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$\newcommand{\XT}{(X,\mathcal{T})}$From Definition 1.2 in Gardner and Jackson (GJ): The $K$‑number $K(\XT)$ of a topological space $\XT$ is the cardinality of the Kuratowski monoid of operators on $\XT$ generated by closure $b$ and complement $a$ under composition. For any $A\subset X,$ the $k$‑number $k(A)$ of $A$ is the cardinality of the family of subsets generated by $A$ under $\{a,b\}.$ The $k$‑number of the space is $$k((X,\mathcal{T}))=\max\{k(A):A\subset X\}.$$

Question. What values does the ordered pair $(K(\XT),k(\XT))$ take?

Partial Answer. Lemma 2.8 in GJ states that $k(\XT)=4$ iff $\XT$ is a non-discrete door space or $\mathcal{T}\setminus\{\varnothing\}$ is a filter in $2^X.$ Each implies $K(\XT)<14,$ hence $(14,4)$ is impossible (GJ p. 25).

The case $(14,6)$ only gets mentioned once in GJ, on p. 28: $$\unicode{x201C}\text{We do not know of ... any Kuratowski space with }k\text{-number }6.\unicode{x201D}$$ (A Kuratowski space is one that satisfies $K(\XT)=14.)$

It is well known that $k(A)$ is always even (GJ p. 15). It follows from the definitions and the identity $bababab=bab$ that $$2\leq k(\XT)\leq K(\XT)\leq14.$$

By Theorem 2.1 in GJ, the only possible values of $K(\XT)$ are 14, 10, 8, 6, 2. Clearly, $k(\XT)=2$ iff $\XT$ is discrete. Thus, besides possibly $(14,6),$ the pair can only be:

$(2,2),$
$(6,4),$ $(6,6),$
$(8,4),$ $(8,6),$ $(8,8),$
$(10,4),$ $(10,6),$ $(10,8),$ $(10,10),$
$(14,8),$ $(14,10),$ $(14,12),$ $(14,14).$

Each occurs in some space $\XT$ with $|X|\leq7$ (this holds for both Kuratowski monoids that satisfy $K(\XT)=10).$ Examples are listed below.

Theorem 2.10 in GJ states that for any $A\subset X,$ the family of subsets generated by $A$ under $\{b,i\}$ where $i$ denotes interior satisfies exactly one of 30 possible collapses of the Hasse diagram:

$\hspace{268px}$

The table below lists these collapses in the same order as they appear in Table 2.1 in GJ. Each entry is labeled by what I am calling the $h$‑number of $A$, or $h(A).$ The identity operator is denoted by $\textsf{id}.$

\begin{array}{|c|c|c|c|} \hline h(A) & h(aA) & \text{collapse} & k(A) \\ \hline 1 & 1 & \varnothing & 14 \\ \hline 2 & 3 & bi=ibi & 12 \\ \hline 3 & 2 & ib=bib & 12 \\ \hline 4 & 5 & bib=b & 12 \\ \hline 5 & 4 & ibi=i & 12 \\ \hline 6 & 6 & ib=ibi,\ bi=bib & 10 \\ \hline 7 & 7 & ib=bib,\ bi=ibi & 10 \\ \hline 8 & 9 & ib=bib,\ ibi=i & 10 \\ \hline 9 & 8 & bi=ibi,\ bib=b & 10 \\ \hline 10 & 11 & bi=ibi=i & 10 \\ \hline 11 & 10 & ib=bib=b & 10 \\ \hline 12 & 12 & bib=b,\ ibi=i & 10 \\ \hline 13 & 13 & ibi=bi=ib=bib & 8 \\ \hline 14 & 16 & ib=ibi=i,\ bi=bib & 8 \\ \hline 15 & 17 & ib=bib,\ bi=ibi=i & 8 \\ \hline 16 & 14 & ib=ibi,\ bi=bib=b & 8 \\ \hline 17 & 15 & ib=bib=b,\ bi=ibi & 8 \\ \hline 18 & 19 & bi=ibi=i,\ bib=b & 8 \\ \hline 19 & 18 & ib=bib=b,\ ibi=i & 8 \\ \hline 20 & 21 & ibi=bi=ib=bib=i & 6 \\ \hline 21 & 20 & ibi=bi=ib=bib=b & 6\\ \hline 22 & 22 & ib=ibi=i,\ bi=bib=b & 6 \\ \hline 23 & 24 & \textsf{id}=b,\ ib=ibi=i,\ bi=bib & 6 \\ \hline 24 & 23 & \textsf{id}=i,\ bi=bib=b,\ ib=ibi & 6 \\ \hline 25 & 25 & ib=bib=b,\ bi=ibi=i & 4,6 \\ \hline 26 & 27 & \textsf{id}=bi=bib=b,\ ib=ibi=i & 4 \\ \hline 27 & 26 & \textsf{id}=ib=ibi=i,\ bi=bib=b & 4 \\ \hline 28 & 29 & \textsf{id}=b,\ ibi=bi=ib=bib=i & 4 \\ \hline 29 & 28 & \textsf{id}=i,\ ibi=bi=ib=bib=b & 4 \\ \hline 30 & 30 & ibi=bi=ib=bib=b=i=\textsf{id} & 4 \\ \hline \end{array}

