Problem with understanding the lower bound of OPT in Greedy Set Cover approximation algorithm
Let's assume for simplicity that $C = \emptyset$; I'll let you work out the general case. Note that there are at most $|U|$ sets in OPT. Let $OPT = \{O_1,\ldots,O_m\}$, where $m \leq |U|$. We have
$$
\sum_{i=1}^m \frac{c(O_i)}{|O_i|} \frac{|O_i \setminus (O_1 \cup \cdots \cup O_{i-1})|}{|U|} \leq \sum_{i=1}^m \frac{c(O_i)}{|O_i|} \frac{|O_i|}{|U|} = \frac{OPT}{|U|}.
$$
On the left-hand side we have a weighted average of the quantities $c(O_i)/|O_i|$, so one of them must be at most $OPT/|U|$. The greedy algorithm selects a set $S$ with $c(S)/|S|$ minimal, so in particular
$$
\frac{c(S)}{|S|} \leq \min\left(\frac{c(O_1)}{|O_1|},\ldots,\frac{c(O_m)}{|O_m|}\right) \leq \frac{OPT}{|U|}.
$$