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Suppose $X$ is a compact Hausdorff topological space with a basis of clopen sets. Let $A$ be a closed subspace of $X$ and let $A^{(0)}=A$, $A^{(1)}=A^\prime$, etc. My question is: it is true that $A^{(n)}=A\cap X^{(n)}$ for all $n\in\mathbb{N}$?

Since $A\subset X$, then $A^{(n)}\subset X^{(n)}$. Also, since $A$ is closed, one can prove by induction on $n$ that $A^{(n)}\subset A$. These two things together prove one of the directions of the problem. But I can’t prove that $A\cap X^{(n)}\subset A^{(n)}$.

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  • $\begingroup$ What is $A'$? I'm unfamiliar with this notation in topology. $\endgroup$
    – kabel abel
    Commented May 25 at 0:32
  • $\begingroup$ In my experience, $A'$ is the set of accumulation points / limit points, sometimes also called the derived set. $\endgroup$ Commented May 25 at 0:33
  • $\begingroup$ @PrincessEev Ah, gotcha. $\endgroup$
    – kabel abel
    Commented May 25 at 0:35
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    $\begingroup$ What if $X$ is perfect and $A$ is not? $\endgroup$ Commented May 25 at 0:45
  • $\begingroup$ Remember to accept one of the answers if you find it satisfactory. $\endgroup$
    – David Gao
    Commented Jun 3 at 14:42

2 Answers 2

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So, here's a counterexample: take the space $X = (\omega + 1)^2$. It is clearly hausdorff, compact, and has a basis of clopen sets. Now, take $A = (\omega + 1) \times \{\omega\}$. We have $A^{(2)} = \emptyset$, but $X' = (\omega + 1) \times \{\omega\} \cup \{\omega\} \times (\omega + 1)$, and thus $X^{(2)} \cap A = \{\omega\} \times \{\omega\}$.

Edit: Actually, there's not even a need to go to the 2nd step. $X' \cap A = A \not \subset A' = \{\omega\} \times \{\omega\}$ is already a valid counterexample.

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    $\begingroup$ $X=\omega+1$ and $A=\{\omega\}$ would work just as well. $\endgroup$ Commented May 25 at 5:07
  • $\begingroup$ What if $A$ is not only a closed set but a clopen set? Does my question have affirmative answer? I think so, but I’m not sure. $\endgroup$
    – Earnur
    Commented May 25 at 7:46
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    $\begingroup$ @Earnur If A is clopen, then any deleted neighbourhood of a point in A can be intersected with A to get another deleted neighbourhood, which proves your theorem. $\endgroup$
    – kabel abel
    Commented May 25 at 11:50
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An easy counterexample: just take $X$ to be the Cantor set and $A$ be a singleton in $X$.

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