$X$ is a normed space and for any $x \in X$ and $r \gt 0$, let $T=\{y\in X: \|y-x\|\le r\}$ and $S=\{y\in X: \|y-x\|\lt r\}$, then $\bar{S}=T$.

Question

Let $X$ be a normed linear space and, for any $x \in X$ and $r \gt 0$, let $T=\{y\in X: \|y-x\|\le r\}$ and $S=\{y\in X: \|y-x\|\lt r\}$.

If $z\in T$ and $z_n=(1-n^{-1})z$, for $n\in \mathbb{N}$, show that $\lim z_n=z$ and hence show that $\bar{S}=T$.

Below is the solution to the problem. But I think it's wrong. It shows that $\|z_n\|\lt r$, but we need $\|z_n-x\|\lt r$, where $x$ is a fixed point in $X$, to show that $z_n \in S$. How can I show this? Perhaps the problem is wrong? I would greatly appreciate any help.

Perhaps your choice for $z_n$ is not actually in $S$. For example let $S$ be the interior of the circle $(x-2)^2+y^2=1$ in the Cartesian plane and let $z$ be the point $(1,0)$ on the circle. What are the coordinates of $z_n=\left(1-\frac{1}{n}\right)z$? Is it inside the circle? — John Wayland Bales, Mar 12 '16 at 06:29
@JohnWaylandBales It's not in the circle, so this problem is wrong then. — nomadicmathematician, Mar 12 '16 at 06:31
@takecare Do you know how to express a point $P$ on the segment connecting $x$ and $z$ in terms of a parameter $t$ between 0 and 1? That might be of use using $\frac{1}{n}$ rather than $t$ and $z_n$ instead of $P$. — John Wayland Bales, Mar 12 '16 at 06:45
@JohnWaylandBales $tx+(1-t)z$, but how is that of use here? We need to use $1/n$ to form a sequence. And are you saying $T=\bar{S}$ is correct here? — nomadicmathematician, Mar 12 '16 at 06:47
And I think you meant $P=tx+(1-t)z$ but use $z_n$ rather than $P$ and $\frac{1}{n}$ rather than $t$. — John Wayland Bales, Mar 12 '16 at 06:50
@takecare You should be able to show that $||x-z_n||<r$ and that $||z-z_n||<\frac{r}{n}$. — John Wayland Bales, Mar 12 '16 at 06:53
@JohnWaylandBales Yes I got it, so this way $z_n\to z$ and $z_n\in S$, hence $z$ is an element of $T$ in $\bar{S}$. Thanks a lot!! — nomadicmathematician, Mar 12 '16 at 06:54

$X$ is a normed space and for any $x \in X$ and $r \gt 0$, let $T=\{y\in X: \|y-x\|\le r\}$ and $S=\{y\in X: \|y-x\|\lt r\}$, then $\bar{S}=T$.

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