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"Let $(E,\tau)$ where $E=\mathbb{N}^*=\{1,2,\ldots\}$ and $\tau=\{\emptyset,E,\{A_n\}_{n\in\mathbb{N}^*}\}$, where $A_n=\{1,\ldots,n\}$

Let $A=\{2n, n\in \mathbb{N}^*\}$ find $A', \overline{A}, \overset{\circ}{A}$

we have by definition : $x\in A'\Longleftrightarrow \forall V \in \mathcal{V}_x, (V\setminus\{x\}) \cap A\neq\emptyset $

for example for $x=2$ , an open containing x in $A_2=\{1,2\}$

$(A_2\setminus\{2\})\cap A= \{1\}\cap \{2,4,6,\ldots\}=\emptyset$ so $2\notin A'$ and also $x=1\notin A'$

If $x=3$ an open containing $3$ in $\{1,2,3\}$ and $\{1,2,3\}\setminus\{3\}\cap \{2,4,6,...\}=\{2\}$ then $3\in A'$

How to obtain A' ? please

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    What is $A_n$? Is $A_n = \{1,…,n\}$? If so: If $τ$ ought to be a topology on $E$, it also needs to contain $E = ℕ^*$ itself, which is not of the form $A_n$ for any $n ∈ ℕ^*$. Also, I’m unsure about how to read “$\{A_n\}_{n ∈ ℕ^*}$”. I think you should write “$τ = \{∅, E\} ∪ \{A_n; n ∈ ℕ^*\}$ with $A_n = \{1, …, n\}$”. – 2017-02-09
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    but $\cup_{n\in \mathbb{N}^*} A_n =\mathbb{N}^*$ – 2017-02-09
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    @k.stm the infinet union of $A_n$ in $\mathbb{N}^*$ – 2017-02-09
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    Yes, $A_n$ for $n ∈ ℕ^*$ is a base for the topology, but by $τ$ you probably want to the denote the topology itself, that is the set of all open sets. Since $E$ is open in $E$, it has to be in $τ$. – 2017-02-09
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    Why you do -1 , i begin the proof , what is this ?????? – 2017-02-09
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    @k.stm you are right i edited the question – 2017-02-09
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    I didn’t downvote. Another thing: Your definition of derived sets $A'$ should have a “$≠$” instead of “$=$” at “$V\setminus \{x\} ∩ A = 0$”. – 2017-02-09
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    ok i don't read a seconde time – 2017-02-09
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    Are you not just looking for all $n$ such that $A_{n-1} \cap A \neq \emptyset$? – 2017-02-09
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    @TheoreticalEconomist i don't understand your question – 2017-02-09
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    I was trying to suggest that the above is a simpler condition for characterising $A^\prime$, but perhaps not. Also, I don't quite see how $1 \in A^\prime$. $A_1 \in \mathcal V_1$, is it not? – 2017-02-09
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    you are right $1\notin A'$ – 2017-02-09

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$A^\prime = \{ n :A_{n-1} \cap A \neq \emptyset \}$. That is, $A^\prime$ contains all the natural numbers greater than some even number. Hence, $A^\prime = \{3,4,5,\ldots \}$.

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    ok, $\overline{A}=\{2,3,4,....\}$ right ? – 2017-02-09
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    and $ \overset{\circ}{A}=\emptyset$ – 2017-02-09
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    $\bar{A} = A \cup A^\prime$, so yes to your first question. Your second comment is also correct, as is easy to show. – 2017-02-09