# Open sets in Y where Y is a subspace of the metric space (X,d) with the induced metric

I want to prove the following: Let $Y\subset X$, where $(X,d)$ is a metric space and $d_Y$ is the induced metric on Y. Show that: $U\subset Y$ is open in Y if and only if there exists $V\subset X)$ such that $U=V\cap Y$.

Quite frankly, I have no clue how to start the proof as I dont properly know what it means exactly for a set to be open in Y. I know what it means for a set to be open in some metric space and it seems that the definition should be quite similar in this case but I can not figure it out.

• A set is open in Y if it is open relative to the induced metric on Y. – Ispil Oct 24 '16 at 19:31
• Do you mean in the third line $U=V\cap Y?$ – mfl Oct 24 '16 at 19:32
• yes thank you, edited it. – Joogs Oct 24 '16 at 19:33

## 1 Answer

Hint

If $U$ is open in $Y$ then for any $u\in U$ there exists $r_u>0$ such that $B_Y(u,r_u)\subset U.$

• Is it $B_X(u,r_u)$ open in $X$?
• Is it $\bigcup_{u\in U}B_X(u,r_u)$ open in $X?$
• What is the intersection of $\bigcup_{u\in U}B_X(u,r_u)$ and $Y?$
• Can you give one more hint for the first question of your hint, you probably spelled it out a lot already but my intuition is completely off here for some reason. – Joogs Oct 24 '16 at 19:49
• The set $B_X(u,r_u)=\{x\in X: d(x,u)<r_u$ is open in $X$ because $(X,d)$ is a metric space. Can you describe $B_X(u,r_u)\cap Y?$ – mfl Oct 24 '16 at 19:50
• I think it is $B_Y(u,r_u)$ – Joogs Oct 24 '16 at 19:53
• You're right. What about $\left(\bigcup_{u\in U}B_X(u,r_u)\right)\cap Y?$ – mfl Oct 24 '16 at 19:59
• It should be $\cup_{u \in U) B_Y(u,r_u)$ – Joogs Oct 24 '16 at 20:01