Closure and Interior Comparison

Question

I need to find a subset $A\subset \Bbb{R}$ such that the following sets are all different.

$$A\qquad \mathring{A} \qquad \overline{A}\qquad \overline{\mathring{A}}\qquad \mathring{\overline{A}}\qquad \mathring{\overline{\mathring{A}}}\qquad \overline{\mathring{\overline{A}}} $$

My best attempt was with the following set:

$A=(1,2)\cup (2,3)\cup \{4\}$

I have:

$$ \mathring{A}=(1,2)\cup(2,3) \\ \overline{A}=[1,3]\cup\{4\} \\ \overline{\mathring{A}}=[1,3] \\ \mathring{\overline{A}}=(1,3). $$

All of these sets are different, but

$$ \overline{\mathring{\overline{A}}}=\overline{\mathring{A}}=[1,3]\quad\textrm{and}\quad \mathring{\overline{\mathring{A}}}=\mathring{\overline{A}}=(1,3).$$

Can I add something to my attempt to fix this?

Answer 1

Indeed, you can add something to fix your attempt:

Consider $B:= A \cup ( \mathbb{Q} \cap [5,6])$. Then we have

• $\mathring{B} = \mathring{A} = (1,2) \cup (2,3)$
• $\overline{B} = [1,3] \cup \{4\} \cup [5,6]$
• $\overline{\mathring{B}} = \overline{\mathring{A}} = [1,3]$
• $\mathring{\overline{B}} = (1,3) \cup (5,6)$
• $\mathring{\overline{\mathring{B}}} = \overline{\mathring{A}} = (1,3)$
• $\overline{\mathring{\overline{B}}} = [1,3] \cup [5,6]$