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 $X^\prime\cap Y^\prime$?


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.


  • 3
    $\begingroup$ Another example is $Y = \left\{ \frac{1}{n} \mid n \in \mathbb{Z}_+ \right\}$. Then $Y' = \{0\}$ and $\overline{Y} = Y \cup \{0\}$. $\endgroup$ – Lucy Yang Feb 13 '18 at 17:36
  • $\begingroup$ Doesn't the comment of @LucyYang show that in fact $X'$ needn't be infinite just because it isn't empty? $\endgroup$ – MPW Feb 13 '18 at 17:45
  • $\begingroup$ Hint for question 2: Observe that if $X \subseteq Y$, then $X' \subseteq Y'$. What can you deduce from $X \cap Y \subseteq X, Y$? Note that the reverse inclusion needn't hold: find $X' \cap Y'$ and $(X \cap Y)'$ when $X = \left\{\frac{1}{2n} \mid n \in \mathbb{Z}_+ \right\}$ and $Y = \left\{ \frac{1}{2n +1} \mid n \in \mathbb{Z}_+ \right\}$ $\endgroup$ – Lucy Yang Feb 13 '18 at 17:50
  • $\begingroup$ @MPW my statement was $X^\prime\neq \phi\Rightarrow X$ is infinite and not $X^\prime\neq \phi\Rightarrow X^\prime$ is infinite. In Lucy Yang example, $Y$ is infinite.. $\endgroup$ – PtF Feb 13 '18 at 18:34
  • $\begingroup$ @LucyYang $(X\cap Y)^\prime=\phi$ since $X\cap Y=\phi$ and $X^\prime\cap Y^\prime=\{0\}$, right? Thanks for the example =) $\endgroup$ – PtF Feb 13 '18 at 18:38
  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.
  • 1
    $\begingroup$ Thanks for your comment. As to your first example, it adds nothing new since $\{x\}\neq \phi$ is discrete and I had pointed out in my question that any non-empty discrete set $F$ satisfies $F^\prime\neq \overline{F}$. As to scattered spaces, I've never heard about, can you give me any reference? $\endgroup$ – PtF Feb 13 '18 at 20:32
  • $\begingroup$ @PtF A web search yields some stuff. Wikipedia for example. $\endgroup$ – William Elliot Feb 13 '18 at 22:23

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