6

I'm reading a paper that says $\bigcap V_{T,X}$ is either empty or a closed $l$-ball where $T \subset S$ is a subset of points $S$ and $\operatorname{card}{T} = m + 1 - l$ where $m$ is the dimension of the smooth manifold $\Sigma$ that the points $S$ are sampled from. My question is what happens when $l=0$, meaning that $\operatorname{card}{T} = m+1$. In this case the intersection must be a closed $0$-ball. How is a $0$-ball defined and what does it mean for an intersection of two topological spaces to be a $0$-ball?

Mario Carneiro
  • 27,438
user104605
  • 73
  • 2
    I guess, $0$-ball must be a point. – Berci Nov 01 '13 at 00:21
  • Please note that you received contradicting answers. That is because you haven't given us the context of your question, and 'balls' are used in many different contexts. Some questions that come to mind: what's $X$ and how does it relate to $\Sigma$? What are the $V_{T,X}$'s? Over what is the intersection taken? – Jonathan Y. Nov 01 '13 at 08:49

2 Answers2

5

A zero-ball is a point.

Interestingly, however, the zero-sphere consists of two points! Indeed, the zero-sphere is the boundary of the $1$-ball, which is an interval...

Bruno Joyal
  • 54,711
3

The closed $0$-ball $\bar B(x,0)$ is the singleton $\{x\}$; the open $0$-ball $B(x,0)$ is empty. (Note that in this case $\bar B$ is just notation, since it is not the closure of $B(x,0)$.)

Edit: If the designation $0$ is for the dimension and not the radius, we generalize the definition of the $n$-ball: $B^n=\{x\in\Bbb R^n:|x|=\sqrt{\sum_i x_i^2}<1\}$ (where the closed ball has weak inequality) to dimension $0$. $\Bbb R^0$ is the set of all $0$-tuples, of which there is only one, $\langle\rangle=\emptyset$ (the empty sequence). we have $\sum_i x_i^2=0$ (the empty sum), so both $\bar B^0$ and $B^0$ contain this point (since $0\le1$ and $0<1$). Thus $\bar B^0=B^0=\{\emptyset\}=\Bbb R^0$.

Mario Carneiro
  • 27,438