# Visualizing $E=Y\cap G$ geometrically for some open set $G\subset X$

I came across this theorem(2.30) in Rudin's text on analysis. Suppose $Y\subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E=Y\cap G$ for some open subset $G$ of $X$. (A set $E\subset Y$ is open relative to $Y$ when for every $p\in E$, there is an $r_p>0$ such that $q\in Y$ and $d(p, q)<r_p$ implies $q\in E$.)

I can understand the proof he has provided. He has tried to construct such a $G$ by defining open sets $V_p$ as follows: $V_p=\{q\in X|d(p,q)\}<r_p$ and has taken $G= \cup_{p\in E}V_p$ which is good enough to prove that $E\subset G\cap Y$.

But I am still unable to understand the geometric intuition behind this; in particular, what is the geometric motivation behind such a construction? I am interested in a better way to visualize this which will allow me to construct a formal proof by looking at the picture.

we have the big space $$X=\mathbb R^2$$, the usual plane, and the subset $$A$$, which is the circle. You also see a point $$P$$ on the circle and the intersection of a neighbourhood of $$P$$ with the circle. This "square""neighbourhood would be given by a distance if the distance used on the plane is given by the norm $$||(x,y)||=\max(|x|,|y|)$$. The picture also shows a point $$Q$$ on $$A$$ and an open set in $$X$$ whose intersection with $$A$$ is not a neighbourhood of $$Q$$.