Closure and complement generate at most fourteen sets from any subset, and $(0,1)\cup (1,2)\cup \{3\}\cup ([4,5]\cap \mathbb{Q})$ attains fourteen

$cc=1$ and $kk=k$ hold by [A3].

For an open $B\subseteq X$: $B\subseteq kB$ by [A3], and $B=iB$ because $B$ is open, so monotonicity of $i$ gives $B=iB\subseteq ikB$ and monotonicity of $k$ gives $kB\subseteq kikB$; conversely $ikB\subseteq kB$ gives $kikB\subseteq kkB=kB$. Hence $kikB=kB$ for every open $B$.

In $\mathbb{R}$, for $a<b$ the interval $(a,b)$ is open: for $x\in (a,b)$ put $r:=\min \{x-a,b-x\}>0$ by [L3]; then $(x-r,x+r)\subseteq (a,b)$. The rays $(-\mathrm{\infty },b)$ and $(a,\mathrm{\infty })$ are open by the same computation with one of the two bounds omitted.

In $\mathbb{R}$, for $a<b$ and any $x\in [a,b]$ and any $r>0$, the interval $J:=(\max \{a,x-r\},\min \{b,x+r\})$ is nonempty: its left endpoint is below its right endpoint because $a<b$, $a\le x<x+r$ and $x-r<x\le b$ and $x-r<x+r$. Every point of $J$ lies in $[a,b]\cap (x-r,x+r)$.

No nonempty open subset of $\mathbb{R}$ is contained in $\mathbb{Q}$: it would contain a ball $(x-r,x+r)$, which is uncountable by [L2], whereas a subset of $\mathbb{Q}$ is at most countable by [L2].

In $\mathbb{R}$, for $a\le b$ the interval $[a,b]$ is closed, its complement being $(-\mathrm{\infty },a)\cup (b,\mathrm{\infty })$, a union of two open sets; likewise $(-\mathrm{\infty },b]$ and $[a,\mathrm{\infty })$ are closed, and a singleton $\{t\}$ is closed, its complement being $(-\mathrm{\infty },t)\cup (t,\mathrm{\infty })$.

By step 1.4 and [L2] the set $J$ contains a rational, and being a nonempty open interval it is uncountable by [L2] while $\mathbb{Q}$ is at most countable, so $J$ also contains a point outside $\mathbb{Q}$. Hence for every $x\in [a,b]$ and every $r>0$ the ball $(x-r,x+r)$ meets both $[a,b]\cap \mathbb{Q}$ and $[a,b]\setminus \mathbb{Q}$.

Claim 1: by step 1.2 applied to the open set $B:=iA$ one gets $kikiA=kiA$ for every $A$, that is $kiki=ki$ as operators; substituting $i=ckc$ turns $ki$ into $kckc$ and gives $kckckckc=kckc$; composing on the right with $c$ and using $cc=1$ gives $kckckck=kck$.

Consequently $\overline{[a,b]\cap \mathbb{Q}}=[a,b]=\overline{[a,b]\setminus \mathbb{Q}}$ for $a<b$: each closure is contained in $[a,b]$, which is closed by step 2.1, and contains $[a,b]$ by step 2.2 together with the neighbourhood criterion for the closure.

Likewise $\overline{(a,b)}=[a,b]$ for $a<b$: the inclusion $\subseteq$ holds because $[a,b]$ is closed and contains $(a,b)$, and $\supseteq$ because for $x\in [a,b]$ and $r>0$ the nonempty interval $J$ of step 1.4 meets $(a,b)$, being contained in $(a,b)$ except possibly for its endpoints, which it excludes. The same argument gives $\overline{(a,\mathrm{\infty })}=[a,\mathrm{\infty })$ and $\overline{(-\mathrm{\infty },b)}=(-\mathrm{\infty },b]$.

Claim 2: using $cc=1$ and $kk=k$, every word in $k$ and $c$ equals an alternating word, one with no two adjacent equal letters. An alternating word of length at least $8$ contains $kckckck$ as a block of seven consecutive letters — the first seven if it begins with $k$, the second through eighth if it begins with $c$ — and replacing that block by $kck$ shortens it by four. Iterating, every word equals an alternating word of length at most $7$. There are exactly two alternating words of each length from $1$ to $7$ and one of length $0$, and the length-$7$ word beginning with $k$ is $kckckck=kck$ by claim 1; the remaining fourteen are those listed in the statement.

