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# Balls and spheres

Let $(X,d)$ be a metric space with distance $d\colon X \times X \to [0,\infty)$. Two important concepts are the ball of radius $r > 0$ centered at $x_0 \in X$, $B_r(x_0) := \{ x \in X \colon d(x,x_0) < r \},$ and the sphere of radius $r > 0$ centered at $x_0 \in X$, $S_r(x_0) := \{ x \in X \colon d(x,x_0) = r \}.$ For normed spaces, or other vector spaces that are also metric spaces, we simply write $B_r := B_r(0)\quad \text{ and }\quad S_r = S_r(0),$ for balls and spheres centered at the origin (zero element). The sets $B_1$ and $S_1$ are called the unit ball and unit sphere, respectively.

Ex.
• The ball of radius $2$ centered at $(1,0)$ in Euclidean space $\R^2$:

$B_2((1,0)) = \{ (x,y) \in \R^2 \colon (x-1)^2 + y^2 < 4\}.$

• Sequence spaces are spaces in which each element $x = \{x_n\}_{n \in \mathbb N} = (x_1,x_2, \ldots)$ is a sequence (usually of real or complex numbers). The most important ones are the so-called $l_p$-spaces:
• Let $l_\infty$ be the space of sequences $\{x_j\}_{j\in\mathbb N} = (x_1, x_2, \ldots)$ for which $\|x\|_{l_\infty} = \sup_{j \in \mathbb N} |x_j| < \infty.$ Then the sequence $(1/2,2/3,3/4, \ldots) \in S_1$ in $l_\infty$ (since $\sup_{j \in \mathbb N}| \frac{j}{j+1} | = 1$).
• For any $p \geq 1$, let $l_p$ be the space of sequences $\{x_j\}_{j\in\mathbb N} = (x_1, x_2, \ldots)$ for which $\|x\|_{l_p} = \big( \sum_{j \in \mathbb N} |x_j|^p \big)^{1/p}< \infty.$ Let further $e_1 = (1,0,0,\ldots), \quad e_2 = (0,1,0,0\ldots) \quad\text{ and }\quad e_j = (\ldots,0,1,0,\ldots), \quad j \in \mathbb N.$ Then $e_j \in S_1 \quad\text{ for all }\quad j \in \mathbb N,$ but $d(e_i,e_j) = \| e_i - e_j \|_{l_p} = (|1|^p + |-1|^p)^{1/p} = 2^{1/p} \geq 1 \quad\text{ whenever }\quad i \neq j.$ Note that such a sequence of elements could never exist in $\mathbb R^n$ (or any other finite-dimensional vector space).
• The unit ball in $BC([0,1],\mathbb R)$ consists of all functions whose graph $y = f(x)$ lies strictly between the lines $y=\pm 1$.

N.b. The unit ball may look quite different depending on the underlying metric/norm. The following illustration captures this in the case of the $l_p$-norm on $\R^2$. Homogeneity and the triangle inequality however imply that a ball in any metric given by a norm will always be a convex set in the underlying space.