TITLE: Difference between 'A open subset of B' and 'A open relative to B'
QUESTION [7 upvotes]: If $A\subset B$ and $B$ is a metric space. What is the difference between "$A$ open subset of $B$" and "$A$ open relative to $B$"?

REPLY [2 votes]: I just came across this definition in Principles of Mathematical Analysis and perhaps this might be helpful to someone as I had to read it over a couple of times and even though this is minor, it helped me.
A small notational adjustment to the definition of the open ball:

Definition: Let $X$ be a metric space and $p \in X$ then let $N_r^X(p) := \{q \in X \mid d(p,q) < r\}$. 

Now, using that $E$ is open if every point of $E$ is an interior point of $E$ which is Definition 2.18, I prefer this way of defining openness:

Definition: A set $E \subset X$ is open in $X$ if for each $p \in E$ there is an associated $r > 0$ such that $N_r^X(p) \subset E.$

So using the above we now get:

Definition: A set $E \subset Y$ is open relative to Y if for every $p \in E$ there is an associated $r >0$ such that $N_r^Y(p) \subset E.$ 

Why do I find this helpful? Well, using the above and working an example you can show that an open ball isn't necessarily a "ball" anymore and I think it's very helpful to realize that quickly as it tends to highlight how our usual notation of open/closed sets are relative. 
For example, take $X = \mathbb{R}, Y = [0,1] \text{ and } E = [0,\frac{1}{2})$. The open ball $N_r^Y(0)$ turns out to be $[0,r)$ and is no longer $(-r,r)$ as we tend to expect it to be. So in this specific example we can readily see that by setting $r =\frac{1}{4}$ that $N_{1/4}^Y(0) = [0,\frac{1}{4}) \subset [0,\frac{1}{2}) = E $. So the "problematic point" turns out not to be a problem at all. 
Hope this may help someone!