Help showing the closure of open neighborhoods is a closed neighborhood in $\mathbb{R}^n$
In $\mathbb{R}^n$ I'm trying to prove that the closure of an open neighborhood $N_r(x) = \{y : d(x,y)<r\}$ is a closed neighborhood where $\bar{N}_r(x) = \{y : d(x,y)\leq r\}$ denotes the closed set and $\overline{N_r(x)}$ denotes the closure.
To show that $\overline{N_r(x)} \subset \bar{N}_r(x)$ I have that as $\overline{N_r(x)}$ is the smallest set containing $N_r(x)$ and $N_r(x) \subset \bar{N}_r(x)$ then it follows that $\overline{N_r(x)} \subset \bar{N}_r(x)$.
However to show $\bar{N}_r(x) \subset \overline{N_r(x)}$ I'm having a difficult time show how to prove this. Should I take a set $N_{r}(x) = \{y:d(x,y) = r\}$ and say that as this set contains all the points on the boundaries, then I just need to show that there exists elements from both sets $\overline{N_r(x)}$ and $\bar{N}_r(x)$ in that set?
Any advice in the right direction would be gratefully appreciated
$\endgroup$ 101 Answer
$\begingroup$You need to do 2 things: note that $\overline{N}(x,r) = \{y : d(x,y) \le r\}$ is indeed closed, by proving the complement is open (using the triangle inequality).
Then $\overline{N(x,r)} \subseteq \overline{N}(x,r)$ follows,as the left hand side is the smallest closed set containing $N(x,r)$ and $\overline{N}(x,r)$ is a closed set containing it.
You then need to argue that any point in $\overline{N}(x,r)$ is in this closure. The only non-trivial case occurs when $d(x,y) = r$, otherwise $y \in N(x,r)$ already . This one can see by showing that such $y$ are limits along the radius of the ball of points from $N(x,r)$.
$\endgroup$
