Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In a metric space $X$ for all $x \in X, r > 0$ following is true: $B(x,r) \subseteq \overline{B(x,r)} \subseteq \overline{B}(x,r)$. Here $\overline{B(x,r)}$ is the closure of the open ball of center $x$, radius $r$ and $\overline{B}(x,r)$ is the closed ball of center $x$, radius $r$. The discrete metric gives an example where $B(x,r) = \overline{B(x,r)} \subsetneq \overline{B}(x,r) \ $ by having $X$ be a set with at least two elements, $x \in X$ and $r = 1$. My question is, is there an example where you have proper inclusion between all three sets?

share|cite|improve this question
up vote 11 down vote accepted

How about $[0,1]\cup\{2\}$, with the standard distance and $B(1,1)$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.