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

Question

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.}

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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.