Why we can assume compactness in Victor Klee's theorem?

Question

Let $\ C $ and $\ C_1,...,C_n$ be closed convex sets in Euclidean space s.t.

1. $\ C \cap \bigcap^n_{i=1,i\neq j} C_i \neq\emptyset $ for $ j=1,2,...,n$

2. $\ C \cap \bigcap^n_{i=1} C_i = \emptyset$

Then

$C \nsubseteq \bigcup^n_{i=1} C_i $ .

The proof starts by assuming that $\ C $ and $\ C_1,...,C_n$ are all compact. Otherwise, we can replace $ C$ by the set A = $ Conv \{y_j : j=1,2,...n \}$ and $ y_j \in C \cap \bigcap^n_{i=1,i\neq j} C_i $ and each $ C_i$ can be replaced by $ B_i = C_i \cap A$

I'm not sure I understand this argument, because I can think of counterexample:

Let $ C \subseteq \mathbb{R} = \{x: x \geq 4 \}$ and $C_1 = \{x: x \geq 7 \} $

so $ Conv\{y_1\} = C_1 = \{x: x \geq 7 \} $ and it is not a compact set. Am I missing something?

Thanks.

Answer 1

Note that the condition $C \cap \bigcap^n_{i=1,i\neq j} C_i \neq\emptyset$ cannot be fulfilled if $n=1$.

For $n \geq 2$, $y_j$ is a single fixed element of $C \cap \bigcap^n_{i=1,i\neq j} C_i$. Therefore, $A$ is the convex hull of a finite set and thus compact.

In your example, we have $Conv \lbrace y_ 1 \rbrace = \lbrace y_1 \rbrace$.