# Brief moment from theorem 8.5 on PMA Rudin

He want to prove that $A$ is open. He take any $x_0\in A$ and showed that exists $\varepsilon:=R-|x_0|$ such that for any $x\in N_{\varepsilon}(x_0)$ we have $f(x)=0$. $A$ is the set of all limit point of $E$ in $S$.

How he conclude that $N_{\varepsilon}(x_0)\subset A$?

Can anyone explain this brief moment please.

• Dear users. Please help! – Raheem Najib Dec 15 '15 at 20:16

He proved that $N_{\epsilon}(x_0)\subseteq E$ and $A$ the set of all limit points of $E$ in $S=(-R,R)$. Hence $N_{\epsilon}(x_0)\subseteq A$.
Let $p\in N_{\epsilon}(x_0)$ then $p\in (x_0-\epsilon,x_0+\epsilon)$ and for any $\delta>0$ we have that $N'_{\delta}(p)\cap N_{\epsilon}(x_0)\neq \varnothing$ $\Rightarrow$ $N'_{\delta}(p)\cap E\neq \varnothing$. Then $p$ is a limit point of $E$ in $S$, i.e. $p\in A$. Thus $N_{\epsilon}(x_0)\subseteq A$.