general topology - Constructing a circle from a square

I have seen a [picture like this] several times:

featuring a "troll proof" that $\pi=4$. Obviously the construction does not yield a circle, starting from a square, but how to rigorously and demonstratively prove it?

For reference, we start with a circle inscribed in a square with side length 1. A step consists of reflecting each corner of figure $F_i$ so that it lies precisely on the circle and yielding figure $F_{i+1}$. $F_0$ is the square with side length 1. After infinitely many steps we have a figure $F_\infty$. Prove that it isn't a circle.

Possible ways of thinking:

1. Since the perimeter of figure $F_i$ indeed does not change during a step, it is invariant. Since it does not equal the perimeter of the circle, $\pi\neq4$, it cannot be a circle.

While it seems to work, I do not find this proof demonstrative enough - it does not show why $F_\infty$ which looks very much like a circle to us, is not one.

2. Consider one corner of the square $F_0$. Let $t$ be a coordinate along the edge of this corner, $0 \leq t \leq 1$ and $t=0, t=1$ being the points of tangency for this corner of $F_0$ and the circle.
   By construction, all points $t \in A=\{ \frac{n}{2^m} | (n,m\in \mathbb{N}) \& (n<2^m)\}$ of $F_\infty$ lie on the circle. I think it can be shown that the rest of the points, $\bar{A}=[0;1] \backslash A$, lie in an $\varepsilon$-neighbourhood $U$ of the circle. I also think that in the limit $\varepsilon \to 0$, points $t\in\bar{A}$ also lie on the circle. Am I wrong in thinking this? Can we get a contradiction from this line of thought?

Any other elucidating proofs and thoughts are also welcomed, of course.

Answer

You have rigorously defined $F_i$, but how do you define $F_\infty$? You cannot say: "after infinitely many steps...".

In this case you could define $F_\infty = \bigcap_i F_i$ (i.e. the intersection of all $F_i$), since $F_i$ is a decreasing sequence this is a good notion of limit. Notice however that $F_\infty$ is a circle! But this does not mean that the perimeter of $F_i$ should converge to the perimeter of $F_\infty$.

You could also choose a metric on subsets of the plane to define some sort of convergence $F_i \to F_\infty$ as $i\to \infty$. In any case, if you choose any good metric you find that either $F_\infty$ is the circle or that the sequence does not converge.

The point here is that the perimeter is not continuous with respect to the convergence of sets... so even if $F_i\to F_\infty$ (in any decent notion of convergence) you cannot say that $P(F_i)\to P(F_\infty)$ (where $P$ is the perimeter).