# Relationship between $(A')^{c}$ and $(A^{c})^\circ$ for all $A$

For the purpose of this question, $$A'$$ is the derived set of set $$A$$, $$A^\circ$$ is the interior of set $$A$$, and $$A^{c}$$ is the complement of set $$A$$.

As we know, $$A$$ is closed if and and only if $$A^{c}$$ is open. To paraphrase,

\begin{align} A' \subset A &\iff A^{c} \subset (A^{c})^\circ \\ A^{c} \subset (A')^{c} &\iff A^{c} \subset (A^{c})^\circ \end{align}

As a result, we have

$$\forall A, (A')^{c} = (A^{c})^\circ$$

This doesn't seem to be right however, because

\begin{align} A' &= \{x : \forall \delta > 0, U^\circ(x, \delta) \cap A \ne \varnothing \} \\ (A')^{c} &= \{x : \exists \delta > 0, U^\circ(x, \delta) \cap A = \varnothing \} \\ (A^{c})^\circ &= \{x : \exists \delta > 0, U(x, \delta) \cap A = \varnothing \} \\ \end{align}

Obviously, $$U^\circ(x, \delta)^\ddagger$$ is not equivalent to $$U(x, \delta)$$, yet according to my reasoning $$(A')^{c}$$ and $$(A^{c})^\circ$$ are equal for all $$A$$. Are these two sets really equal?

$$\ddagger$$: $$U^\circ$$ stands for a deleted neighbourhood.

$$A' \subseteq \overline A.$$ So $$(\overline A)^c \subseteq (A')^{c}.$$ Now we know that $$(\overline A)^c = (A^c)^{\circ}.$$ So we have $$(A^c)^{\circ} \subseteq (A')^{c}.$$

$$\forall A, (A')^{c} = (A^{c})^\circ$$ does not follow from $$A^{c} \subset (A')^{c} \iff A^{c} \subset (A^{c})^\circ$$. The two sets, $$(A')^{c}$$ and $$(A^{c})$$, can be two different supersets of $$A^c$$.

No. These sets are different. Plese think of $$A$$ which has a isolated point. $$(A^{'})^c$$ have a isolated point, but $$(A^c)^\circ$$ doesn't have.

$$A^c$$ doesn't have it, so the outcome you wrote, that is \begin{align} A' \subset A &\iff A^{c} \subset (A^{c})^\circ \\ A^{c} \subset (A')^{c} &\iff A^{c} \subset (A^{c})^\circ \end{align} happens. But as you suspect, these are different sets.

Your argument is invalid.

If $$x \in (A^\complement)^\circ$$ then $$U(x,r) \subseteq A^\complement$$ for some $$r>0$$. This implies that $$x \notin A'$$ or $$x \in (A')^\complement$$. So we always have that $$(A^\complement)^\circ \subseteq (A')^\complement$$

for all $$A$$. But if we think about the reverse inclusion, start by reasoning about a point $$x\in (A')^\complement$$. This can mean two things: there is a neighbourhood $$U(x,r)$$ of $$x$$ that misses $$A$$ entirely (and then this neighbourhood does witness that $$x \in (A^\complement)^\circ$$) or we have such $$U(x,r)$$ such that $$U(x,r) \cap A = \{x\}$$ (or otherwise put, $$x$$ is an isolated point of $$A$$). In that case $$x \notin A^\complement$$ at all, let alone in its interior! So if we want a counterexample to the identity we look at $$A$$'s with isolated points:

The simplest is $$A=\{0\}$$ (in the reals, standard topology). Then $$A'=\emptyset$$ and $$(A')^\complement = \mathbb{R}$$, while $$A^\complement = \mathbb{R}\setminus\{0\} = (A^\complement)^\circ$$ as the complement of the closed $$A$$ is already open.