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Suppose we have a metric space $(M,D)$, and let $A\subset M$. Let $U$ be an open set such that $U\subset M$. How do I prove that $B:=U\cap A$ is open in $A$?

My attempt: If $B=\emptyset$, it is open, so we assume $B\ne\emptyset$.

Let $x\in B$, and define $S_r(x)=\{y\in B |D(x,y)0$ s.t. $S_r(x)\subset B$ we are done. I'm not really sure how to do that, though.

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    What is you definition of open of $A$? – 2017-02-09
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    In most treatments, that is the *definition* of an open subset of $A$. – 2017-02-09
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    From my textbook (Kaplansky I.) Let $U$ be a subset of a metric space $M$. We say that $U$ is open in $M$ if for every $x\in U$ an entire open ball $S_r(x)$ is contained in $U$. – 2017-02-09

2 Answers 2

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If $x\in B$, then $x\in U$, as $U$ is open in $M$, there exist $r>0$, such that $\{y\in M | D(x,y)

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By definition, $C$ is open in $A$ if exists an open set $E$ in $M$ such that $C = E \cap A$. Then $B$ is an open set in $A$ because $U$ is an open set in $M$: the topology in $A$ is $$ {\tau}_A = \{E \cap A : E \mbox{ is open in } M\}\mbox{.} $$ Further, it is valid for all topology $\tau$ in $M$; you don't need to consider a metric $D$ in $M$.