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Show that for any $x\in \mathbb R^n$, $\overline{B}_{r}(x) = \bigcap_i^\infty B_{r+\frac1n}(x)$

For $r< r+ \frac1n$ we have that $B_r(x) \subset\overline{B}_r(x) \subset B_{r+\frac1n}(x)$, But I don’t really see how to go about this. I tried to approach this by inclusion exclusion as follows.

$”\subset”$ Take $y \in \bigcap_i^\infty B_{r+\frac1n}(x)$. This means that $d(x,y) < r+ \frac1n$ for every $n$. But I don’t see how we can get this to $d(x,y) \le r$ which is what we would need. I’m tempted to say that since it holds for every $n$, in particular there would exists $n$ such that $d(x,y) \le r + \frac1n$, but I’m not sure this is true.

For the other direction I didn’t manage to proceed. Any hints for this?

Arctic Char
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Wondera
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    Try to argue by contradiction: if instead $d(x, y) > r$, then .... – Arctic Char Sep 08 '21 at 13:05
  • Perhaps consider the sequence of balls as converging, what does this say about the set where the limit points are: https://math.stackexchange.com/questions/2846902/convergence-of-a-sequence-implies-closed-set – mavavilj Sep 08 '21 at 13:14
  • Also consider that infinite intersectinons and limits are related: https://math.stackexchange.com/questions/535321/infinite-intersection-and-limits – mavavilj Sep 08 '21 at 13:20

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If $n\in\Bbb N$, then $\overline{B_r(x)}\subset B_{r+1/n}(x)$, and therefore $\overline{B_r(x)}\subset\bigcap_{n\in\Bbb N}B_{r+1/n}(x)$.

On the other hand, if $y\notin\overline{B_r(x)}$, you must have $d(y,x)>r$. But then $d(y,x)>r+\frac1n$ for some $N\in\Bbb N$, and so $y\notin\bigcap_{n\in\Bbb N}B_{r+1/n}(x)$.

José Carlos Santos
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  • How come $d(y,x) > r \implies d(y,x) > r + \frac1n$? – Wondera Sep 08 '21 at 13:43
  • I did not write that (which is good, since it makes no sense). What I wrote was that $d(y,x)>r\implies d(y,x)>r+\frac1n$ for some $n\in\Bbb N$. – José Carlos Santos Sep 08 '21 at 13:58
  • I’m trying to determine if this can be done without the contradiction. By taking $y \in \bigcap_n B_{r+1/n}(x)$ I have that $$d(x,y) < r+ \frac1n$$ for some $n$. And now I want to get to $d(x,y) \le r$. From $d(x,y) < r+ \frac1n $ I have that $n > \frac{1}{d(x,y)-r}$, but this doesn’t seem to help... – Wondera Sep 08 '21 at 14:53
  • If $d(y,x)<r+\frac1n$ for every $n\in\Bbb N$, then$$d(y,x)\leqslant\lim_{n\to\infty}r+\frac1n=r.$$ – José Carlos Santos Sep 08 '21 at 15:24