Let be $A \subset \mathbb{R}^n$ an open set. Show that $B := \{ x \in A \ ; \ d(x,\partial A) \geq \frac{1}{n}, n \in \mathbb{N} \}$ is compact.
I don't have idea how to prove this, I would like to a hint. Thanks in advance!
EDIT1: $n$ is fixed.
EDIT2: I was reading "Calculus on Manifolds" by Spivak and there he assumes that $A_i := \{ x \in A \ ; \ |x| \leq i \ and \ distance \ of \ $x$ \ to \ boundary \ of \ A \geq \frac{1}{i} \}$ is compact for $A$ an open set in $\mathbb{R}^n$. I supposed that $B$ is compact, so the intersection of $A$ and $\{ x \in A \ ; \ |x| \leq i$ would be compact and I thought that the reason for $A_i$ being compact, sorry for not put the full question from the beginning, but nevertheless how to prove that the intersection of this sets is compact?
In general $B$ is not bounded, so it cannot be compact except in particular cases. Just to make an example, let $A$ be an open half-space, say the points having the positive first coordinate and the other coordinates are arbitrary: for any $s>0$ you can find $x\in A$ so that $d(x,\partial A)>s$.
On the other hand $$ A_n=\{x\in A:|x|\le n, d(x,\partial A)\ge\tfrac{1}{n}\} $$ is compact. Indeed the set $$ A'_n=\{x\in A:d(x,\partial A)\ge\tfrac{1}{n}\} $$ is closed.
Suppose $(x_k)$ is a sequence in $A'_n$ which converges to $x$. Then $x$ belongs to the closure of $A$ so it either belongs to $A$ or to $\partial A$. However, the latter case cannot happen, because otherwise the distance $d(x_k,x)$ could be made smaller than $1/n$.
Therefore $x\in A$ and $d(x,\partial A)\ge 1/n$ by continuity of $z\mapsto d(z,\partial A)$.
Hence $A_n$ is the intersection between the closed set $A'_n$ and of the closed sphere of radius $n$, which is compact. So $A_n$ is closed and bounded, hence compact.