Metric spaces proof, open and closed sets.

In my book, I have proof of open and closed sets theorem. I think author made a mistake, I highlighted it red.

($\langle X,\rho \rangle$ is a metric space, $X \setminus A$ is complement of $A$ and $B_r(x)$ is open ball of radius $r$ and center $x$)

Let $A$ be closed set and $x \in X \setminus A$. If $x$ wasn't interior point of $X \setminus A$, then for each $r > 0$ exists such $\color{red}{y \not \in A}$ that $y \in B_r(x)$.

I think it should be $\color{green}{y \not \in X\setminus A}$. By definition,
$x_0$ is interior point of $A$ if and only if $\exists r>0 \quad B_r(x_0) \subseteq A$
Inclusion can be written as $\forall x(x \in B_r(x_0) \Rightarrow x \in A)$. So $x_0$ is not an interior point of $A$ if,
$\neg(\exists r>0 \space \forall x(x \in B_r(x_0) \Rightarrow x \in A)) \equiv \forall r>0 \space \exists x (x \in B_r(x_0) \land x \not \in A)$
In our case we are looking at set $X\setminus A$, so it should be $x \not \in X\setminus A$.

Am I right?

• Yes, it should be one of $y\in A$ or $y\notin X\setminus A$. – Daniel Fischer Dec 29 '16 at 18:27
• @DanielFischer Okay, thanks for response! – chandx Dec 29 '16 at 18:40