Closure and Complements, Oh my!

Let $\overline{X}$ be the closure of a set $X$ , and let $X^\circ$ be the interior of the set $X$ . Note that $\overline{\overline{X}} =\overline{X}$ and that $(X^\circ)^\circ = X^\circ$ . If we consider the operations of taking the closure and the interior, we can only obtain new sets by applying them alternately.

In addition, if $Y = (\overline{X})^\circ$ , then $(\overline{Y})^\circ = Y$ . To see this note that $Y$ is an open set contained in $\overline{Y}$ , and hence $Y \subseteq (\overline{Y})^\circ$ . On the other hand $Y \subseteq \overline{X}$ , so $\overline{Y} \subseteq \overline{X}$ , and hence $(\overline{Y})^\circ \subseteq (\overline{X})^\circ = Y$ .

Similarly, if $Y = \overline{X^\circ}$ , then $\overline{Y^\circ} = Y$ .

Thus there are at most $7$ sets that can be formed from any set $A$ by the operations of taking the interior and the closure, namely $\big(\overline{A^\circ}\big)^\circ\;,\; \overline{A^\circ}\;,\; A^\circ\;,\; A\;,\; \overline{A}\;,\;\big(\overline{A}\big)^\circ\;,\; \overline{\big(\overline{A}\big)^\circ}$ If we are allowed to take complements as well, we can also construct the complements of each of these $7$ sets, making a total of $14$ sets in all. Note that the interior of a set is the complement of the closure of the complement of that set, and so being able to take closures and complements allows us to take interiors as well.

However, in this case, while the sets $\big(\overline{A^\circ}\big)^\circ$ , $\overline{A^\circ}$ , $A^\circ$ , $A$ , $\overline{A}$ are distinct, we see that $\big(\overline{A}\big)^\circ = \big(\overline{A^\circ}\big)^\circ = (-\infty,0)\cup(2,4) \hspace{2cm} \overline{\big(\overline{A}\big)^\circ} = \overline{A^\circ} = (-\infty,0] \cup [2,4]$ Thus we can form $5$ different sets using the operations of interior and closure. These sets are all bounded above, while their complements are unbounded above. Including complements gives us another $5$ sets. Thus we have a total of $\boxed{10}$ different sets.

The largest number of sets that can be formed is $14$ . This number can be achieved by starting, for example, with the set $A \; = \; (-\infty,0) \cup \{1\} \cup [2,3) \cup (3,4] \cup \big((5,6)\cap\mathbb{Q}\big)$

Relevant wiki: Closed Sets

I will denote "The closure of A" as $\overline{A}$ and "The complement of A" as $\mathbb{R} - A$

First note that the closure of the closure of a set is equal to the closure of a set (since the closure is itself closed), so applying a closure twice in a row does not produce a new set. Additionally note that the complement of the complement of a set is equal to the original set, so applying a complement twice in a row does not produce a new set either. All distinct sets that we can form can be created by applying the closure and the complement operations alternately.

So, starting with applying a closure and alternating between closure and complement, we have:

1. A itself
2. $\overline{A} = (-\infty, 0] \cup \lbrace 1 \rbrace \cup [2, 4]$
3. $X - \overline{A} = (0, 1) \cup (1, 2) \cup (4, \infty)$
4. $\overline{X - \overline{A}} = [0, 2] \cup [4, \infty)$
5. $X - \overline{X - \overline{A}} = (-\infty, 0) \cup (2, 4)$
6. $\overline{X - \overline{X - \overline{A}}} = (-\infty, 0]\cup [2, 4]$
7. $X - \overline{X - \overline{X - \overline{A}}} = (0, 2) \cup (4, \infty)$

Note that applying the closure operation returns us to step 4 after which we will be in an infinite loop, alternating closure and complement.

But we can arrive at a few more sets by starting by taking the complement, and then continuing closure / complement / closure:

1. $X - A$ = $[0,1)\cup(1,2)\cup\lbrace 3 \rbrace \cup (4, \infty)$
2. $\overline{X - A}$ = $[0,2]\cup\lbrace 3 \rbrace \cup [4, \infty)$
3. $X - \overline{X - A}$ = $(-\infty, 0)\cup(2,3)\cup(3,4)$

Following this, taking a closure will lead to the same set as #6, and we are done.

First, if $X$ is one of the sets then its complement $X' \not= X$ is also one of the sets. Therefore we can divide the sets into complementary pairs. This implies that the answer is an even number/

Next, since $(X')^- = (X^\circ)'$ , we are essentially allowed to make both closures and interiors of the pairs of sets.

Also, taking two closures successively or two interiors successively has no effect.

Therefore we study the two sequences $X, X^-, (X^-)^\circ, ((X^-)^\circ)^-, \dots\ \ \ \ \text{and}\ \ \ \ X, X^\circ, (X^\circ)^-, ((X^\circ)^-)^\circ, \dots$

Thus we have $X = \langle -\infty,0\rangle \cup \{1\} \cup [2,3\rangle \cup \langle 3, 4]; \\ X^- = \langle -\infty,0] \cup \{1\} \cup [2,4]; \\ (X^-)^\circ = \langle -\infty,0\rangle \cup \langle 2,4 \rangle; \\ ((X^-)^\circ)^- = \langle -\infty,0] \cup [2,4];$ continuing this does not generate any new sets. And $X^\circ = \langle -\infty,0\rangle \cup \langle 2,3\rangle \cup \langle 3, 4\rangle; \\ (X^\circ)^- = \langle -\infty,0] \cup [2,4];$ but we already generated this set in the first sequence. Therefore these are the five different sets that can be obtained by closure and interior; add their five complements, and everything is also closed under taking complements. Thus the answer is five pairs of two sets, or $\boxed{10}$ sets.