Is it possible for a Jordan curve in the plane to enclose a set with area zero? I read about the Isoperimetric Inequality the other day. It says that for any Jordan curve,
$$
\frac{4 \pi A}{L^{2}} \leq 1,
$$
where $ L $ is the length of the curve and $ A $ is the area of the region that it encloses. Is it possible for this quotient to be zero (i.e., $ A = 0 $)?
 A: It is not possible. The interior region of any Jordan curve in $ \mathbb{R}^{2} $ is a non-empty open subset of $ \mathbb{R}^{2} $, so has area $ > 0 $. The Jordan Curve Theorem makes this quite clear:

Jordan Curve Theorem Let $ \gamma $ be a Jordan curve in $ \mathbb{R}^{2} $. Then its complement $ \mathbb{R}^{2} \setminus \gamma $ consists of exactly two non-empty connected open subsets of $ \mathbb{R}^{2} $, one of which is bounded and the other unbounded. We call the bounded connected component the interior and the unbounded connected component the exterior.

Let me add to what Chris has mentioned in his valuable comment above. If we consider only rectifiable Jordan curves (i.e., those Jordan curves for which $ L < \infty $), we can make the ratio $ \dfrac{4 \pi A}{L^{2}} $ as close to $ 0 $ as desired.

Example (Koch Snowflake)
Let $ (\gamma_{n})_{n \in \mathbb{N}_{0}} $ be the well-known sequence of Jordan curves whose pointwise limit is the Koch Snowflake. The first four iterations, $ \gamma_{0} $, $ \gamma_{1} $, $ \gamma_{2} $ and $ \gamma_{3} $, are displayed below.

Let $ (A_{n})_{n \in \mathbb{N}_{0}} $ and $ (L_{n})_{n \in \mathbb{N}_{0}} $ denote, respectively, the sequence of enclosed areas and the sequence of lengths corresponding to $ (\gamma_{n})_{n \in \mathbb{N}_{0}} $. Then
\begin{align}
\forall n \in \mathbb{N}_{0}: \quad
A_{n} &= \frac{\sqrt{3}}{20}
\left[ 8 - 3 \left( \frac{4}{9} \right)^{n} \right] s^{2}, \\
L_{n} &= 3 \left( \frac{4}{3} \right)^{n} s,
\end{align}
where $ s $ is the length of one side of the initial equilateral triangle $ \gamma_{0} $. As
$$
\lim_{n \to \infty} A_{n} = \frac{2 \sqrt{3}}{5} \cdot s^{2} \quad \text{and} \quad
\lim_{n \to \infty} L_{n} = \infty,
$$
we see that $ \displaystyle \lim_{n \to \infty} \frac{4 \pi A_{n}}{L_{n}^{2}} = 0 $.
