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I came up with this problem during shortening my road to an university on a car park.

We will build a sequence of geometrical figures. As sequence's first element let's consider a square with a side equal to $1$. Each $n+1$ element of a sequence is created from $n$-th element by dividing each side facing "up" or "right" on half and turning each "up-right" corner into "down-left" (check the image).

4 first elements

We know that each element has a cuircuit equal to $4$. But the field is constantly decreasing and it's easy to spot that the lower bound of the field is equal to $\frac{1}{2}$.

The question is: can we say that the limit of these figueres is equal to rectangular triangle (can we say about a limit of figures at all)?

If yes, then if we build a numerical sequence $a_k = $ cuircuit of $k$-th figure, we will get a $a_k = 4, \forall k \in \mathbb{R} $, but $\lim_{k\rightarrow \infty} a_k = 2+\sqrt2$ ?

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1 Answers1

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Clearly, $a_k=4\implies \lim_{k\to\infty}a_k=4$.

You just shown that the limit of the length is not the length of the limit. (No theorem states this.)

  • But if consider e.g. Euclidean metric we have $\forall \epsilon >0, \exists n : d(x,y_n) < \epsilon$, where $x$- square diagonal, $y_n$- $n$-th figure "diagonal" curve (to ommit domain problems let's rotate the figures $45^{\circ}$). In Euclidean metric limit is equal to diagonal? Where do I make a mistake? – vermator Mar 15 '19 at 10:00
  • I meant $L_2$ metric. – vermator Mar 15 '19 at 10:08
  • Analogically $1/n$ is never equal to $0$ but it convergences to it. As I said it appears that in $L_2$ the $y_n$ has a limit and it is $x$ (as the condition of convergence is met). – vermator Mar 15 '19 at 10:32
  • @vermator: I changed my answer. –  Mar 15 '19 at 10:40