Let $C$ be a compact subset of $R^2$.
Let $D$ be the set of all pairs of points $(P,Q)$ from $C$, such that the open segment between $P$ and $Q$ is contained in $C$:
$$D = \{(P,Q)|P\in C, Q\in C, segment(P,Q)\subseteq C\}$$
Is $D$ a compact set?
My current conjecture is yes. The proof goes like this:
$C\times C$ is compact by Tychonoff's theorem. Since this is a metric space, it is also sequentially compact. Hence, every infinite sequence of pairs $\{(P_i,Q_i)\}_{i=1}^{\infty}$ has a limit, which is a pair $(P,Q)\in C\times C$.
It remains to prove that if every element in the sequence is in $D$, then the limit is also in $D$.
$(P_i,Q_i)\in D$ implies that the open segment between $P_i$ and $Q_i$ is in $D$. That is, for every $t\in(0,1)$, the point $t P_i + (1-t) Q_i$ is in $C$.
Since $(P,Q)$ is a limit point of the sequence $\{(P_i,Q_i)\}_{i=1}^{\infty}$, for every $t$ the point $t P + (1-t) Q$ is a limit point of the sequence $\{t P_i + (1-t) Q_i\}_{i=1}^{\infty}$.
Because $C$ is compact, it contains all its limit points. Hence, it contains $t P + (1-t) Q$.
This is true for every $t\in(0,1)$, which means that the entire segment between $P$ and $Q$ is contained in $C$. Hence $(P,Q)\in D$ as required. $\square$
My first question is: is this proof correct?
My second question is about generalizations of this result (if it is correct). I could think of the following generalizations:
- $D = \{(P,Q)|P\in C, Q\in C, rectangle(P,Q)\subseteq C\}$ (the axis-parallel rectangle having $P$ and $Q$ as opposite corners is contained in $C$)
- $D = \{(P,Q)|P\in C, Q\in C, circle(P,Q)\subseteq C\}$ (the circle having $P$ as center and $Q$ as a peripheral point is contained in $C$)
- etc.
Is there a more general result which has all these as special cases?