# Derived set of a closed subspace

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 $$A\cap X^{(n)}\subset A^{(n)}$$.

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

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.
• $X=\omega+1$ and $A=\{\omega\}$ would work just as well. Commented May 25 at 5:07
• 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. Commented May 25 at 7:46
An easy counterexample: just take $$X$$ to be the Cantor set and $$A$$ be a singleton in $$X$$.