Let $(X, \| \|)$ be an NLS, $x\in X$ and $0<r<s$. Show that, $B(x,r)\subsetneq B(x,s)$.

Question

It is an exercise given in my text book of "Topology of Metric spaces by S. Kumaresan"

Note that since in an Norm Linear Space $B(x,r)=x+rB(0,1)$, if we can prove that $B(0,r)\subsetneq B(0,s)$ for $0<r<s$, we are done.
I can prove the above result for $X=\Bbb{R}^2$, where the norm is defined by $\|(x,y)\|:=\sqrt{x^2+y^2}$.(Just take the point $\left(\frac{r+s}{2\sqrt 2},\frac{r+s}{2\sqrt 2}\right)\in B(0,s)\backslash B(0,r)$)
Now I have to generalize it for any NLS $(X,d)$ i.e. we have to find a point $x\in B(0,s)\backslash B(0,r)$ or equivalently $r<\|x\|<s$.
But can't find such an element $x$. How can I solve it?
Thanks for help in advance.

Answer 1

Take $y\in B(x,r)$ such that $y\neq x$. Then $0<\lVert y-x\rVert<r$. Let $z=x+\lambda\frac{y-x}{\lVert y-x\rVert}$, where $\lambda\in(r,s)$. Then $\lVert z-x\rVert=\lambda\in(r,s)$ and therefore $z\in B(x,s)\setminus B(x,r)$.