Show that in a Normed linear Space $X,\overline {B(x,r)}=B[x,r]$

Question

where $\overline {B(x,r)}$ is closure of the set $\{y\in X:||y-x||<r\}$ and $B[x,r]=\{y\in X:||y-x||\leq r\}$

$\overline {B(x,r)}\subseteq B[x,r]$ follows easily but how to do the converse?

score 4 · Accepted Answer · answered Mar 07 '15 at 10:35

4

For any point $y\in B[x,r]$, let $y_n=x+(1-\frac{1}{n})(y-x)$. Then $\{y_n\}$ is a sequence in $B(x,r)$ converging to $y$. Therefore $y$ must be in $\overline{B(x,r)}$ by definition of the closure.

answered Mar 07 '15 at 10:35

AMPerrine

3,043

good answer man! – Learnmore Mar 07 '15 at 13:14

Show that in a Normed linear Space $X,\overline {B(x,r)}=B[x,r]$

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