When $h(A)=25,$ if $A$ is dense with empty interior (e.g., $\mathbb{Q}$ in $\mathbb{R}$ under the usual topology) then $k(A)=4,$ otherwise $k(A)=6.$

Conjecture. Let $A$ and $B$ be subsets of a topological space. If $$\tag1h(A)\in\{22,23,24,26,27\}\text{ and }h(B)=25,$$ then $$\tag2\min\{h(A\cup B),h(A\cap B),h(A\cup aB),h(A\cap aB)\}<20.$$

If true, the conjecture implies $(14,6)$ is impossible. For, suppose $\XT$ is a Kuratowski space with $k$‑number 6. Since $K(\XT)=14,$ there exist subsets $A$ and $B$ of $X$ such that $biA\neq ibiA$ and $ibB\neq ibiB.$ Since $k(\XT)=6,$ it follows from the table that $(1)$ holds. The conjecture then contradicts $k(\XT)=6.$

Computer experiments have verified the conjecture for all 2450 inequivalent $\text{non-}T_0$ spaces such that $1\leq|X|\leq7$ (these were recently posted here) and roughly 40,000 others such that $8\leq|X|\leq16.$ We ignore $T_0$ spaces because finite ones satisfy $K(\XT)\leq10$ by Theorem 3 in Herda and Metzler. We also ignore sets $A$ such that $h(A)\in\{24,27\}$ because our C program scours entire power sets and De Morgan's laws imply that $A$ provides a counterexample iff $aA$ does.

Our list of 136 known (at the time of writing) $4$‑tuples $$(h(A\cup B),h(A\cap B),h(A\cup aB),h(A\cap aB))$$ satisfying $h(A)\in\{22,23,26\}$ and $h(B)=25$ can be found here.*

A few weeks ago I posted a question here based on the case $(26,25,30,12),$ where $h(A)=26.$ Currently no answer has appeared.

Here is the list promised above. It is surely complete, but we lack proof. The cardinality of each space is minimal. It is assumed that $X=\{1,\ldots,n\}$ when $|X|=n.$

\begin{array}{|c|c|c|c|} \hline \text{pair} & |X| & \text{base for }\mathcal{T} & h\text{-numbers that occur} \\ \hline (2,2) & 1 & \{\{1\}\} & 30 \\ \hline (6,4) & 2 & \{\{1,2\}\} & 25,30 \\ \hline (6,6) & 3 & \{\{1\},\{2,3\}\} & 25,30 \\ \hline (8,4) & 2 & \{\{1\},\{1,2\}\} & 28\text{-}30 \\ \hline (8,6) & 3 & \{\{1\},\{1,2\},\{1,2,3\}\} & 20,21,28\text{-}30 \\ \hline (8,8) & 4 & \{\{1\},\{2\},\{1,3\},\{2,4\}\} & 13,28\text{-}30 \\ \hline (10,4) & 3 & \{\{1\},\{2\},\{1,2,3\}\} & 26\text{-}30 \\ \hline (10,6) & 4 & \{\{1\},\{2\},\{1,2,3\},\{1,2,4\}\} & 22,26\text{-}30 \\ \hline (10,8) & 4 & \{\{1\},\{2\},\{1,3\},\{1,2,4\}\} & 14,16,23,24,26\text{-}30 \\ \hline (10,10) & 5 & \{\{1\},\{2\},\{1,3\},\{2,4\},\{1,2,5\}\} & 6,14,16,23,24,26\text{-}30 \\ \hline (14,8) & 4 & \{\{1\},\{2,3\},\{1,2,3,4\}\} & 18,19,26\text{-}30 \\ \hline (14,10) & 5 & \{\{1\},\{2,3\},\{1,4\},\{1,2,3,5\}\} & 10,11,14,16,18,19,23,24,26\text{-}30 \\ \hline (14,12) & 6 & \{\{1\},\{2\},\{3,4\},\{1,5\},\{1,2,6\}\} & 4,5,12,14\text{-}17,23\text{-}30 \\ \hline (14,14) & 7 & \{\{1\},\{2\},\{3,4\},\{1,5\},\{2,6\},\{1,2,7\}\} & 1,4,5,6,12,14\text{-}17,23\text{-}30 \\ \hline \end{array} * Update. (added Apr 30 2019)

When $h(A)=23,$ there exists $x\in A$ such that $h(A\setminus\{x\})=14.$ A proof is given here.

