Difference between 'A open subset of B' and 'A open relative to B' 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 $B$"?   
 A: 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!
A: There is only a difference if $B$ itself is a subspace of some larger (say metric) space $X$: i.e., $A \subset B \subset X$.
Then, "$A$ is an open subset of $B$" means that (i) $A \subset B$ and (ii) $A$ is open in $X$.  However "$A$ is open relative to $B$" means that (i) $A \subset B$ and (ii) $A$ is open when viewed as a subspace of $B$, i.e., given any $x \in A$, there exists an $\epsilon > 0$ such that every element of $B$ of distance less than $\epsilon$ from $x$ lies in $A$.  (However it need not be the case that every element of X with distance at most $\epsilon$ from $x$ lies in $A$.)
Here is a simple example of this: suppose that $X = \mathbb{R}^2$, $A = (0,1) \times \{0\}$ and $B = [0,1] \times \{0\}$ -- that is, $A$ is an open interval on the horizontal line $y = 0$ and $B$ is the corresponding closed interval.  Then $A$ is open relative to $B$ but is not an open subset of $B$.
(Note though that $B$ is not just closed in itself -- which is trivial -- or closed in $\mathbb{R} \times \{0\}$ but is actually closed in $X = \mathbb{R}^2$.  Not coincidentally, $B$ is compact.  Compactness can be viewed -- among other ways -- as a sort of "absolute closedness": if a subset $X$ of a metric space $Y$ is compact, then it is not only closed in $Y$ but in fact in every metric space $Z$ containing $Y$ as a subspace.)
A: If $(X,\rho)$ is a metric space, then we say that $A\subseteq X$ is an open subset of $X$ if to each $a\in A$, there exists $\delta>0$ such that $b\in A$ whenever $\rho(a,b)<\delta$. 
If $(X,\rho)$ is a metric space and if $Y\subseteq X$ is a subspace of $X$, then we say that $A\subseteq Y$ is open relative to $Y$ if to each $a\in A$, there exists $\delta>0$ such that $b\in A\cap Y$ whenever $\rho(a,b)<\delta$ and $b\in Y$.
Exercise 1: If $X$ is a metric space and if $A\subseteq X$, prove that $A$ is an open subset of $X$ if and only if $A$ is open relative to $X$.
Exercise 2: Let $X$ be a metric space and let $Y\subseteq X$. Prove that if $A$ is an open subset of $X$, then $A\cap Y$ is open relative to $Y$.
Exercise 3: Let $X$ be a metric space and let $Y\subseteq X$. If $A\subseteq Y$, then is it true that $A$ is an open subset of $X$ if $A$ is open relative to $Y$? Prove or give a counterexample.
Exercise 4: Let $X$ be a metric space and let $Y$ be an open subset of $X$. Prove that if $A\subseteq Y$ is open relative to $Y$, then $A$ is an open subset of $X$.
Exercise 5: Let $X$ be a metric space and suppose $X=\bigcup_{n=1}^{\infty} X_n$. If $A\subseteq X$ is such that $A\cap X_n$ is open relative to $X_n$ for all positive integers $n$, then is it true that $A$ is an open subset of $X$? Prove or give a counterexample.
Research Project: Let $\{X_{\alpha}\}_{\alpha\in A}$ be a collection of subspaces of the metric space $X$. We say that the topology of $X$ is coherent with the subspaces $\{X_{\alpha}\}_{\alpha\in A}$ if the following property is satisfied:

A subset $A$ of $X$ is open in $X$ if and only if $A\cap X_{\alpha}$
  is open relative to $X_{\alpha}$ for all $\alpha\in A$.

Investigate thoroughly necessary and sufficient conditions for the topology of $X$ to be coherent with a collection of subspaces. More specifically, state as many results and examples as you can in this connection. After you have thought about this problem sufficiently deeply, you may also refer to the internet and textbooks for further discussion.
Warning: Solutions to Exercises
Solution to Exercise 1: Since $A\cap X=A$, a careful examination of the definitions above shows that $A$ is an open subset of $X$ if and only if $A$ is open relative to $X$.
Solution to Exercise 2: Let $a\in A\cap Y$. Since $A$ is an open subset of $X$, there exists $\delta>0$ such that $b\in A$ whenever $\rho(a,b)<\delta$. In particular, we have $b\in A\cap Y$ whenever $\rho(a,b)<\delta$ and $b\in Y$. Therefore, $A\cap Y$ is open relative to $Y$.
I hope this helps!
