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Let $A\subseteq X$. If $d$ is a metric for the topology of $X$, show that $d\restriction_{A\times A}$ is a metric for the subspace topology on $A$.

I've shown that $d'=d\restriction_{A\times A}$ is a metric on $A$.

I am letting $\tau_B$ denote the subspace topology on $A$ that is induced by the metric topology on $X$ and $\tau_{B'}$ denote the topology on $A$ induced by the metric $d'$. My goal is to show $\tau_B=\tau_{B'}$.

Let $B=\{A \cap B_\varepsilon^d(x) \mid x \in X, \varepsilon>0\}$, a basis for $\tau_B$ and $B'=\{B_\varepsilon^{d'}(a) \mid a \in A\}$, a basis for $\tau_{B'}$. I've shown $B' \subseteq B$ and so $\tau_{B'} \subseteq \tau_B$.

For showing $\tau_B \subseteq \tau_{B'}$ I've picked some $a \in A$ and $A\cap B_\varepsilon^d(x)$, a basis element of $\tau_B$ such that $a \in A\cap B_\varepsilon^d(x)$. My goal is to find some basis element $C$ of $\tau_{B'}$ such that $a \in C \subseteq A\cap B_\varepsilon^d(x).$ If $x \in A$, then $$A\cap B_\varepsilon^d(x)=B_\varepsilon^{d'}(x)$$ and so I have my needed basis element $C$. The part where I have been having trouble is if $x \notin A$. In this case, what I have tried so far is to see if I can find some $\delta >0$ such that $$a \in B_\delta^{d'}(a) \subseteq A\cap B_\varepsilon^d(x)$$ but in searching for a delta and trying to show via set containment that the $\delta$-ball is contained in the $\varepsilon$-ball intersected with $A$ has been difficult. I've tried using triangle inequality but I am running into a problem of not knowing how to show that if $y \in B_\delta^{d'}(a)$, meaning $d'(a,y)$ that $y \in A\cap B_\varepsilon^d(x)$, mainly that $d(x,y) < \varepsilon$. It seems with the different $\delta$'s I have tried, I can use the triangle inequality to show something like $d(x,y) < \frac{3}{2}\varepsilon$ but not quite $\varepsilon$.

2 Answers2

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Suppose that $a\in A\cap B_\epsilon^d(x)$ for some $x\in X$. You know that $\{B_r^{d'}(a):r>0\}$ is a local base at $a$ in $A$, so there must be some $\delta>0$ such that $B_\delta^{d'}(a)\subseteq B_\epsilon^d(x)$, and hence $B_\delta^{d'}(a)$ is a $d'$-nbhd of $a$ contained in $A\cap B_\epsilon^d(x)$. Since you can do this for each $a\in A\cap B_\epsilon^d(x)$, $A\cap B_\epsilon^d(x)\in\tau_{B'}$.

Yulia Alexandr
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Brian M. Scott
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Let be $(X,d)$ a metric space and for any $Y\in\mathcal P(X)$ let be $\delta:Y\times Y\rightarrow\Bbb R$ the restriction on $Y\times Y$ of $d$ that is $$ \delta(x,y):=d(x,y) $$ for any $x,y\in Y$. So clearly the function $\delta$ is a metric so that we let to prove that the topology $\mathcal T_\delta$ generated by $\delta$ is just the subspace topology $\mathcal T_d|_Y$ induced by $d$ on $Y$ and exactely we let do this showing that a base for $\mathcal T_\delta$ is even a base for $\mathcal T_d|_Y$ but this is trivially because if $B_\delta(x,\epsilon)$ and $B_d(x,\epsilon)$ are two open balls of radius $\epsilon$ centered at $x\in Y$ with respect $\mathcal T_\delta$ and $\mathcal T_d$ then $$ B_\delta(x,\epsilon):=\{y\in Y:\delta(x,y)<\epsilon\}=\{y\in Y:d(x,y)<\epsilon\}=B_d(x,\epsilon)\cap Y $$ and it is a well know result that if $\mathcal B$ is a base for a topology $\mathcal T$ on $X$ then $$ \mathcal B_Y:=\{B\cap Y:B\in\mathcal B\} $$ is a base for the induced topology on any $Y\subseteq X$.

Antonio Maria Di Mauro
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