The cases that remain are $h(A)=22$ and $h(A)=26.$ The updated list of found $4$‑tuples appears below. The newest entry got added over a million randomly-generated spaces ago, so it may be complete.

\begin{array}{|c|c|c|c|c|} \hline h(A)\! & h(A\cup B) & h(A\cap B) & h(A\cup aB) & h(A\cap aB) \\ \hline \phantom{22,26\,}\llap{22,26} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\,\phantom{iiiii}} \\ \hline \end{array} \begin{array}{|c|c|c|c|c|} \hline \phantom{22,26\,}\llap{22,26} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{18\phantom{((((}} & \phantom{h(A\cup aB)}\llap{19\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\,\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22,26} & \phantom{h(A\cup B)}\llap{19\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{18\,\phantom{iiiii}} \\ \hline \end{array} \begin{array}{|c|c|c|c|c|} \hline \phantom{22,26\,}\llap{22,26} & \phantom{h(A\cup B)}\llap{19\phantom{((((}} & \phantom{h(A\cup B)}\llap{18\phantom{((((}} & \phantom{h(A\cup aB)}\llap{19\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{18\phantom{iiiii}} \\ \hline \end{array} \begin{array}{|c|c|c|c|c|} \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{22\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{22\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{22\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{22\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \end{array} \begin{array}{|c|c|c|c|c|} \hline \phantom{22,26\,}\llap{22,26} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{25\phantom{((((}} & \phantom{h(A\cup aB)}\llap{25\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22,26} & \phantom{h(A\cup B)}\llap{25\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{25\phantom{iiiii}} \\ \hline \end{array} \begin{array}{|c|c|c|c|c|} \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{22\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{26\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{26\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{22\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{22\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{26\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{26\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{22\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \end{array} \begin{array}{|c|c|c|c|c|} \hline \phantom{22,26\,}\llap{26\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{26\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{26\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{26\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{26\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{26\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \end{array} \begin{array}{|c|c|c|c|c|} \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{22\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{27\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{27\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{22\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{22\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{27\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{27\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{22\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \end{array} \begin{array}{|c|c|c|c|c|} \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{26\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{27\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{27\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{26\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{26\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{27\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{27\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{26\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \end{array} \begin{array}{|c|c|c|c|c|} \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{30\phantom{((((}} & \phantom{h(A\cup aB)}\llap{25\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{22\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{22\phantom{((((}} & \phantom{h(A\cup B)}\llap{25\phantom{((((}} & \phantom{h(A\cup aB)}\llap{30\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{25\phantom{((((}} & \phantom{h(A\cup B)}\llap{22\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{30\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{22\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{30\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{22\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{25\phantom{iiiii}} \\ \hline \end{array} \begin{array}{|c|c|c|c|c|} \hline \phantom{22,26\,}\llap{26\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup B)}\llap{30\phantom{((((}} & \phantom{h(A\cup aB)}\llap{25\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{26\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{26\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{25\phantom{((((}} & \phantom{h(A\cup B)}\llap{26\phantom{((((}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{30\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{26\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{26\phantom{((((}} & \phantom{h(A\cup B)}\llap{25\phantom{((((}} & \phantom{h(A\cup aB)}\llap{30\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{12\phantom{iiiii}} \\ \hline \phantom{22,26\,}\llap{26\phantom{\,\,\,\,}} & \phantom{h(A\cup B)}\llap{30\phantom{((((}} & \phantom{h(A\cup B)}\llap{12\phantom{((((}} & \phantom{h(A\cup aB)}\llap{26\phantom{iiiii}} & \phantom{h(A\cup aB)}\llap{25\phantom{iiiii}} \\ \hline \end{array}

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  • $\begingroup$ There was some discussion of related topics at math.stackexchange.com/questions/186017/… though I don't think it reached the level of the question here. $\endgroup$ – Gerry Myerson Apr 30 '19 at 23:18
  • $\begingroup$ A silly typo: surely $\mathcal T \setminus \varnothing$ should be $\mathcal T \setminus \{\varnothing\}$? $\endgroup$ – LSpice Jun 2 '19 at 23:55
  • 1
    $\begingroup$ @LSpice thanks, good catch. It's been corrected. $\endgroup$ – mathematrucker Jun 3 '19 at 0:41
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$\newcommand{\XT}{(X,\mathcal{T})}\newcommand{\vsp}{\raise7pt\hbox{$\phantom-$}}$As expected, the answer is no.

