0

Suppose $(X,d)$ is a metric space. Take two points $x_1, x_2 \in X$ and take $r_1, r_2 \in \mathbb{R}^+$. Then $B_1(x_1, r_1)$ and $B_2(x_2, r_2)$ are open balls. Suppose that $B_1(x_1, r_1) \subseteq B_2(x_2, r_2)$. Does this imply that $r_1 \le r_2$?

Luke Poeppel
  • 439
  • 1
  • 5
  • 15
Dagny Taggart
  • 1
  • 1
    What have you tried? It's not that difficult to make progress, draw a picture. – Qi Zhu Apr 17 '20 at 18:46
  • 1
    The answer is no, you can look here for several examples: https://math.stackexchange.com/questions/734248/example-of-two-open-balls-such-that-the-one-with-the-smaller-radius-contains-the – Salech Alhasov Apr 17 '20 at 18:48

1 Answers1

3

It is not true in general. For example, let us consider the discrete metric space $(X,d)$, where $d$ is the discrete metric on the set $X$ defined by $d(x,y)=\delta_{xy}$ ($\delta$ stands for Dirac delta function). Then for any $0<r_1,r_2<1,\,B(x,r_1)=B(x,r_2)=\{x\}$ for each $x\in X$.

Surajit
  • 1,216
  • Is the proposition true if I have a normed space? – John Mars Apr 27 '21 at 13:54
  • 1
    @JohnMars If $X$ is a nonzero normed linear space over the scalar field $\mathbb{R}$ or $\mathbb{C}$, then the proposition is true, because then diameter of an open ball $B(x,r)$ is $2r$. See https://math.stackexchange.com/q/1142549/528430 – Surajit Apr 29 '21 at 05:21