> Let $A=\{(x,y)\in \mathbb R^2|x^2+y^2\ge 1\}\cup\{(0,0)\}$, and let $d$ be a usual metric on $A$. Find the closure of $B_d((0,0),1)$

Question

Let $A=\{(x,y)\in \mathbb R^2|x^2+y^2\ge 1\}\cup\{(0,0)\}$, and let $d$ be a usual metric on $A$. Find the closure of $B_d((0,0),1)$and int$\{(0,0)\}$. prove your result.

$B_d((0,0),1)=\{(x,y)\in A:x^2+y^2<1\}=\{(0,0)\}$ Case1:- Any neighbourhood of $(0,0)$ intersects $\{(0,0)\}$. So, $(0,0)\in \overline{B_d((0,0),1)}$. $x$ can not use for inspection. Since $x\notin A$. $B_d(y,.5)\cap \{(0,0)\}=\phi$.So, $y \notin \overline{B_d((0,0),1)}$. similarly $B_d(z,.5)\cap \{(0,0)\}=\phi$. Hence $\{0,0\}$ is the only limit point. int$\{(0,0)\}=\{0,0\}$. Since, $B_d((0,0),.5)\subset \{(0,0)\}$. Am I right?

Answer 1

Other than showing {(0,0)} is open,
you've accomplished nothing.

Exercise. Prove for all p in a
metric space, that {p} is closed.

With that you're basically done.

Answer 2

As @William Elliot points out, in any metric space points are closed.

Note that $B_d((0,0),1)=\{(0,0)\}\subset A$.

So you are taking the closure of a closed set (consisting of one point).

Conclusion?..

Secondly, since you have proved $p=(0,0)$ is open, use the fact that the interior is the largest open set contained in the set of interest. Or, the interior of an open set is the set itself.