Proof about a subset of a metric space Prove that a subset $A$ of metric subspace $(P, p')$ of metric
space $(M, p)$ is open in subspace $(P, p')$, regarded as a metric space in its own
right, if and only if there exists an open set $U$ in $M$ such that $A$ is the intersection of $U$ and $P$.
I'm not really sure if you're supposed to prove that $A$ is its own metric space or just that it's open, but I'm guessing it's the former.
Open sets are defined as either the null set or the union of a set of open spheres.
Thanks so much for your help!
 A: $(P,p')$ being a metric subspace of $(M,p)$ means that $P \subseteq M$ and $p'$ is the restriction of $p$ to $P \times P \subseteq M \times M$. Note also that this means that for $x \in P$, $B_{p'}(x, r) = B_{p}(x,r) \cap P$, where $B_{p'}$ denotes the ball in $P$. Simply because a point is in the $p'$-ball if it is in $P$ (of course) and the $p'$-distance to $x$ is less than $r$, but this is just equal to the $p$-distance, as $p'$ is the restriction. So it is in the intersection, and the reverse is also obvious. So for open balls the fact is already clear.
As open sets are unions of balls, the same relation holds for open sets: e.g. suppose that $A \subset (P,p')$ is open. Then $A = \cup \{B_{p'}(x, r_x): x \in A\}$ (for some choice of radii $r_x, x \in A$, and then if we define $A' = \cup \{B_{p}(x, r_x): x \in A\}$, then $A'$ is open in $(M,p)$ as a union of open balls, and 
$$A' \cap P = \cup \{B_{p}(x, r_x): x \in A\} \cap P = \cup \{B_{p}(x, r_x) \cap P: x \in A\} = \cup \{B_{p'}(x, r_x): x \in A\} = A$$
so that indeed $A$ is the intersection of $P$ and an open set in the surrounding space.
On the other hand, if $A = A' \cap P$, where $A'$ is open in $(M,p)$, for each $x \in A$ we have that $x \in A'$, so there is some ball $B_p(x, r_x) \subset A'$. But then $B_{p'}(x, r_x) = B_p(x, r_x) \cap P \subset A' \cap P = A$, showing that $x$ is an interior point of $A$ in $(P,p')$, so $A$ is open, as $x$ was arbitrary in $A$.  
