[Math] Closure vs Derived set in $\mathbb R$

general-topologyreal-analysis

Let $X\subset \mathbb R$ a subset and $a\in \mathbb R$.

i) We say $a$ is an adherent point of $X$ if $X\cap
(a-\varepsilon, a+\varepsilon)\neq \emptyset$ for every $\varepsilon>0$.
The set of all adherent points of $X$, denoted by $\overline{X}$, is
called the closure of $X$.

There is a similar notion:

ii) We say $a$ is a limit point of $X$ if $X\cap ((a-\varepsilon, a+\varepsilon)-\{a\})\neq \emptyset $ for every
$\varepsilon>0$. The set of all limit points of $X$, denoted by
$X^\prime$, is called the derived set of $X$.

From the definition it is clear that:

$$X\subset \overline{X}\quad \textrm{and}\quad X^\prime\subset \overline{X}.$$ Indeed, it holds $$\overline{X}=X^\prime\cup X.$$

Question 1. Can anyone provide me examples where $\overline{X}\neq X^\prime$?

I thought of the following:

  • $X=(0, 1)\cup \{2\}$. Then $X^\prime=[0, 1]$ and $\overline{X}=[0, 1]\cup \{2\}$. In particular, $X^\prime$ does not contain $X$.

  • If $F\neq \emptyset $ is discrete then $F^\prime=\emptyset $ and $\overline{F}=F\neq F^\prime$.

Furtheremore, I know that $$(X\cup Y)^\prime=X^\prime\cup Y^\prime, \quad\overline{X\cup Y}=\overline{X}\cup \overline{Y}\quad \textrm{and}\quad \overline{X\cap Y} \subset \overline{X}\cap \overline{Y}.$$

Question 2. What is the relationship between $(X\cap Y)^\prime$ and $X^\prime\cap Y^\prime$?

Finally:

Question 3. Are there other interesting properties of the derived set?

I already know that:

  • $X$ is closed if and only if $X^\prime\subset X$;

  • If $X^\prime\neq \emptyset $ then $X$ is infinite.

Thanks.

Best Answer

  1. $\overline {\{x\}}\neq \{x\}^\prime$;
  2. As $A \subseteq B$ implies $A^\prime \subseteq B^\prime$,
    $(A' \cap B') \subseteq A' \cap B'.$
  3. Limit points are a nusiance, of no use topologically except for scattered spaces (a minor topic), and a terrible way to define the closure of a set. $A = \{ 1/n : n\in \mathbb N \}$ is a scattered space of scattering height two; namely $A^{\prime\prime}=\emptyset$ but $A^\prime\neq \emptyset$. Every ordinal is scattered. A discrete space is scattered.
Related Question