openness of induced metric space

Let $\ (M, d)$ be a metric space, $\ A ⊂ M$. Consider the metric space $\ (A, d_A)$, where $\ d_A$ is the induced metric on $\ A$. Is it true that $\ U'$ is open in $\ A$ if and only if $\ U$ is open in $\ M$ with $\ U' = U ∩ A$?

• where are you stuck? Commented May 2, 2017 at 15:12
• @masacroso below is where I am: 1) proof of sufficiency: In case U' is open in A, suppose U is closed in M with U′=U∩A. 1-1) If A is closed in M, U' is also closed in M since intersection of closed set is closed. However, if U' is closed in M, M\U' is open, then A\U' is also open since A is in M. so it is contradict to openness of U' in A 1-2) If A is open in M and if U is closed in M, i.e. M/U is open, then (this is where I am now. I want to induce contradiction) Commented May 2, 2017 at 15:18
• what definition of topological subspace you knows? From what theory this question comes? I mean: I cant help you because Idk the context from where this question comes. To me the above "question" is the definition of topological subspace (from a pure topological point of view) for metric spaces. Commented May 2, 2017 at 15:24
• Assumption that $U$ is closed in $M$ and obtaining contradiction doesn't mean that $U$ need to be open. Commented May 2, 2017 at 15:25
• Closed means when its complement is open. Could it simultaneously exist? Commented May 3, 2017 at 2:00

Note, that for $x\in A$ if $B(x,r)=\{y\in M:d(x,y)<r\}$ denote open ball of $x$ inside $M$, and $B_A(x,r)=\{y\in A:d_A(x,y)<r\}$ is open ball in $A$, then $B_A(x,r)=B(x,r)\cap A$.
Let $U\subset A$ be open in $A$. Every $x\in U$ has some $r_x>0$ such that $B(x,r_x)\subset U$. This means that $U=\bigcup\limits_{x\in U}B_A(x,r_x)$. Now take $U'=\bigcup\limits_{x\in U}B(x,r_x)$. $U'$ is open subset of $M$, and $U'\cap A=U$.
Now take any $V'$ open in $M$. Proof of $V'\cap A$ being open in $(A,d_A)$ metric space goes almost the same.
• Look at the definition of induced metric. It follows that for all $x\in A,y\in A\ d(x,y)=d_A(x,y)$. $B_A(x,r)=\{y\in A:d_A(x,y)<r\}$ $=\{y\in A:d(x,y)<r\}$ $=\{y\in M:d(x,y)<r\land y\in A\}$ $=\{y\in M:d(x,y)<r\}\cap A$ $=B(x,r)\cap A$. Now you have all the details. Try to practice set operations more. Commented May 3, 2017 at 9:53