Pi

Assume it converges to the circle. In every step you get 4 more points that intersect with the square, so after infinte iterations you would still have countable intersections, but this is a contradiction to the fact that R is uncountable.

This would be so easy to explain with Landau symbols. Why does nobody use Landau symbols?

use them and explain, then

I don't understand what this math thing is. Like... why math?

Here's an explanation, brainlets.
Imagine a different example, where instead you placed points around a circle and made tiny tiny triangles with roughly negligible area. You then take these triangles, and stretch them out far away from the circumference of the circle, but at the same time adjust the closeness of the base points such that the area of the triangles remains constant as they stretch out (similar to the dirac delta function). You would end up with a shape that has a massive perimeter, but almost identical area to before. This is why this doesn't work. Pi is by definition, the ratio of circumference to diameter for a circle. This is not a circle.

You have a sequence of paths [math]\gamma_n(t)[/math] representing the perimeter with each time corners removed. E.g [math]\gamma_1(t)[/math] is a square and [math]\gamma_2(t)[/math] and [math]\gamma_3(t)[/math] represents the perimeters as in the pictures 3 and 4. Now for each natural number [math]n[/math], the length of [math]\gamma_n(t)[/math] is [math]\int_0^1|\gamma_n'(t)| dt=4[/math]. But if we let [math]n[/math] approach to infinity, we get a circle [math]\gamma_(t)[/math] which lenght is [math]\int_0^1|\gamma'(t)| dt=\pi[/math].

so in this case [math] \lim_{n \rightarrow \infty} \int_0^1 |\gamma_n'(t)|dt= 4 \neq \int_0^1 |\frac{d}{dt} \lim_{n \rightarrow \infty} \gamma_n(t)|dt=\pi[/math]

Correct answer.