distance. it will in fact only ever be able to intersect the real line (they are infinitely thin remember) at N+1 points at most (where N is the number of turns you make when following it along the grid in manhattan distance). because of this you will always have a very large portion of your approximation lying off of the line.
  - ewjordan 16y ago
    Put another way, the right angles approximation as N->infinity puts a countable infinity of points exactly on the line. The true line has an uncountable infinity of points on the line, and the difference between an uncountable infinity and a countable one is an uncountable infinity, so we're still way the hell off.
    - RiderOfGiraffes 16y ago

neilk 16y ago

On a less trollish note, I had always been puzzled by the related question you discuss: why Manhattan Distance does not converge to the Euclidean distance. I'm pretty uneducated about math, so please let me know what trap I'm falling into. Your answer isn't yet helping me because you just note that Manhattan Distance remains the same no matter how many twists and turns there are. Yes, this is the premise of the question! But turn up the twistiness to infinity. Now the Manhattan Distance line is identical to the hypotenuse. There is no point on the Manhattan distance line isn't also on the hypotenuse and vice versa. They both have an infinite number of points, of the same aleph-number, I think. You mention a hypothetical path that doubled back on itself -- it is easy to see why that would result in a different answer. But the Manhattan Distance line is not doubling back on itself. Every point is a step towards the goal, and it doesn't cross itself. All of its deviations from the simple hypotenuse are infinitely small. So why are the distances still different?

simcop2387 16y ago
The reason it won't converge is that a very large number of the points will ALWAYS lie off of the line. While you might be able to make something that looks like a line on a computer screen when you make it smaller and smaller segments of manhattan distance. it will in fact only ever be able to intersect the real line (they are infinitely thin remember) at N+1 points at most (where N is the number of turns you make when following it along the grid in manhattan distance). because of this you will always have a very large portion of your approximation lying off of the line.
- ewjordan 16y ago
  Put another way, the right angles approximation as N->infinity puts a countable infinity of points exactly on the line. The true line has an uncountable infinity of points on the line, and the difference between an uncountable infinity and a countable one is an uncountable infinity, so we're still way the hell off.
  - RiderOfGiraffes 16y ago