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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

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  • $\begingroup$ There's a confusion between $\bar N_r (x)$ and $\overline {N_r (x)}$. If you can tell me which one of these is the closure, and which is not, then we can clarify things together and answer your question. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Mar 3, 2017 at 8:51
  • $\begingroup$ @астонвіллаолофмэллбэрг $\bar{N}_r(x)$ is the closed neighborhood while $\overline{N_r(x)}$ is the closure. I'll make an edit and be more specific. $\endgroup$
    – ghost
    Mar 3, 2017 at 8:52
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    $\begingroup$ Ok, thank you for the clarification. The rest of the question is excellent. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Mar 3, 2017 at 8:53
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    $\begingroup$ I apologize, I am unable to understand what you have done so far. I think you took $y$ on the boundary of the open ball, and if you take any ball around $y$, it will intersect the open ball at some point. This shows that $y$ is a boundary point. Would you like me to elaborate? $\endgroup$
    – Sarvesh Ravichandran Iyer
    Mar 3, 2017 at 9:15
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    $\begingroup$ Excellent. So we've got through this together. Nevertheless, if you have any doubts, do clarify them. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Mar 3, 2017 at 9:23

1 Answer 1

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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 $N(x,r)$.

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