Every bounded set $E\subseteq\mathbb{R}^2$ is contained in a disc of minimal radius

Question

Consider the following proof from the book "The Banach-Tarski Paradox" by Stan Wagon:

In the second paragraph he seems to implicitely use that every bounded set $E\subseteq\mathbb{R}^2$ is contained in a unique disc of minimal radius. Now the uniqueness of such a disc is actually easy to see, but I don't see why it must necessarily exist.

Answer 1

Let's suppose that $D(p_0,r_0)$ is a closed disc with center $p_0$ and radius $r_0$ that contains $E$.

Let $CR \subset \mathbb R^2 \times [0,\infty) \subset \mathbb R^3$ be the set of ordered pairs $(q,s)$ such that the closed disc $D(q,s)$ contains $E$ and such that $s \le r_0$.

Once you believe that the set $CR$ is closed and bounded, it follows that $CR$ is compact, and therefore the projection function $CR \mapsto [0,\infty)$ has compact image and hence has a minimum.

The set $CR$ is bounded, because if $(p,r) \in CR$ then $r \le r_0$, and then by picking any $q \in E$ one sees that $d(p,p_0) \le d(p,q) + d(q,p_0) \le r + r_0 \le 2 r_0$.

The set $CR$ is closed, because if $(p,r) \in \mathbb R^2 \times [0,\infty)$ is a limit of a sequence $(p_i,r_i) \in CR$ then for each point $q \in E$ we have $d(q,p_i) \le r_i$, and since distance is continuous we may pass to the limit to deduce that $d(q,p) \le r$.

Answer 2

First note, that C and its interior is a closed disc.

This is a little bit messy, but: You take the closure. (Call this F.) This must be a subset of the smallest (closed) disc, if one exists. Now the closure is closed and bounded, so it's compact, and there are two points that realize the diameter. (Call these x and y.) If you take the closed disc for which this diameter is the diameter of the disc, I claim it would be such a smallest closed disc.

First if you had a smaller closed disc, then it's subset, F, would have a bigger diameter, a contradiction. Second, if there were points outside of this disc in F, you could use triangle inequality to argue that that point along with either x or y form a longer diameter in F.