Definition of an open set confusion How is the definition of an open set related to the definition of a set being open in another set?
Saying $S \subseteq D$ is open in $D$ means that $S$ open in the subspace topology on $D$. That is, $S = U \cap D$ for some open set $U \subseteq \mathbb{R}^n$. But how is this related in any way to S being an open set definition?
I cannot find any definition else online for being open in another set which is weird so any explanation or links would be appreciated.
 A: If you know that functions induce topologies, there is a short cut we can take. We can characterize the subspace topology as follows: let $S\subseteq D$. Then, there is a $\textit{set}$ map $i:S\to D$ which is just the inclusion; that is, it sends $s\in S$ to $s$ in $D$. We can check that your definition of an open set in the subspace topology coincides exactly with the claim that $i$ is continuous. More precisely, the open sets in $S$ are $\textit{defined}$ to be those that are inverse images of open sets in $D:\ A$ is open in $S$ if and only if there exists a open set $B$ in $D$ such that $i^{-1}(B)=A$. And since $i^{-1}(B)=B\cap S$, the two definitions coincide. 
A: 
how is this related in any way to S being an open set definition?

To be able to tell whether $S$ is open or not, you need a topology on $D$. By definition, if $D$ is a subset of $X$, where $X$ is $\mathbb R^n$ with its standard topology (or any topological space), then the subspace topology on $D$ is exactly 
$$O_D = \{ U\cap D\; |\; U \textrm{ open in } X \}$$
What you have to do is to show that $O_D$ is a topology, that is:


*

*$\emptyset$ and $D$ are elements of $O_D$

*finite intersections of elements of $O_D$ are also in $O_D$ 

*arbitrary unions of elements of $O_D$ are also in $O_D$.
All of this is easy using the properties of open sets in $X$, and set theory.
