Show that an irregular 1-set in the plane is totally disconnected

Question

A $1$-set is a Borel set such that $0 < \mathcal{H}^1(A) < \infty$, where $\mathcal{H}^s$ is the Hausdorff measure.

Let $A$ be an irregular $1$-set in the plane. Deduce from the theorem below that $A$ is totally disconnected.

Definition of $Proj_{\theta}$ is

and definition of irregularity is

and here is my definition of totally disconnected.

A set $F$ is totally disconnected if and only if $\forall x \in F$, if $U$ is connected subset of $F$ and $x \in U$ this implies that $U=\{x\}$.

I think that you should be able to do this by contradiction. So...

Suppose for contradiction that $\exists x \in F$ and $U$ which is connected subset of $F$ such that $x \in U$ and that $U$ contains at least two points with $x \neq y$. Let us denote one of the other points by $y$. Now the projection of any connected set is a connected set and any connected subset of $\mathbb{R}$ is an interval. Now as F is an irregular $1$-set in $\mathbb{R}^2$ we may apply Theorem 5.4.2 so the $proj_{\theta}$ has zero length. Therefore as $U \neq \emptyset$ $U=\{x\}$ and $x=y$. This is contradiction as $x$ and $y$ were distinct, therefore $F$ is totally disconnected.

However I am not sure how to choose my $\theta$ (or how to apply Theorem 6.4). I feel that the last half of my attempted proof fades in rigour.

Answer 1

Let $F,U,x,y$ be as in your attempt.

1. For every $\theta$ the projection map $\operatorname{proj}_\theta$ is continuous; hence, $\operatorname{proj}_\theta (U)$ is connected.

2. A connected subset of a line with two distinct points has positive length.

3. For all $\theta\in [0,\pi)$ except one (the one parallel to the line segment $xy$), $\operatorname{proj}_\theta(x)\ne \operatorname{proj}_\theta(y)$.

4. Combining 1,2, and 3, we have a contradiction with Theorem 6.4.