Exercise 2.6

Let $x$ be a limit point of $E^{\prime }$. Any neighborhood $N$ of $x$ contains an $x^{\prime }\in E^{\prime }$ for which $x^{\prime }\ne x$. By theorem 2.22, $N\setminus \{x\}=N\cap {\{x\}}^c$ is open, and therefore contains a neighborhood $N^{\prime }$ of $x^{\prime }$. $N^{\prime }$ intersects $E$ since $x^{\prime }\in E^{\prime }$, and consequently so does $N$. This implies that $x\in E^{\prime }$, and since this holds for any limit point, $E^{\prime }$ is closed.

Since $E\subseteq Ē$, a limit point of $E$ is a limit point of $Ē$. If $x$ is a limit point of $Ē=E\cup E^{\prime }$, $N_{\frac{1}{n}}(x)$, must intersect either $E$ for an infinite number of values of $n\in N$, or it must intersect $E^{\prime }$ for an infinite number of values of $n$. (Otherwise, there would be an $N$ for which $N_{\frac{1}{n}}(x)$ did not intersect either whenever $n>N$, a contradiction.) Consequently, since $N_{𝜖}(x)\subseteq N_{{𝜖}^{\prime }}(x)$ whenever $𝜖<{𝜖}^{\prime }$, all neighborhoods of $x$ intersect either $E$ or $E^{\prime }$. In both cases, $x\in E^{\prime }$. This finishes the proof.

Let $X$ be a metric space for some distance $d$, and $E\subset X$.

We note first that, for all subsets $A\subset X,B\subset X$,

Indeed, suppose $A\subset B$ . If $a\in A^{\prime }$, then for all $𝜀>0$, there is some $b\in A,b\ne a$ such that $b\in N_{𝜀}(a)=\{x\in X\mid d(x,a)<𝜀\}$. Since $A\subset B$, $b\in B$. This shows that for all $a\in A^{\prime }$,

so $a\in B^{\prime }$. We conclude that $A^{\prime }\subset B^{\prime }$.

(a)

We show first a little more (to facilitate the writing of part (b)): $$\begin{array}{lll}\hfill {(Ē)}^{\prime }\subset E^{\prime }.& & \hfill \text{(2)}\end{array}$$

Let $p$ be any point in ${(\bar{E})}^{\prime }$. For any $𝜀>0$, consider

$$N_{𝜀}(p)=\{x\in X\mid d(x,p)<𝜀\}.$$

Since $p$ is a limit point of $E^{\prime }$, there is some $q\in \bar{E}$ such that $q\in N_{𝜀}(p)$ and $q\ne p$. Put

$$\eta =\min <!--FUNCTION\; APPLICATION-->(d(p,q),𝜀-d(p,q)).$$

Since $p\ne q$, $d(p,q)>0$ and since $q\in N_{𝜀}(p)$, $𝜀-d(p,q)>0$, so $\eta >0$. Since $q\in \bar{E}$, there is some $r\in E$ such that $d(q,r)<\eta$.

Then

$$d(p,r)\le d(p,q)+d(q,r)<d(p,q)+\eta \le d(p,q)+(𝜀-d(p,q))=𝜀,$$

thus $r\in N_{𝜀}(p)$. Moreover $d(p,r)\ge d(p,q)-d(q,r)>d(p,q)-\eta >0$, so $p\ne r$.

This shows that every neighborhood $N_{𝜀}(p)(𝜀>0)$ contains some $r\in E$, where $r\ne p$, so $p\in E^{\prime }$. Therefore ${(\bar{E})}^{\prime }\subset E^{\prime }$.

To conclude we note that $E^{\prime }\subset \bar{E}$, thus by (1) and (2),

$${(E^{\prime })}^{\prime }\subset {(\bar{E})}^{\prime }\subset E^{\prime }.$$

So every point limit of $E^{\prime }$ is in $E^{\prime }$. By definition 2.18(d), $E^{\prime }$ is closed.

(b)

First $E\subset \bar{E}$, thus by (1), $E^{\prime }\subset {(\bar{E})}^{\prime }$. By (2), ${(\bar{E})}^{\prime }\subset E^{\prime }$, so $${(Ē)}^{\prime }=E^{\prime }.$$

This shows that $E$ and $\bar{E}$ have the same limit points.

(c)

As a counterexample, take $E=\{1∕n\mid n\in {\mathbb{N}}^{\ast }\}$ in the metric space $ℝ$. Then $E^{\prime }=\{0\}$ and ${(E^{\prime })}^{\prime }=\varnothing \ne E^{\prime }$. So $E$ and $E^{\prime }$ have not always the same limit points