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Let $X$ be a normed space, let $B_X$ be a closed unit ball of X, and let $C$ be a countable dense subset of $B_X$. Then is that possible $C$ is finite? If possible why?

Thank you!!

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    $\begingroup$ This is possible if and only if $X$ is zero-dimensional. Note that a finite subset of a Hausdorff space is closed, so if it's dense in the unit ball then it must be equal to the unit ball. $\endgroup$ – Qiaochu Yuan Dec 4 '17 at 23:33
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    $\begingroup$ @QiaochuYuan If the unit ball is defined by norm less than or equal to $1$ then this is false even for one dimensional space. $\endgroup$ – Ethan Bolker Dec 4 '17 at 23:37
  • $\begingroup$ Yes, sorry, I had the unit ball confused with the unit sphere. $\endgroup$ – Qiaochu Yuan Dec 4 '17 at 23:38
  • $\begingroup$ I am sorry guys. I don't understand. Could you be more specific? Thank you. $\endgroup$ – Answer Lee Dec 4 '17 at 23:43
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$X$ is a normed space, so it is Hausdorff. In particular, $C$ finite implies $C$ is closed.

$C$ dense means by definition $C = \bar C = B_X $. Thus $B_X$ is finite.

For any non-trivial vector-space, any subset with non-empty interior is infinite. Hence it is only possible when $X$ has dimension $0$.

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  • $\begingroup$ I am sorry I don't get one thing why for any non-trivial vector-space, any subset with non-empty interior is infinite. $\endgroup$ – Answer Lee Dec 7 '17 at 23:12
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Only if $X=\{0\}.$ If $X\ne \{0\}$ let $p\in X$ and let $\overline {B(p,1)}$ be a closed unit ball. Let $0\ne q\in X.$ Let $q'=q/\|q\|.$ Let $n\in \Bbb N.$ Then $$F(n)= \{B(p+q'(4j+1)/4n\; , 1/4n) : j\in \{0,...,n-1\}\}$$ is a family of pair-wise disjoint sets (by the Triangle Inequality) with $n$ members. Members of $F(n)$ are open in $X$ and are subsets of $B(p,1),$ so if $C$ is dense in $B(p,1)$ then $C$ intersects each member of $F(n).$ So for every $n\in \Bbb N,$ there are at least $n$ members of $C.$

Remark: We could also take $F=\{B(p+4^{-n}q',4^{-n-1}): n\in \Bbb N\},$ an infinite pair-wise disjoint family of open balls that are subsets of $B(p,1)$ and observe that $C$ must intersect every member of $F.$

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