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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?)?

Styles
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    @WilliamElliot $a$ is not a limit point of $S$ here. – Paulo Mourão Jun 29 '19 at 08:17
  • @WilliamElliot:https://math.stackexchange.com/questions/490968/non-t1-space-is-the-set-of-limit-points-closed/490976#comment6736864_490976 look at the answer by Stefan Hamcke. – Styles Jun 29 '19 at 08:18

1 Answers1

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

$a \notin A'$ 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.

Henno Brandsma
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