0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ A set is open in Y if it is open relative to the induced metric on Y. $\endgroup$ – Ispil Oct 24 '16 at 19:31
  • $\begingroup$ Do you mean in the third line $U=V\cap Y?$ $\endgroup$ – mfl Oct 24 '16 at 19:32
  • $\begingroup$ yes thank you, edited it. $\endgroup$ – Joogs Oct 24 '16 at 19:33
1
$\begingroup$

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 $Y?$
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Joogs Oct 24 '16 at 19:49
  • $\begingroup$ 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 $B_X(u,r_u)\cap Y?$ $\endgroup$ – mfl Oct 24 '16 at 19:50
  • $\begingroup$ I think it is $B_Y(u,r_u)$ $\endgroup$ – Joogs Oct 24 '16 at 19:53
  • $\begingroup$ You're right. What about $\left(\bigcup_{u\in U}B_X(u,r_u)\right)\cap Y?$ $\endgroup$ – mfl Oct 24 '16 at 19:59
  • $\begingroup$ It should be $\cup_{u \in U) B_Y(u,r_u)$ $\endgroup$ – Joogs Oct 24 '16 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.