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Let $G$ be open set and $a\in G$. If $\partial G$ is not empty set, can we prove $$\operatorname{dist}(a,\partial G) = \sup \{r: B(a,r) \subset G\}?$$

Daniel Fischer
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Sushil
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1 Answers1

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Let $X = \mathbb{R}\setminus \{0\}$ with the usual metric, and let

$$G = (0,3),\quad a = 1.$$

Then

$$\sup \{ r : B(a,r) \subset G\} = 1 < 2 = \operatorname{dist}(a,\partial G).$$

We always have the inequality

$$\sup \{ r : B(a,r) \subset G\} \leqslant \operatorname{dist}(a,\partial G),$$

but in general, the inequality can be strict.

We have equality in spaces where all balls are connected: Then suppose

$$s = \sup \{ r : B(a,r) \subset G\} < +\infty.$$

For every $\rho > s$, we need to show $B(a,\rho) \cap \partial G \neq \varnothing$. By definition, $B(a,\rho) \not\subset G$. If $B(a,\rho)\cap \partial G$ were empty, then

$$B(a,\rho) = (B(a,\rho)\cap G) \cup (B(a,\rho)\setminus \overline{G})$$

would be a partition of the ball $B(a,\rho)$ into two nonempty open sets, contradicting the connectedness of the ball.

In particular, we have equality in $\mathbb{R}^n$ for all $n$ (with any of the usual metrics), since balls there are connected.

Daniel Fischer
  • 206,697
  • ya got it. Thanks but why it is true in R or R^n? – Sushil Aug 10 '15 at 11:47
  • I just added a sufficient condition for equality to the answer. I don't think there is a manageable non-tautological condition that is necessary and sufficient. – Daniel Fischer Aug 10 '15 at 11:55
  • One more question: Does in a metric space paces all balls are connected implies space is connected? – Sushil Aug 15 '15 at 09:24
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    Yes, for any $a\in X$ we have $$X = \bigcup_{r = 1}^\infty B(a,r),$$ so $X$ is a union of a family of connected sets with nonempty intersection. Such unions are themselves connected. – Daniel Fischer Aug 15 '15 at 09:29