(a) We use the following notation here: we define four operations as follows, $iA=Int(A)$ , $cA=\overline{A}$ , $bA=Bd(A)$ , $xA=X-A$ . So that, for example, $xicA=X-Int(\overline{A})$ and $cA=iA\cup bA$ (Exercise 19(a)). First, several statements.

1. $i$ and $c$ are monotone. If $A\subset B$ then $i(A)\subset i(B)$ and $c(A)\subset c(B)$ . Indeed, every open set contained in $A$ is contained in $B$ as well, and every closed set containing $B$ contains $A$ as well.
2. $ccA=cA$ , $xxA=A$ , $iiA=iA$ , but $bbA\neq bA$ in general. Indeed, $cA$ is closed, $xxA=X-(X-A)=A$ , and $iA$ is open, but $bb(\mathbb{Q})=b(\mathbb{R})=\emptyset$ .
3. $bA=bxA$ , and $cA=iA\cup bA$ and $X=iA\cup bA\cup ixA$ are partitions (the sets on the right are disjoint). This follows from the definition of the boundary of a set, Exercise 19(a), and the following facts: $X=A\cup xA\subset cA\cup cxA$ , $iA\subset A$ , and $ixA\subset xA$ .
4. $xcA=ixA$ , $xiA=cxA$ , and $iA=xcxA$ . These follow from property 3 immediately, indeed, $ixA=x(iA\cup bA)=xcA$ . Hence, using property 2, $cxA=xxcxA$ $=xixxA$ $=xiA$ , and $iA=xxiA=xcxA$ .
5. $iA\subset icA$ , $ciA\subset cA$ . Indeed, $iA\subset cA$ where $iA$ is open, and $cA$ is closed. Both inclusions can be proper: consider $A=\{0\}\cup[1,2)\cup(2,+\infty)$ .
6. $icicA=icA$ , $ciciA=ciA$ . Using properties 5 and 2, $cicA\subset ccA=cA$ , but $icA\subset cicA$ , therefore, using property 1, $icA=iicA\subset icicA\subset icA$ . Hence, $icicA=icA$ . Now, using this and properties 2 and 4, $ciciA=$ $cicixxA=$ $xicicxA=$ $xicxA=$ $cixxA=$ $ciA$ .
7. $icA$ , $ciA$ and $A$ are, in general, not comparable by inclusion. Consider the following set: $A=[0,1]\cap\mathbb{Q}\cup(2,+\infty)$ . Then $icA=(0,1)\cup(2,+\infty)$ and $ciA=[2,+\infty)$ , and each of the following points belongs to one set only: $0$ , $1/\sqrt{2}$ and $2$ .

Property 4 shows that even if we take $Int$ as an additional operation we are not going to get more sets.

Property 2 shows that in a sequence of operations of closure and complementation we can substitute any two operations of the same type by one or zero operations of the same type. Therefore, we may construct only two sequences of sets: starting with two initial sets $A$ and $xA$ we apply the sequence of operations $c$ , then $x$ , then $c$ etc.

Further, we may use properties 4 and 2 to move all operations $x$ to the right (by switching it with another operation on the right: $xc=ix$ and $xi=cx$ ) and to cancel two $x$ ’s in a row ($xxA=A$ ).

Accordingly, the first few members of the sequence starting with $A$ are $A$ , $cA$ , $xcA=ixA$ , $cxcA=cixA$ , $xcxcA=icA$ , $cxcxcA=cicA$ , $xcxcxcA=icixA$ , $cxcxcxcA=cicixA$ , and the second sequence starts with $xA$ , $cxA$ , $xcxA=iA$ , $cxcxA=ciA$ , $xcxcxA=icxA$ , $cxcxcxA=cicxA$ , $xcxcxcxA=iciA$ , $cxcxcxcxA=ciciA$ . Note that, according to property 6, in each sequence the 8th member equals the 4th member, therefore, there can be at most 7 distinct sets in each sequence, and 14 sets overall.

(b) Every set of each sequence presented above starting from the second one is open or closed. Let $A$ be neither open nor closed. This ensures that no other set in the sequence equals $A$ or $xA$ . The following table divides the sets in the sequences into 4 different groups according to whether a set is a sequence of operations $i$ and $c$ applied to $A$ or $xA$ , and whether it is closed or open. The table also shows inclusion relations among the sets within one group, using properties 1, 2 and 5 (property 7 tells us that at least some sets from different groups can be made incomparable).

	$A$	$xA$
open	$iA\subset iciA\subset icA$	$ixA\subset icixA\subset icxA$
closed	$ciA\subset cicA\subset cA$	$cixA\subset cicxA\subset cxA$

The sets in different groups can easily be made different. For example, closed sets and open sets are different in the standard topology on the real line (with the exception for the empty set and whole space). Also, we can easily have some different points in sequences starting from $A$ and $xA$ (both operations $i$ and $c$ preserve interior points). So, the main challenge is to try to make all sets within one group different.

For example, to make $ciA$ and $cicA$ different, we need

a) make $iA$ and $icA$ different, so $A$ must have a limit point such that it is an interior point of $cA$ but not an interior point of $A$ , which can be easily done by removing a point from an interval;

b) make $iA$ and $icA$ so different, that their closures would differ as well, so we might want to remove a whole dense set of points from an interval, but such that the closure of the remaining points is still the closed interval, and the set we used to illustrate property 7 is an example of how this can be done.

But that example will not work for $cicA$ and $cA$ . We also need to add an isolated point to the set, which will disappear in $icA$ .

Having a set like this, we already have all sets in the group of closed sets derived from $A$ different. This also implies that $icA$ is different from $iA$ and $iciA$ . So, additionally, we only need to ensure that $iA$ and $iciA$ are different, but this can be easily done if we include a punctured interval, that we would need for $xA$ anyway.

So, $A$ consists of an isolated point, an interval with a dense subset removed, and an interval with a point removed.

Now, the same should work for the complement $xA$ as well, but the isolated point of $A$ becomes an interval with a point removed in $xA$ , and vice versa, and the interval with a dense set of points removed becomes the dense set. So, basically, we don’t need any additional constructions for $xA$ .

Here is an example:

	$A$	$xA$
	$\{-2\}\cup([-1,1]\cap\mathbb{Q})$ $\cup(1,2)\cup(2,+\infty)$	$(-\infty,-2)\cup(-2,-1)$ $\cup([-1,1]-\mathbb{Q})\cup\{2\}$
$c$	$\{-2\}\cup[-1,+\infty)$	$(-\infty,1]\cup\{2\}$
$xc=ix$	$(-\infty,-2)\cup(-2,-1)$	$(1,2)\cup(2,+\infty)$
$cxc=cix$	$(-\infty,-1]$	$[1,+\infty)$
$xcxc=ic$	$(-1,+\infty)$	$(-\infty,1)$
$cxcxc=cic$	$[-1,+\infty)$	$(-\infty,1]$
$xcxcxc=icix$	$(-\infty,-1)$	$(1,+\infty)$

In both cases the next closure equals the 4th member of the sequence. So, there are maximum of 14 sets that can be obtained from a given set by taking closure and complement (and interior, as the latter can be expressed in terms of the former two operations), and above is an example of a set for which all 14 sets are different.