definition of open sets in metric spaces For metric spaces, the definition of an open set $U\subset X$ is that it is a set which for any point $u\in U$ in the set there exists some $\epsilon&gt;0$ such that the open ball $B_\epsilon(u)\subset U$.
My question is where is the effect of the mother metric space X?
The set of points in the open ball $B_\epsilon(u)$ depends on the ambient space $X$. That is, if we considered $Y \subset X$ with $U \subset Y$, then it is not necessary that $B_{\epsilon,X}(u) = B_{\epsilon,Y}(u)$. For example, if $Y = U$, then $B_{\epsilon,Y}(u)$ is by definition contained in $U$.
Consider the euclidean metric with 


$U =[0,1)$ ; $X= \Bbb R_+$ including $0$
$U =[0,1)$ ; $X= \Bbb R$


In first case $U$ is open because of $X$. We cannot move outside $0$ to find the neighborhood around $0$ because of $X$. Thus there exist a positive $\varepsilon$ such that $( 0, +\varepsilon)$ lies entirely in $U$: this can be checked for other points in $U$ as well.

Second case, $U$ is not open as there is no $\varepsilon$ such that $(-\varepsilon,+\varepsilon)$ lies entirely in $U$.