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.

No it is not means the same, every subset contains an open subset that this the empty set, but is not always open like the closed ball. For the classical topology on $R$, you can take the example of the closed interval $[a,b]$ which contains the open interval $(a,b)$ but the closed interval is not open.