Let $X=\mathbb R$, with the usual metric on $\mathbb R$ and $A=((0,1)\cap \mathbb Q)\cup$ {$2,3$}. Find the limit points of $A$, exterior points of $A$, $A^o$, $\overline A$ and $\partial A$.
Can anyone please explain how to answer this as I don't understand how to go about this? Any help would be appreciated.
$\endgroup$2 Answers
$\begingroup$$A$ is the union of all rational numbers in $(0,1)$ and isolation points of {$2,3$}. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, any real number in $[0,1]$ is the limit point of $A$. Also note that no isolation point is the limit point. Denote $A^{'}$ as the limit points set of $A$. Then
$A'=[0,1]$
$A^o=\emptyset$ (for $\mathbb{Q}$ has no open set within it)
$\overline{A}=A\cup A'=[0,1]\cup \{2,3\}$ (Closure is the union of limit points set and self.)
$\operatorname{exterior}(A)=(\overline{A})^c=(-\infty,0)\cup (1,2)\cup (2,3)\cup (3,\infty)$ (Exterior set is the complement of closure.)
$\partial A=\overline{A}\cap\overline{A^c}=\overline{A}-A^o=\overline{A}=[0,1]\cup \{2,3\}$ (Boundary is the intersection of closure and complement of closure, or closure exclude interior.)
$\endgroup$ 4 $\begingroup$You need to find those points that are outside of $A$ but still are infinitely close. By that I mean, there exists an infinite set of points in $A$ that get arbitrarily close to that point. e.g. 0 is a limit point. This is because $1/2^n$ is in $A$ for all $n$ and they get arbitrarily close. You could make the same argument regarding irrationals in $[0,1]$ since the rationals are dense in them.
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