Playing with closure and interior to get the maximum number of sets

Question

Can you find $A \subset \mathbb R^2$ such that $A, \overline{A}, \overset{\circ}{A}, \overset{\circ}{\overline{A}}, \overline{\overset{\circ}{A}}$ are all different?

Can we get even more sets be alternating again closure and interior?

Answer 1

According to Kuratowski's closure-complement problem, the monoid generated by the complement operator $a$ and the closure operator $b$ has $14$ elements and is presented by the relations $a^2 = 1$, $b^2 = b$ and $(ba)^3b = bab$. Now you are interested by the submonoid generated by the closure operator $b$ and by the interior operator $i = aba$. This submonoid has only $7$ elements: $1$, $b$, $i$, $bi$, $ib$, $bib$ and $ibi$. You can use Kuratowski's example $$K = {]0,1[} \cup {]1,2[} \cup \{3\} \cup ([4,5] \cap \mathbb{Q})$$ to generate the $14$ sets and hence the $7$ sets you are interested in. This is an example in $\mathbb{R}$, but $K \times \mathbb{R}$ should work for $\mathbb{R}^2$.

Answer 2

You can start by looking at $B = [0,1]^2\cap \Bbb Q^2$. In that case $$ \overline B = [0,1]^2\\ \overset{\circ} B = \emptyset\\ \overset{\circ}{\overline{B}} = (0,1)^2\\ \overline{\overset{\circ}{B}} = \emptyset $$ If you let $A$ be the union of $B$ and, say, the square $[3, 4]^2$, this would let you tell the difference between $\overset\circ A$ and $\overline{\overset{\circ}{A}}$, since the latter contains $(3,3)$ and $(4,4)$, while the former does not.

Answer 3

Well, given $A$, we know that $\overline A$ is not equal to $A$ whenever $A$ is not closed, and we that ${\rm int}\, A\neq A$ whenever $A$ is not open (where ${\rm int}\, A$ is the interior of $A$ in $\mathbb R$). So the set we are looking for is neither open nor closed.

Now, one can quite easily construct a non-open, non-closed set in the plane. But can you arrange for the other two sets to be different?

Note: arranging for ${\rm int}\, A=\emptyset$ is not good enough if you require $\overline{{\rm int}\, A}\neq{\rm int}\, A$. So the interior of $A$ must have some interior.

I leave you to think about the final two properties.