In $\mathbb{R}$ with $A=(0,1)\cup (1,2)\cup \{3\}\cup ([4,5]\cap \mathbb{Q})$: by [A2] the closure of the four-term union is the union of the four closures, which by steps 2.1, 3.1 and 3.2 are $[0,1]$, $[1,2]$, $\{3\}$ and $[4,5]$; hence $kA=[0,2]\cup \{3\}\cup [4,5]$.

$cA=(-\mathrm{\infty },0]\cup \{1\}\cup [2,3)\cup (3,4)\cup ([4,5]\setminus \mathbb{Q})\cup (5,\mathrm{\infty })$. Here $\overline{[2,3)}=[2,3]$, because $[2,3]$ is closed and contains $[2,3)$ while monotonicity gives $[2,3]=\overline{(2,3)}\subseteq \overline{[2,3)}$; the other five closures are $(-\mathrm{\infty },0]$, $\{1\}$, $[3,4]$, $[4,5]$ and $[5,\mathrm{\infty })$ by steps 2.1, 3.1 and 3.2. So by [A2] the closure of the six-term union is $(-\mathrm{\infty },0]\cup \{1\}\cup [2,3]\cup [3,4]\cup [4,5]\cup [5,\mathrm{\infty })=(-\mathrm{\infty },0]\cup \{1\}\cup [2,\mathrm{\infty })$, that is $kcA=(-\mathrm{\infty },0]\cup \{1\}\cup [2,\mathrm{\infty })$.

$ckA=(-\mathrm{\infty },0)\cup (2,3)\cup (3,4)\cup (5,\mathrm{\infty })$, and its closure is $(-\mathrm{\infty },0]\cup [2,4]\cup [5,\mathrm{\infty })$ by [A2] and steps 3.2 and 2.1; so $kckA=(-\mathrm{\infty },0]\cup [2,4]\cup [5,\mathrm{\infty })$.

$ckcA=(0,1)\cup (1,2)$, whose closure is $[0,2]$ by [A2] and step 3.2; so $kckcA=[0,2]$. Then $ckckcA=(-\mathrm{\infty },0)\cup (2,\mathrm{\infty })$, whose closure is $(-\mathrm{\infty },0]\cup [2,\mathrm{\infty })$ by [A2] and step 3.2; so $kckckcA=(-\mathrm{\infty },0]\cup [2,\mathrm{\infty })$.

$ckckA=(0,2)\cup (4,5)$, whose closure is $[0,2]\cup [4,5]$ by [A2] and step 3.2; so $kckckA=[0,2]\cup [4,5]$.

The seven sets $A$, $kA$, $kckA$, $kckckA$, $kcA$, $kckcA$, $kckckcA$ have the following membership pattern at the five test points $0$, $1$, $3$, $9\mathrm{/}2$, $6$, writing $1$ for "belongs" and $0$ for "does not": $A$ gives $(0,0,1,1,0)$, since $9\mathrm{/}2$ is a rational in $[4,5]$; $kA$ gives $(1,1,1,1,0)$; $kckA$ gives $(1,0,1,0,1)$; $kckckA$ gives $(1,1,0,1,0)$; $kcA$ gives $(1,1,1,1,1)$; $kckcA$ gives $(1,1,0,0,0)$; $kckckcA$ gives $(1,0,1,1,1)$.

The remaining seven words of claim 2 are the complements of these seven — the list of fourteen words consists of the seven above and those seven preceded by $c$ — so their patterns are the bitwise complements $(1,1,0,0,1)$, $(0,0,0,0,1)$, $(0,1,0,1,0)$, $(0,0,1,0,1)$, $(0,0,0,0,0)$, $(0,0,1,1,1)$, $(0,1,0,0,0)$. The fourteen patterns are pairwise distinct, so the fourteen sets are, and claim 3 holds.