Can a ball not containing the origin contain its negatives?

Question

My question is a bit more general than I was able to describe in the title:

Let $(X,\|\cdot\|)$ be a real normed space and $B_r(x)$ be a ball not containing the origin. That is, $0 < r < \|x\|$. I am curious if for all $t \in B_r(x)$ and $a > 0$, we must have $–at \notin B_r(x)$.

Geometrically: If a ball does not contain the origin, can it contain any points on the "directly opposite side" of the origin? Geometric intuition suggests this is impossible, and I was able to prove it for real inner product spaces (see proof below). But I cannot prove it assuming only a normed space, using basic tools like the triangle inequality. Please provide a proof or counterexample.

---

Proof assuming $X$ is a real inner product space:

Let $t\in B_r(x).$ Then

$\|t-x\|<r<\|x\|$

$\Rightarrow\|t-x\|^2<\|x\|^2$

$\Rightarrow\|t\|^2-2\langle t,x\rangle+\|x\|^2<\|x\|^2$

$\Rightarrow\langle t,x\rangle>\frac12\|t\|^2>0$

Hence $\langle t,x\rangle$ is positive for all $t\in B_r(x)$, so $-at\in B_r(x),a>0$ would give $\langle-at,x\rangle<0$, a contradiction.

Answer 1

This follows by showing balls are convex.

If $y$ and $-ay$ are in a convex set $S,$ for some $a>0,$ then letting $t=\frac{1}{1+a}$ we have $0<t<1$ and $1-t=\frac{a}{1+a},$ and: $$t(-ay)+(1-t)y=\frac{-a}{1+a}y+\frac{a}{1+a}y=0,$$ is in $S.$

So you only need to show tha $B_r(x)$ is convex.

If $v_1,v_2\in B_r(x)$ and $t\in [0,1]$ we need to show $v=tv_1+(1-t)v_2\in B_r(x).$ Use:

$$v-x =t(v_1-x)+(1-t)(v_2-x)$$

Answer 2

It is easily shown that a ball is convex. It's immediate from the definition that a ball centered at the origin is convex, and then a general ball is convex since it's a translate of a ball centered at the origin.

Therefore, a ball containing both $at$ and $-at$ contains the origin.