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Let $(S,d)$ be a metric space, $\mathcal{S}$ the induced topology. $A\subset S$ a subset. It is easy to see that $A\cap\mathcal{S}=\mathcal{A}$, i.e., the topological subspace on $A$ is the topological space induced by the metric restricted to $A$, by arguing that each open set in $\mathcal{A}$ (topology induced by $d$ restricted to $A$) is union of open balls in $A$ which by definition are open balls in $S$ intersecting $A$.

In Kallenberg's Foundations of Modern Probability, He has a simple argument that in fact, any $B\in\mathcal{A}$ can be written as $B=A\cap(B\cup A^c)^\circ$, where $^c$ means complement, and $^\circ$ means interior, i.e., explicitly as intersection of $A$ with some open set in $S$. While it makes intuitive sense by drawing a picture, I have trouble proving that this is the case. Any suggestions are welcome.

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  • $\begingroup$ By $A^c$ do you mean the complement of $A$ or the closure of $A$? $\endgroup$ – Tim Raczkowski Aug 27 '15 at 0:10
  • $\begingroup$ $A^c$ means the complement, while $A^\circ$ is the interior $\endgroup$ – user138668 Aug 27 '15 at 0:12
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Let $d'$ be the metric restriced to $A$, $B_d(x,\epsilon)=\{y\in S:d(x,y)<\epsilon\}$ and $B_{d'}(x,\epsilon)=\{y\in A:d'(x,y)<\epsilon\}$

$\underline{A\cap(B\cup A^c)^\circ\subseteq B}$

Suppose $x\in A\cap(B\cup A^c)^\circ$. So $x\in A$ and $x\in(B\cup A^c)^\circ$. Hence there is an $\epsilon>0$ such that $B_d(x,\epsilon)\subseteq B\cup A^c$. Because, $x\notin A^c$, we must have $x\in B$.

$\underline{B\subseteq A\cap(B\cup A^c)^\circ}$

Let $x\in B$. Since $B$ is open in $A$ there is an $\epsilon>0$ such that $B_{d'}(x,\epsilon)\subseteq B$. This implies $B_d(x,\epsilon)\subseteq B\cup A^c$. So we see that $x\in A$ and $x\in (B\cup A^c)^\circ$.

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  • $\begingroup$ Thanks. It is crystal clear now. $\endgroup$ – user138668 Aug 27 '15 at 1:02
  • $\begingroup$ Glad I could help. $\endgroup$ – Tim Raczkowski Aug 27 '15 at 1:05

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