What is a 0-ball?
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?
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$\begingroup$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...
$\endgroup$ $\begingroup$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 $\bar B^0=B^0=\{\emptyset\}=\Bbb R^0$.
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