Topology induced by metric and subspace topology Let $(X, d)$ and $(Y, d_Y)$ two metric spaces, where $Y \subset X$ and $d_Y=d_{|Y}$. On $Y \subset X$ there are two topologies, which are the subspace topology $T_{d}^Y$ induced by topology $T_d$ on $X$, and topology induced by metric on $Y$, $T_{d|Y}$. 
Can I conclude that $T_{d|Y}=T_{d}^Y$? how can I prove it?
thank you
 A: The subspace topology induced by the topology on $Y$ consists of members $U\cap Y$ such that $U$ is open in the metric topology $T_d$ on $X$. With your notation, this is $T_d^Y=\{U\cap X\ |\ U\ \in T_d\}$. 
The metric topology $T_{d|Y}$ on $Y$ is given by the collection consisting of all unions of all $d|Y$ balls. 
It's sufficient to prove that for each basic open set (i.e. a ball in $X$ intersected with $Y$) $U$ in $T_d^Y$, and $y \in U$, there is an open set $V$ in $T_{d|Y}$ that is contained in $U$ and for each basic open set (i.e. an open ball in $Y$ with the restricted metric) $B$ and $a \in B$, there is an open set $W$ of the subspace topology such that $a \in W \subset B$. 
(I'll leave it to you to see why this is sufficient, if you don't know that already.)
If $U$ is a non-empty basis element in the subspace topology on $Y$, then $U=Y\cap B_d(x,\varepsilon)$ for some $x \in X$ and $\varepsilon >0$. 
Now, for $y \in U$, there is an $\delta > 0$ such that $B_d(y,\delta) \subset B_d(x,\varepsilon)$. Then, it follows that $B_{d|Y}(y,\delta)\subset U$.
For the other direction, assume that $a \in B_{d|Y}(z,\gamma)$ for some $z,a \in Y$ and $\gamma >0$. Then,  $W=:B_{d}(z,\gamma)\cap Y$ is the desired open set in the subspace topology that contains $a$ and is contained in $B_{d|Y}(z,\gamma)$. 
A: That's true. Take an element $y_0\in Y$ and $\varepsilon>0$ and note that
$$B_{d_Y}(y_0,\varepsilon)=\{y\in Y:\,d_Y(y_0,y)<\varepsilon\}=\{y\in Y:\,d(y_0,y)<\varepsilon\}=Y\cap B_d(y_0,\varepsilon)$$
So take an open set $O$ in de metric topology $T_{d_Y}$ and $y\in O$. Then there exists some $\varepsilon>0$ such that $B_{d_Y}(y_0,\varepsilon)\subseteq O$ and by the preceding argument $Y\cap B_d(y_0,\varepsilon)\subset O$, thus $O$ is open in the subspace topology. To prove the reverse inclusion it's enough to apply the same argument starting from a subspace open set $O$. 
