Containing an open set = being an open set? In stats courses - particularly the Casella and Berger text - reference is made to theorems being satisfied if and only if some $\Theta \subset \mathbb{R}^p$ "contains an open set," $p \geq 1$.
Isn't this just the same as $\Theta$ being an open set? Suppose there were some $\Theta^{\prime} \subset \Theta$ that were open. Then $\Theta^{\prime}$ would contain some open rectangles. But $\Theta$ would contain these open rectangles as well. Hence $\Theta$ is open.
I'm not at all familiar with topology, so I apologize for any informal terminology.

Saying $\Theta'$ is open is much stronger as saying that it contains some open rectangles. It says that for all $\theta \in \Theta'$, there is an open rectangle $R$ with $\theta \in R \subset \Theta'$. For example take $[a,b] \subset \Bbb R$. It isn't open, but contains the open set $(a,b)$. 
This is important in the context of statistics, in the cases where $\Theta$ is you parameter space, and you want to be able to derivate with respect to the parameter. You can do that only if the parameter is contained in an open set. 
