If $x\in X$ and $r > 0$, then for every $y\in Y\cap B^{X}_{r}(x)$ there is $s > 0$ such that $y\in B^{Y}_{s}(y)\subset Y\cap B^{X}_{r}(x)$.

Question

Exercise

If $x\in X$ and $r > 0$, then for every $y\in Y\cap B^{X}_{r}(x)$ there is $s > 0$ such that $y\in B^{Y}_{s}(y)\subset Y\cap B^{X}_{r}(x)$.

My attempt

So far it has not been discussed the characterization of open sets as sets that each point is contained in an open ball contained in the set. Having said, could anyone give me some hint to solve such exercise? It is not homework. I am really interested in understanding the theory properly.

Notation

The notation $B^{X}_{r}(x)$ represents the open ball centered in $x$ with radius $r > 0$ within the set $X$.

Answer 1

The proof with some comments: It's enough to prove is that any open ball is open set in a metric space, i.e., for every $\epsilon>0$ and every $x\in X$, given an $y \in B^{X}_{\epsilon}(x)$ there exist some $\delta>0$, $$B^{X}_{\delta}(y) \subset B^{X}_{\epsilon}(x),\tag{*}$$ since you can take the intersection by $Y$ in both sides without changing the $\subset$ sign. Why this? Because, for every $\epsilon>0$ and every $x\in X$, given an $y \in B^{X}_{\epsilon}(x)$, taking intersection in both sides of $(*)$ yields that $$B^{Y}_{\delta}(y) = Y \cap B^{X}_{\delta}(y) \subset B^{X}_{\epsilon}(x) \cap Y. $$

Now, Let us denote the distance between two points by $d(x,y) = \|x-y\|$ by simplicity. Let us start the proof of (*). Let $x \in X$, $r>0$, and $y \in B^{X}_{\epsilon}(x).$ Since $y \in B^{X}_{\epsilon}(x)$, We have that $\|y-x\|<\epsilon$. Hence, $$0 < \epsilon - \|y-x\|.$$ This way, let $\delta = \epsilon - \|y-x\|.$ Given an arbitrary $z \in B^{X}_{\delta}(y),$ we have that $$ \|z-x\| \leq \|z-y\| + \|y-x\| < \delta + \|y-x\| = \epsilon. $$ Hence, $z \in B^{X}_{\epsilon}(x).$ This means that $ B^{X}_{\delta}(y) \subset B^{X}_{\epsilon}(x),$ since every $z \in B^{X}_{\delta}(y)$ is in $B^{X}_{\epsilon}(x)$ as well.

---

You can understand this result as it saying that the distance between $X$ and $D$ is the sum of the distances between $X$ and $Y$ summed with the distance of $Y$ and $D$, since the radius of the circumference centered at $Y$ is actually the distance of $Y$ and $D$. Here, $D$ is the point that intersects the prolonged segment of $X$ and $Y$ with the boundary of the circle.

Answer 2

There is $s>0$ so that $B^X(y,s) \subset B^X(x,r)$ because $B^X(x,r)$ is open in the metric topology on $X$ and $y$ is in it. Next note that $Y \cap B^X(y,s) = B^Y(y,s)$ by definition and take the intersection with $Y$ on both sides of the inclusion. QED.