If we have two-point indiscrete space $X=\{a,b\}$ & let $A=\{a\}$,then what is the derived set of $A?$ If we have two-point indiscrete space $X=\\{a,b\\}$ & let $A=\\{a\\}$,then what is the derived set of $A?$ **Source:** This question was discussed in my topology class.Proffessor hinted us that its derived set is $A'=\\{b\\}$,Which is not closed!! **Queries:** 1.How does closed sets in $T_0$ space look like? 2.What is the derived set of $A$(proof?)?

The closed sets are $\emptyset$ and $X$, the complements of the open sets.

$a \
otin A'$ because a neighbourhood of $a$ is $X$ and $X \cap A =\\{a\\}$ does not contain a point of $A$ different from $a$!

$b \in A'$ because again the **only** neighbourhood of $b$ is $X$ and $X \cap A= \\{a\\}$ contains a point of $A$ different from $b$ (namely $a$).

So $A'=\\{b\\}$. Which indeed is not closed.

In general, if $A=\\{a\\}$ in **any** space $a$ is not a limit point of $A$, because there is no other point than $a$ in $A$ (the definition of $x \in A'$ says that every open set that contains $x$ must contain a point of $A$ different from $x$, and this cannot happen for $x=a$).

Closed sets in any space (also a $T_0$ space) are just the complements of the open sets.