This is a problem from my class, with the first part of the solution. I'd like to ask something about the set $S \subset V$ as defined below.
Problem: Given $G$ on $n$ vertices with $\delta(G) \geq \delta,$ how large can its smallest dominating set be?
$\cdot$ Let $S \subseteq V$ be a random set
$\cdot v \in S$ with probability $p$, independently$\cdot$ For $u \in V$, define the event $E_{u}=\{u \text{ not dominated by } S\}$
$\cdot$ For $E_{u}$ to hold, need: $u \notin S$, and $v \notin S$ for all neighbours $v$ of $u$
$\Rightarrow \mathbb{P}\left(E_{u}\right)=(1-p)^{d(u)+1}$$\cdot \{S \text { not dominating }\}=\cup_{u \in V}E_{u}$
$\cdot \mathbb{P}\left(\cup_{u \in V} E_{u}\right)<\sum_{u \in V} \mathbb{P}\left(E_{u}\right)=\sum_{u \in V}(1-p)^{d(u)+1}$
$\cdot \sum_{u \in V}(1-p)^{d(u)+1} \leq n(1-p)^{\delta+1} \leq n > e^{-p(\delta+1)}$
$\Rightarrow$ if $p=\frac{\ln n}{\delta+1},$ then $ \mathbb{P}(S \text { not dominating })<1$
$\Rightarrow S$ is dominating with positive probability
The line I don't understand is "$\{S \text { not dominating }\}=\cup_{u \in V}E_{u}$". Shouldn't the union be over $V \setminus S$, with the stipulation that it is not empty? The way the union is presented, I interpret it as saying that none of the vertices in $G$ can be dominated by $S$. But then $S$ can only be the empty set?
The union corresponds to an "or" of events: the event $\bigcup_{v \in V} E_u$ holds if any of the events $E_u$ holds. As a result, assuming we accept that the event $E_u$ means "the vertex $u$ is not dominated by $S$", then the event $\bigcup_{v \in V} E_u$ means "some vertex is not dominated by $S$".
We can't take the union over $V \setminus S$, because $S$ is a random variable: we don't know what $V \setminus S$ is. However, when $u \in S$, then $E_u$ will never hold, because $u \notin S$ is one of the conditions in $E_u$. So we do in fact have $$\bigcup_{v \in V} E_u = \bigcup_{v \in V\setminus S} E_u.$$