I am confused about closeness and openness in subspace. My definition says: let Y be a subset of X, and A a subset of Y, then A is closed/open in Y if A is the intersection of Y with a closed/open set B in X.
My question is, if we view Y as the metric space with the restricted metric from X and get rid of X, can we say the intersection of B (closed/open) with Y is closed/open in Y?
Best Answer
The topology $T_d(X)$ induced (or generated) by a metric $d$ on X is, of course, the set of all subsets of $X$ that are unions of open $d$-balls of $X.$ And when $Y \subset X,$ the subspace topology on $Y,$ induced by $T_d(X)$ is $\{Y\cap t:t\in T_d\},$ which I will denote as $T_Y.$
The Q is whether $T_Y=T_d(Y),$ where $T_d(Y)$ is the topology on $ Y$ induced by the metric $d$ restricted to $Y.$
The answer is YES.
(I). Any $B^Y_d(y,r)=\{y'\in Y: d(y',y)\le r\},$ that is, any open $d$-ball of $Y,$ belongs to $T_Y$ because $B^Y_d(y,r)=Y \cap \{x\in X: d(x,y)<r\}\in T_Y.$ So any union of open $d$-balls of $Y,$ that is, any member of $T_d(Y),$ also belongs to $T_Y.$
Therefore $T_d(Y)\subset T_Y.$
(II). But there may be an open $d$-ball in $X$ whose intersection with $Y$ is not an open $d$-ball of $Y.$ So the reverse inclusion $T_d(Y)\supset T_Y$ may not be obvious.
Any $ V\in T_Y$ is equal to $Y\cap V'$ where $V'$ is an open subset of $X.$ So for any $y\in V$ there is an open $d$-ball $B(y)$ of $X, $ centered at $y,$ such that $y\in B(y)\subset V'.$ And $Y\cap B(y)$ is an open $d$-ball of $Y,$ that is, $Y\cap B(y)\in T_d(Y).$
So $V=\cup_{y\in V} (Y\cap B(y))$ is a union of members of $T_d(Y),$ so $V\in T_d(Y).$
Therefore $T_Y\subset T_d(Y). $
Example. Let $X=\Bbb R$ with $d(x,x')=|x-x'|.$ Let $Y=(0,2)\cup (4,6).$ Then $(1,5)$ is an open $d$-ball of X but its intersection with $Y$, which is $(1,2)\cup (4,5), $ is not an open $d$-ball of $Y$. But $(1,2)$ and $(4,5) $ $ are $ open $d$-balls of $Y.$