Show there exists an $f$ s.t. $f(x)>0$ for $x \in C$ where $C$ is compact

Question

I thought to use the same functions $h$ as we used in (c) but the problem is the $h$ is positive on $(a^1-\epsilon,a^1+\epsilon) \times ... \times (a^n - \epsilon, a^n + \epsilon)$ which is not compact. Is there any way to change this?

Answer 1

A set is compact if each open cover has a finite subcover, i.e. from $$ C \subset \bigcup_{a \in C}(a^1-\epsilon,a^1+\epsilon) \times \cdots \times (a^n - \epsilon, a^n + \epsilon) $$ it follows that there is a finite set $S \subset C$ such that $$ C \subset \bigcup_{a \in S}(a^1-\epsilon,a^1+\epsilon) \times \cdots \times (a^n - \epsilon, a^n + \epsilon) $$ For each $a \in S$, define $h_a$ as the function defined in (c) for the "cube" $(a^1-\epsilon,a^1+\epsilon) \times \cdots \times (a^n - \epsilon, a^n + \epsilon)$.

Then $f := \sum_{a \in S} h_a$ has the desired properties.