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let $(X,d)$ be a metric space . Let $A \subset X$ be bounded

$1.)$ Is $d(A) = d( \bar A)$ ?

$2.)$ Is $d(A) = d(A^{o})$ ?

Note : Here $\bar A$ denotes closure of A and $A^{o}$ denotes the interior of $A$

My attempt : If i take $A= [0,1]$ or $(0,1)$ then both $1)$ and $2)$ are true

$d([0,1]) = d([0,1])$ and $d(0,1) = d(0,1)$

jasmine
  • 14,457

1 Answers1

1

$d(A) \leq d(\overset {-} {A})$ is obvious. To prove the reverse inequality take $x,y \in \overset {-} {A}$. There exist $x_n, y_n \in A$ with $d(x_n,x) \to 0$ and $d(y_n,y) \to 0$. Since $d(x_n,y_n) \leq d(A)$ for all $n$ we get $d(x,y) \leq d(x,x_n)+d(x_n,y_n)+d(y_n,y) \leq d(x,x_n)+d(A)+d(y_n,y) \to d(A)$.

2) is false. If $X$ is the real line with usual metric an d$A =\{1,2\}$ then $d(A)=1$ but $A^{0}$ is empty.

Kavi Rama Murthy
  • 311,013