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The problem:

Let $Y$ be a subspace of a metric space $X$, and let $A$ be a subset of the metric space $Y$. Show that $A$ is open as a subset of $Y$ $\Leftrightarrow$ it is the intersection with $Y$ of a set which is open in $X$.

My work so far:

$\Leftarrow$ If $A$ is the intersection with $Y$ of a set which is open in $X$, then given any $x\in A$ there exists $r>0$ such that $S_r(x)\subseteq A$, let $Q=S_r(x) \cap Y$ then for all $q\in Q$, $d(x,q)<r$ thus $Q$ is an open sphere in $Y$.

$\Rightarrow$ If $A$ is open as a subset of $Y$ then for $x\in A$ there is $r>0$ such that $S_r(x)\subseteq Y$. Suppose there exists $z\in X$ such that $d(x,z)<r$, $z\notin Y$, ...

My question:

Does this argument make sense, and how can I complete the proof of $\Rightarrow$? I find the notion of "an intersection with $Y$ of a set which is open in $X$" confusing since I'm not sure how to construct this set.

[For reference, this is Problem 10-7 of Introduction to Topology and Modern Analysis by G.F. Simmons.]

Suzu Hirose
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  • Basically what you're trying to show is that the topology induced by the metric $d$ restricted to $A$ is the same as the subspace topology on $A$. Since any open set in a metric space can be written as the union of open balls, it suffices to only consider open balls - that should help. – Math1000 Nov 13 '14 at 12:52

1 Answers1

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Your argument for the direction $\Leftarrow$ is correct. So let's do the reverse direction.

$\Rightarrow$ Suppose $A$ is an open subset of $Y$. Then, for every $x\in A$, one can find $r_{x}>0$ such that $$ \{y\in Y: d(y, x)<r_{x}\}\subset A $$ As a result, $$ A=\bigcup_{x\in A} \{y\in Y: d(y, x)<r_{x}\} $$ Now let $O=\displaystyle\bigcup_{x\in A} \{y\in X: d(y, x)<r_{x}\}$. Then $O$ is open in $X$, and satisfies $A=Y\cap O$.

Prism
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