Open Balls in ℝ with the Chebyshev Metric

# Open Balls in ℝ with the Chebyshev Metric

Recall from the Open and Closed Balls in Metric Spaces page that if $(M, d)$ is a metric space and $a \in M$ then the open ball centered at $a$ with radius $r > 0$ is defined to be the set:

(1)
\begin{align} \quad B(a, r) = \{ x \in M : d(x, a) < r \} \end{align}

Thus, the open ball $B(a, r)$ is the set of all points in $M$ that have a distance of less than $r$ from $a$.

It is extremely important to note that the open balls may not actually be "balls" geometrically. In fact, if $M$ is a rather abstract set then there may be no geometric intuition behind the open balls in the corresponding metric space. If we instead look at metric spaces on Euclidean space $\mathbb{R}^n$ then there's usually some sort of geometric intuition but once again, the open balls in this metric space may not be geometric balls.

A great example of this is the metric space $(\mathbb{R}^2, d)$ where $d$ is the Chebyshev metric defined for all $\mathbf{x} = (x_1, x_2) , \mathbf{y} = (y_1, y_2) \in \mathbb{R}^2$ by:

(2)
\begin{align} \quad d(\mathbf{x}, \mathbf{y}) = \max \{ \mid x_1 - y_1 \mid, \mid x_2 - y_2 \mid \} \end{align}

We proved that this function $d$ is indeed a metric on the The Chebyshev Metric page.

Now consider the points $\mathbf{x} = (0, 0)$ and $r = 1 > 0$. Then the open ball centered at $\mathbf{x} = (0, 0)$ with radius $1$ is defined to be the set of all points that are of a distance less than $1$ from $\mathbf{x}$, i.e.,:

(3)
\begin{align} \quad B((0, 0), 1) = \{ \mathbf{y} \in \mathbb{R}^2 : \max \{ \mid y_1 \mid, \mid y_2 \mid \} < 1 \} \end{align}

Note that if $\max \{ \mid y_1 \mid, \mid y_2 \mid \} < 1$ then either $\max \{ \mid y_1 \mid, \mid y_2 \mid \} = \mid y_1 \mid$ or $\max \{ \mid y_1 \mid, \mid y_2 \mid \} = \mid y_2 \mid$. If $\max \{ \mid y_1 \mid, \mid y_2 \mid \} = \mid y_1 \mid$ then this implies that $-1 < y_1 < 1$. If $\max \{ \mid y_1 \mid, \mid y_2 \mid \} = \mid y_2 \mid$ then this implies that $-1 < y_2 < 1$. In either case we see that the $B((0, 0), 1)$ is geometrically an open box centered at $\mathbf{x}$: So, even though we define $B(a, r)$ to be an "open ball" it is important to note that this is just terminology! This will be important later on.