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Denote $X$, the space of all sequences $\in$ $\mathbb R$. I have a metric $$d(x,y):=\sum_{n=1}^\infty 2^{-n}\frac{| x_n-y_n|}{1+| x_n-y_n|}$$

and $(X,d)$ is a metric space.

How would I show that the closed (metric) ball $B[0,\frac{1}{3}]$ with centre $0$ and radius $1/3$ is not convex?

I tried working from the definition that a set $A$ in a vector space $X$ is convex if for all $x,y \in A$, and all $t$ in the interval $[0,1]$, $(1 − t ) x + t y \in A,$ but I cant seem to show its not convex.

I was looking up on this, and I read somewhere it might have something to do with summable sequences like $x=(1/2,1/2,1/2,...)?$ Im not sure though, I might be wrong.

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1 Answer 1

Let $x=\langle 1,0,1,0,\dots\rangle$, and let $z$ be the zero sequence. Then

$$\begin{align*} &d(x,z)=\sum_{n\ge 0}2^{-(2n+1)}\left(\frac12\right)=\sum_{n\ge 0}\left(\frac14\right)2^{-2n}=\\ &=\sum_{n\ge 0}\left(\frac14\right)4^{-n}=\sum_{n\ge 1}4^{-n}=\frac{1/4}{1-1/4}=\frac13\;, \end{align*}$$

so $x$ is on the boundary of the ball $B\left[z,\frac13\right]$.

Let $y_r=\langle 0,r,0,r,\dots\rangle$, where $r>0$; then

$$d(y_r,z)=\sum_{n\ge 1}2^{-2n}\left(\frac{r}{1+r}\right)=\frac{r}{1+r}\sum_{n\ge 1}4^{-n}=\frac{r}{1+r}\cdot\frac13\le\frac13\;,$$

so $y_r$ is in the ball for all $r>0$.

Now let $$u_r=\frac12(x+y_r)=\left\langle \frac12,\frac{r}2,\frac12,\frac{r}2,\dots\right\rangle\;.$$ Then

$$\begin{align*} d(u_r,z)&=\sum_{n\ge 0}2^{-(2n+1)}\left(\frac{1/2}{1+1/2}\right)+\sum_{n\ge 1}2^{-2n}\left(\frac{r/2}{1+r/2}\right)\\ &=\frac13\sum_{n\ge 0}2^{-2n}\left(\frac12\right)+\frac{r}{r+2}\sum_{n\ge 1}2^{-2n}\\ &=\frac16\sum_{n\ge 0}4^{-n}+\frac{r}{r+2}\sum_{n\ge 1}4^{-n}\\ &=\frac16\cdot\frac43+\frac{r}{3(r+2)}\\ &=\frac{5r+4}{9(r+2)}\;, \end{align*}$$

so $d(u_r,z)>\frac13$ iff $5r+4>3(r+2)$ iff $r>1$. In other words, for any $r>1$ you have $x,y_r\in B\left[z,\frac13\right]$ but $u_r=\frac12 x+\frac12 y_r\notin B\left[z,\frac13\right]$. Thus, $B\left[z,\frac13\right]$ is not convex.

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