Existence of maximum area triangle among all whose vertices are in a compact subset of $\mathbb{R}^2$

Prove that among all triangles whose vertices are in a compact subset $K$ of $\mathbb{R}^2$, there exists at least one with maximum area.

I am at a loss as to how to rigorously show that you can generate a finite open cover of $K$ using triangles at least one of which has maximum possible area. Below is my first attempt. Any help would be greatly appreciated.

Define the set $T=\{(x,y,z) : x,y,z\in K, x\neq y, x \neq z, y \neq z\}$, and let $\operatorname{tri}(t)$ represent the triangle with vertices at $t=(x,y,z) \in T$. Moreover, define $\operatorname{tri}(a) \cap \operatorname{tri}(b)$ to be empty if the two triangles overlap only on an edge.

Choose $t_0 \in T$. If there exists $u \in T$ such that $\operatorname{tri}(t_0) \subset \operatorname{tri}(u)$, then set $t_0 = u$. Repeat this process until no such $u\in T$ exists. Since $K$ is closed and bounded, this process will terminate. (I have no proof of this.)

Choose $t_i \in T$ such that

$$\operatorname{tri}(t_i) \bigcap \left( \bigcup_{k=0}^{i-1} \operatorname{tri}(t_k) \right) = \emptyset$$

and repeat the same process used to find $t_0$ above.

Since $K$ is compact, the $\cup_{i} \operatorname{tri}(t_i)$ forms a finite open cover of $K$ (I have no proof of this) where the area of each $\operatorname{tri}(t_i)$ is greater than that of all other $t\in T$ such that $\operatorname{tri}(t)\subset \operatorname{tri}(t_i)$. Thus, since we have a finite cover, there exists $\operatorname{max}_i \operatorname{area}( \operatorname{tri}(t_i))$.

The above is just shoddy and unconvincing, and most likely needlessly complicated and incorrect.

-
add comment

1 Answer

Let $A(x,y,z)$ denote the area of a triangle with vertices $x,y,z$ in $K$. It is easy enough to show that $A$ is continuous, for example by noting that $$A(x,y,z)=\frac12\|(y-x)\times (z-x)\|$$ where $\times$ denotes the cross-product. Furthermore, since $K$ is compact, $K^3$ is compact by Tychonoff's theorem. Since $K^3$ is the domain of $A$, and continuous functions on compact domains attain their supremum, we conclude that $A$ obtains its supremum, i.e. that we have some $x_0,y_0,z_0\in K$ such that $A(x_0,y_0,z_0)$ is maximal. Thus the triangle with vertices $x_0,y_0,z_0$ has maximal area.

-
@AnonSubmitter85 I gave a formula for $A$ which shows its continuous, assuming you know the basic properties of continuous functions. I'm not sure what you mean by "$A$ being discrete", but I assure you $A$ is certainly continuous. –  Alex Becker Jun 22 at 22:06
Sorry. I was thinking that $\operatorname{A}$ had to be continuous on $K$ to attain a maximum, but it doesn't. It just has to be continuous on the superset of which $K$ is a compact subset. –  AnonSubmitter85 Jun 22 at 23:06
@AnonSubmitter85 It definitely has to be continuous on $K$! But any function which is continuous on a superset of $K$ is also continuous on $K$. –  Alex Becker Jun 23 at 1:46
But how can that be if $K$ is composed of isolated points? For instance, $K = \{(1/m,1/n) : m,n \in \mathbb{N} \} \cup (0,0)$ is a compact subset of $\mathbb{R}^2$, but, unless I am forgetting something fundamental about the definition of continuity, there is no way $\operatorname{A}$ could be continuous on it. However, since $\operatorname{A}$ is continuous on $R=\{(x,y) : 0 \leq x,y \leq 1\}$ and $K \subset R$, it should still hold that $\operatorname{A}$ attains a maximum value on $K$ since it is compact. –  AnonSubmitter85 Jun 23 at 1:54
@AnonSubmitter85 You are forgetting something fundamental about the definition of continuity. What definition are you working with? –  Alex Becker Jun 23 at 1:56
show 2 more comments