III.5 · Metric Spaces — Problem 5 #

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Problem. Find a metric space $(S,\rho)$, points $x,y \in S$, and radii $r_1 > r_2 > 0$ such that $B_{r_1}(x) \subseteq B_{r_2}(y)$ — a ball of larger radius contained in a ball of smaller radius.

Hint #

Look for a metric in which all sufficiently large balls are the same set. The discrete metric does exactly this.

Solution #

The discrete metric is useful for once! Let $(S, \rho)$ be a metric space with the discrete metric, that is,

$$\rho(a,b) = \begin{cases} 0 & a = b, \\ 1 & a \ne b. \end{cases}$$

For any $p \in S$, both $B_2(p)$ and $B_3(p)$ equal $S$ (every point is within distance $1 < 2$ of $p$). So with $x = y = p$, $r_1 = 3$ and $r_2 = 2$, we have $B_3(p) \subseteq B_2(p)$ and $r_1 = 3 > 2 = r_2$, as desired.

Long solution #

In the discrete metric every distance is $0$ or $1$, so enlarging the radius past $1$ adds nothing to a ball.

Discrete metric space: all points at distance 1 from p, with the balls of radius 2 and 3 both equal to S.

Every point of $S$ is at distance $\leq 1$ from $p$, so $B_2(p)$ (orange) and $B_3(p)$ (dashed) are both all of $S$.

Let $S$ be any nonempty set with the discrete metric $\rho$ above. Fix $p \in S$ and $r > 1$. For every $s \in S$ we have $\rho(s,p) \le 1 < r$, so $s \in B_r(p)$; hence $B_r(p) = S$ whenever $r > 1$. Taking $x = y = p$, $r_1 = 3$, $r_2 = 2$,

$$B_{r_1}(x) = S = B_{r_2}(y),$$

so certainly $B_{r_1}(x) \subseteq B_{r_2}(y)$ with $r_1 > r_2$. $\blacksquare$