Show that $\tau_A=\{U\cap A: U\in\tau\},A\subset X$, where $\tau$ is all open sets in $(X,d)$ is equal to the topology induced by $d|_{A\times A}$

Question

Let $(X,d)$ be a metric space and let $A\subset X$. Show that the topology $\tau_A=\{U\cap A: U\in\tau\}$ induced by the metric topology of $X$, $\tau$ (meaning $\tau$ is a collection of all subsets of $X$ open with respect to $d$), is equal to the metric topology of $A$ with the restriction $d|_{A\times A}: A\times A\to \Bbb{R}_{+}$.

(If I used any term incorrectly, or wasn't clear in the way I put it, it was due to a lack of the relevant terminology in English.).

I almost thought I can handle this question, that is, until I realized such a subset, $A\subset X$, can be closed. Suppose there is some $V\in \tau_A$. It is an intersection of $A$ and some open set $U$ in the metric topology $\tau$ of $X$. Before I can show it can be represented as a union of open balls with respect to the restriction of $d$ to $A\times A$, I have to clarify to myself that it is actually open in $A$. I mean, what if $\partial A$ intersects $V$? I am totally confused and could use some guidance.

Answer 1

Let $V \in \tau_A$, and choose $x \in V$.

By the definition of $\tau_A$, there must exist some $U \in \tau$ such that $V = U \cap A$. In particular, since $\tau$ is the topology induced by $d$, there is some $r>0$ such that $d(x,y)

Fix some $x\in A$, and let $r>0$. Notice that

$$ \{ y \in A : d\vert_{A \times A}(x,y) < r \} = \{ y \in X : d(x,y)

This means that the open ball (with respect to $A$) around $x$ is in fact a member of $\tau_A$.

To (hopefully) answer the concern you outline at the end of your question: your normal intuitions about what sets are open and closed in the ambient space don't necessarily always apply.

To be more specific, consider the topology induced on $[0,1]$ by the standard topology on $\mathbb R$. Sets of the form $(a,1]$ where $a\ge0$ are open in $[0,1]$ (but clearly not in $\mathbb R$).