I want to prove that we can find planar simple closed curves with arbitrarily small area and '$infinite$' perimeter, but not the other way around, i.e. we cannot find a simple closed curve with finite perimeter with '$infinite$' area (on a plane).
The motivation behind this is The staircase paradox, or why $\pi\ne4$.
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Claim 1: We can find a simple closed curve on a plane with area $A$ and perimeter $P$ such that $0<A < \varepsilon$ and $P>G$, $\forall \epsilon, G>0$.
Proof: We consider the rectangle with width $x$ and height $\frac{1}{x^2}$.
Evidently, $A(x)=\displaystyle\frac{1}{x}$ and $P(x)=2x+2\displaystyle\frac{1}{x^2}$.
For an arbitrary $\varepsilon>0$, we can find $G_0>0$ such that $0<A(x)=\displaystyle\frac{1}{x}< \varepsilon$, $\forall x>G>G_0 $, where $G$ is an arbitrary quantity $>G_0$.
Clearly, $P(x)>G$ and $A(x) < \varepsilon$ for any $x>G$.
However, we have $finite \ perimeter \implies finite \ area$ from the Isoperimetric Inequality $A \leq \displaystyle\frac{L^2}{4\pi}$ in case of planar simple closed curve. Thereby, $L \to 0^+ \implies$ $A \to 0^+$
Now, can we conclude from the above discussion that (in case of the simple closed curves lying on a plane) "$approximating \ the \ area$" does not necessarily imply that the "$perimeter$" is also being approximated in that process but when the perimeter is being taken arbitrarily close to the limiting curve, the area must also approach towards that of the limit curve as well ?
Best Answer
By the Jordan curve theorem, a simple closed curve separates the plane into two path-connected regions: a bounded region (the points interior to the curve) and an unbounded region (the points exterior to the curve). You are asking if the bounded region can have infinite area. The answer is no since the curve is a compact and hence bounded subset of the plane. So you can find a rectangle that includes the curve and hence includes its interior, implying that the area of the interior is at at most equal to the (finite) area of the rectangle.