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I know it is possible, for instance if we consider a non empty set $X$ with the discrete metric, then for each $x \in X$ the balls $b[x;r)$ for $r \in (0,1]$ are equal to the singleton set $\{x\}$. Also the balls $b[x;r)$ for $r \in (1,\infty)$ are equal to $X$ for all $x \in X$.

What is the idea behind two balls with different radii and centre's being equal? What I don't understand is, based on the above example, in what sense are the two balls equal?

What is the meaning of equality of two balls in a metric space?

In this example one ball has only singleton element $\{x\}$ and the other one is the whole metric space $X$ then how are they equal?

I am a little confused!

Michael Albanese
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johny
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2 Answers2

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They are equal in the usual way two sets are equal: they have the same members.

Robert Israel
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You're getting confused. In a discrete metric space

$$b(x, r) = \begin{cases} \{x\} &\ \text{if}\ r \in (0, 1]\\ \\ \,\,X &\ \text{if}\ r \in (1, \infty). \end{cases}$$

The claim is not that $b(x, r_1) = b(x, r_2)$ for any $r_1 \in (0, 1]$ and any $r_2 \in (1, \infty)$; this is indeed false. Instead the claim is that for any $r_1, r_2 \in (0, 1]$, $b(x, r_1) = b(x, r_2)$, and, for any $R_1, R_2 \in (1, \infty)$, $b(x, R_1) = b(x, R_2)$.

Michael Albanese
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  • @ Albanese Well i just wanted to clarify one point that has been in my mind for some time now, when we say that, for each x∈X the balls b[x;r) for r∈(0,1] are equal to the singleton set {x}, then the equality is only in terms of these sets being singleton, and not being the same singleton set, right ? – johny Oct 11 '13 at 11:16
  • No, they are the same singleton set. All the balls have the same centre, just different radii. So $b(x, 1) = {x}$, $b(x, \frac{1}{2}) = {x}$, and so on. – Michael Albanese Oct 12 '13 at 14:11