Let $U\subset Y$ be open in $(Y,d_Y)$, where $d_Y:Y\times Y \rightarrow \mathbb{R}$ is the metric induced by $(X,d)$. Is $ U$ also open in $X$? Let $(X,d)$ be a metric space and $Y\subset X$.
Let $U\subset Y$ be open in $(Y,d_Y)$, where $d_Y:Y\times Y \rightarrow \mathbb{R}$ is the metric induced by $(X,d)$.
Is $ U$ then also inevitably open in $X$? If not, why?
Thoughts:
If $U$ is open in $(Y,d_Y)$ but not open in $(X,d)$ then there exists an $a\in U$ such that for every $r>0$:
$U_r(a)|_{d_Y}\subset U$ and $U_r(a)|_{d_X}\not\subset U$
with $U_r(a)|_{d_Y}=\{ y\in Y| d_Y(a,y)<r\}$ and $U_r(a)|_{d_X}=\{ x\in X| d(a,x)<r\}$
 A: A necessary (and in fact sufficient) condition for every open subset of $Y$ to be open in $X$ is that $Y$ itself be open in $X.$
A: This holds only if $Y$ is an open subset of $X$.
To see it, consider any point $a \in U$. Since $U$ is open in $Y$, there exists $\varepsilon>0$ such that $U$ contains the open $\varepsilon$-ball  in $Y$ about $a$, i.e.
$$
  U \supset B_{\varepsilon,Y}(a) = \{y\in Y: d(y,a)<\varepsilon\}.
$$
However, as $Y$ is open in $X$ and $a\in Y$, there exists $\varepsilon' >0$ such that
$$
  Y \supset B_{\varepsilon',X}(a) = \{x\in X: d(x,a)<\varepsilon\}.
$$
So for any $z\in X$ with $d(z,a)<\delta:=\min\{\varepsilon,\varepsilon'\}$, $z \in U$. In summary, for every $a\in U$ there is $\delta>0$ such that $U$ contains the open ball in $X$ about $a$.  Consequently, $U$ is open in $X$.
To see that your claim does not hold without other assumptions, consider $U=Y={x} \in X$. Then $U$ is open in $Y$, but unless there exists $\varepsilon>0$ such that $B_{\varepsilon,X}(x)=\{x\}$ (which is true for all the standard metrics except for the discrete metric), then $U$ is not open in $X$.
A: No. Let $X = \mathbb R$ with $d(a,b) = \lvert a - b \rvert$ and $Y =\{0\}$. Then $Y$ is open in $Y$, but not open in $X$.
