# May a 'ball' that has been 'cut off' still be called a 'ball'?

Consider the metric subspace $[0,1] \subseteq \mathbb{R}$ with the metric defined in the usual sense, and the ball $B(0,1)$, defined to be the ball centred at $x=0$ with radius $1$.

Now since only the right half of the ball 'exists' within our subspace, then can we still say that the ball $B(0;1)$ exists?

• Side note: sounds pretty harsh in its current form. – barak manos May 25 '16 at 12:50
• Shouldn't you be using the word sphere, and therefore the answer is it's a hemisphere? – Mazura May 25 '16 at 20:08
• @Mazura Using centre $0$ and radius $1$, I imagine sphere ($S$) would fit the set $S(0,1)=\{x\in \Bbb{R} \mid |\,x\,|=1 \}$, rather than ball ($B$), which is $B(0,1)=\{x\in \Bbb{R} \mid |\,x\,|<1 \}$. – snulty May 25 '16 at 21:05

I'd say so yes. The ball just denotes the set $B(0;1):=\{x\in X\mid d(0,x)\leq 1\}$. It has nothing to do with the geometric intuition one might have about balls

Yes, in any metric space, you may define the ball of any center and radius.

In your example, the ball of center $0$ and radius $1$ in the metric space $[0,1]$ is $[0,1)$.

I think expecting the same intuitive result for open balls in different metric spaces isn't a good idea in general.

For example, take $\Bbb R$ with the discrete metric $d(x,y)=0$ iff $x=y$ and $d(x,y)=1$ otherwise. Then $B(0,r)=\Bbb R$ if $r>1$ and $\{0\}$ if $r\leq 1$.

Of course it's quite unintuitive that what's in the ball doesn't depend on the radius except whether it is bigger or smaller than $1$. They're still the open balls in that space.

You can however expect to find the same open sets in two equivalent metrics, like $\Bbb R^2$ with $d_E$ the standard Euclidean metric and $d_\infty(x,y)=\max\{|x_1-y_1|, |x_2-y_2|\}$.

The Euclidean metric gives open balls that look like discs, and I believe that with $d_\infty$ they look like squares, but all of the open sets are the same.

Lastly if you took $(0,1)$ as a metric subspace of $\Bbb R$, you would have a ordinary open intervals as open balls. However the fact that some of the open balls are 'missing' half of their usual elements could either be intuitively attributed to the fact that you're using a subspace hence missing elements, or in another way to the fact that you're now using a bounded metric space, so after some finite radius you get no new elements to an open ball.

Again though that intuition breaks down in other examples.

• Another example of a metric where balls have counterintuitive shapes: let $d_E$ be the Euclidean distance function, and let $d(x,y) = d_E(x,y)$ IFF $d_E(x,y) \leq 1$ and $d(x,y) = \frac{1}{2} + \frac{1}{2d_E(x,y)}$ otherwise. (This does in fact meet the requirements for a distance-function defining a metric in $\mathbb{R}^n$.) Then, for balls where $\frac{1}{2} < r < 1$, the shape of the ball is the entire space minus a "thick shell." – Kyle Strand May 25 '16 at 22:37
• That's a very interesting example, thanks for that! – snulty May 25 '16 at 23:07
• Thank you! I came up with it as part of a thesis paper on metric spaces I wrote in college. I was always kind of afraid I'd be the only person who found the concept of a continuously-decreasing distance-function interesting. – Kyle Strand May 25 '16 at 23:08