Let $(M,d) $ be metric space and $x$ is in $M$. Can we find some open ball $B(x,r)$ such that closure of each balls $ B(x,s)$ contained in $B(x,r)$ whenever $ s <r$?
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Can we show that closure of $B \left( x, s \right)$ is $B \left[ x, s \right]$ which is a closed ball? – Aniruddha Deshmukh Nov 13 '18 at 19:06
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No. Not in general metric space (take discrete space). But true in normed space.. – CHOUDHARY bhim sen Nov 13 '18 at 19:08
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Ok. Working on it! – Aniruddha Deshmukh Nov 13 '18 at 19:09
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@user439199 I think the statement is true for discrete space since $B[x,r]=B(x,r)$. – Levent Nov 13 '18 at 19:10
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@user439199 In a discrete metric space, every subset is both open and closed. So, the statement mentioned above is also true! – Aniruddha Deshmukh Nov 13 '18 at 19:16
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Well, you may proceed this way.
Let $B \left( x, r \right)$ be given. Let $s < r$. Then, we need to show that $\overline{B\left( x, s \right)} \subseteq B \left( x, r \right)$. So, let $y \in \overline{B \left( x, s \right)}$.
Then, $\forall \epsilon > 0, \exists z \in B \left( x, s \right)$ such that $d \left( y, z \right) < \epsilon$. Hence,
$$d \left( x, y \right) \leq d \left( x, z \right) + d \left( y, z \right) < s + \epsilon$$
Hence, $d \left( x, y \right) \leq s < r$ and $y \in B \left( x, r \right)$. Hence the proof!
Aniruddha Deshmukh
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Ok i got that ..becoz closure of any open ball can not be larger than closed ball – CHOUDHARY bhim sen Nov 13 '18 at 19:34