What is a plane figure covering all sets of diameter $1$? Problem : (1) Show that there is a plane figure $F$ of least area
which is capable of covering any plane figure of unit diameter.
(2) Try to guess what is $F$.
Proof of (1) : Define $\mathcal{H}$ to be set of all compact convex set in unit
ball. Here $\mathcal{H}$ is a compact metric space and an area function on $\mathcal{H}$ is continuous, which is crucial in the proof of well-known isoperimetric problem.
We will mimic that if we can do : Note that a set of diameter $\leq 1$ can be covered by a convex set of diameter $1$, which can be provved.
If $F'$ is cover for all convex set of diameter $1$, then we define $\mathcal{I}$ to be a set of all such $F'$.
Clealy $\mathcal{I}$ is compact. But we don't know whether the area function is continuous or not.
Proof of (2) : Consider a regular $(2n+1)$-gon $Q_{2n+1}=(p_1\cdots p_{2n+1})$ s.t. $$|p_1p_n|=|p_1p_{n+1}| =1 $$
Each triangle $p_1p_np_{n+1}$ define a circular sector $A$ in unit-ball.
Hence we attach $(2n+1)$-copies of $A$ to $Q_{2n+1}$ so that we have $Q_{2n+1}'$. I guess that $F$ is from overlapping all such $Q_{2n+1}'$.
 A: Here is a sketch of a proof for (1). (Take all sets to be closed in this argument, to avoid some convergence/boundary issues.)
I will assume for this proof that we can place some bound on an optimal region; say, that it is contained inside a closed disc of radius $10^{100}$. This seems patently obvious given that we are looking for a cover of shapes that have diameter $1$, but I haven't found a great argument for this yet; if I see a proof, I will edit it in. Suggestions welcome in the comments.
First, note that we can restrict to convex sets of unit diameter, since taking the convex hull doesn't increase the diameter.
Now consider the set of convex polygons with rational coordinates of diameter $\le1$ - this is a countable set, so we can enumerate them $P_1, P_2,\ldots$.
Observe that any (compact) cover $F$ for all $P_i$ in our list is a cover for all unit-diameter convex sets. This is because, given convex $S$ of unit diameter, the space of possible placements of scaled copies of $S$ is a compact set. But any copy of $S$ scaled by a factor less than $1$ fits inside one of our $P_i$ and therefore inside our cover. So the continuous function on our space of placements measuring the scale of $S$ has a supremum of $1$, and by compactness attains it somewhere in the set.
So how do we find a cover of minimal area for the $P_i$? Let $A$ be the infimum of possible areas; this infimum obviously exists, since a circle of area $\pi$ suffices as a cover.
We place $P_1, P_2,\ldots$ on top of each other iteratively via an inductive process: after placing $P_1,\ldots,P_n$, we choose $P_n$ such that for all $\epsilon>0$, there is a placement of all further $P_i$'s which attains area at most $A+\epsilon$. (Intuitively, we place each new polygon so as to never "box ourselves into a corner" and place a suboptimal lower bound on the area of our cover.)
Why can we choose such a location for $P_n$? By the inductive hypothesis, there is one such placement for every $\epsilon>0$, but by our boundedness assumption, we can restrict our attention to a compact set of possibilities, so one placement will attain this. (The base case is trivial; any placement of $P_1$ will do.)
Repeating this process, we end up with a series of sets ordered by inclusion: $P_1\subset (P_1\cup P_2)\subset (P_1\cup P_2\cup P_3)\subset\ldots$, each of which has area at most $A$. Thus, their union also has area at most $A$, as does its closure. The result gives us a cover $F$ of area exactly $A$ for all planar sets of unit diameter.
Edit: Actually, I'm not sure I can justify that the area of the closure is equal to the area of $A$. It seems obvious enough, but there are some pathological cases in which it fails. Suggestions for how to remedy this are welcome, but in the meantime this proof is certainly incomplete.

In the case where $F$ is convex, this is Lebesgue's universal covering problem; it is an open problem, and the best-known solutions thus far are not particularly obvious or guessable. So I would be quite surprised if (2) can be answered very well.
In practice, I don't think any non-convex covers are known that improve on the best-known convex case? I'm not sure if much work has been done on that side of things, though.