Overview. The proof makes heavy use of characterizations given in [1] and Theorem 2.10 in [2]. It begins by showing that every Kuratowski space contains a subset $A$ such that $h(A)=26$. Our space also contains a subset $B$ satisfying a condition that allows us to assume $h(B)=25$ (for otherwise, $k(B)\geq8$).

The $4$‑tuple in the conjecture turns out to be an effective place to look for subsets with $k$‑number greater than $6$. Which set works depends on the value of $h(A\cup B)$:

$$\eqalign{h(A\cup B)<20\;&\Longrightarrow\;k(A\cup B)\geq8,\cr h(A\cup B)\in\{20,25\}\;&\Longrightarrow\;k(A\cup aB)\geq8,\cr h(A\cup B)=26\;&\Longrightarrow\;k(A\cap aB)\geq8,\cr h(A\cup B)=30\;&\Longrightarrow\;k(A\cap B)\geq8.}$$

As will be shown, this exhausts all possible values of $h(A\cup B)$, hence every Kuratowski space has $k$‑number greater than $6$.

Theorem. Every topological space $\XT$ such that $K(\XT)=14$ satisfies $k(\XT)\geq8$.

Proof. Let $\XT$ be a topological space.

Lemma 1. For each $E\subset X$, let $g(E)$ denote the number of distinct subsets $E$ generates under $b$ and $i$. When $h(E)\neq25$, we have $$k(E)=2g(E).$$ When $h(E)=25$, if $bE=X$ and $iE=\varnothing$, then $k(E)=4$, otherwise $k(E)=2g(E)=6$.

Proof. This is proved in the second paragraph of Section 2.2 in [2], on page 15.

The following lemma implies that every Kuratowski space contains a subset $A$ such that $h(A)=26$.

Lemma 2. For each $E\subset X$,

$\ \ \ \ \ \ \ b(biE)=(biE)=bi(biE)=bib(biE)$ and

$\ \ \ \ \ \ \ ib(biE)=i(biE)=ibi(biE)$.

Proof. This holds immediately by idempotence of $b$, $bi$ and $ib$.

From here on, it will be assumed that $K(\XT)=14$.

There exists $E\subset X$ such that $i(biE)\neq(biE)$. By Lemma 2 this implies $h(biE)=26$, so we let $A=biE$.

There exists $B\subset X$ such that $ibiB\neq ibB$. Since we can assume $h(B)\geq20$, it follows that $h(B)=25$.

The sets $A$, $B$ and $aB$ satisfy the following Hasse diagrams, where sets are equal iff they have the same color.

Hasse diagrams Kuratowski closure complement 14 sets

Since $h(A)=26$, it follows by Theorem 1 in [1] that $$\tag1A=iA\cup V$$ where $A$ is closed, $iA\neq\varnothing$, $V\neq\varnothing$, and $iA\cap V=\varnothing$.

Since $h(B)=h(aB)=25$, it follows by Theorem 5 in [1] that $$\tag2B=iB\cup Y$$ and $$\tag3aB=iaB\cup Z$$ where $iB$ and $iaB$ are clopen (possibly empty), $Y\neq\varnothing$, $Z\neq\varnothing$, and $iB\cap Y=iaB\cap Z=\varnothing$.

Note: Theorems 1 and 5 in [1] supply further properties that have been omitted since they are not needed.

Claim 1. $Y\cup Z$ is clopen and $bY=bZ=Y\cup Z$.

Proof. We have $$\tag4X=iB\cup iaB\cup Y\cup Z$$ where the sets on the right are pairwise disjoint. Since $bB$ and $baB$ are each clopen, the set $Y\cup Z=a(iB)\cap a(iaB)=baB\cap bB$ is clopen. Thus, $bY\subset b(Y\cup Z)=Y\cup Z$. Since $iB$ is clopen, we have $$iB\cup bY=biB\cup bY=b(iB\cup Y)=bB=a(iaB)=iB\cup(Y\cup Z).$$ Since $(Y\cup Z)\cap iB=\varnothing$, this implies $Y\cup Z\subset bY$. Hence, $bY=Y\cup Z$. The equation $bZ=Y\cup Z$ holds similarly. This proves Claim 1.

The next claim will be used often.

Claim 2. $i(A\cup B)=iA\cup iB$ and $i(A\cup aB)=iA\cup iaB$.

Proof. Since $A$ is closed and $i(A\cup B)\subset A\cup B$, the set $i(A\cup B)\setminus A$ is an open subset of $B$. Thus $$i(A\cup B)=[i(A\cup B)\cap A]\cup[i(A\cup B)\setminus A]\subset A\cup iB.$$ Since $iB$ is clopen and $i(A\cup B)\subset A\cup iB$, the set $i(A\cup B)\setminus iB$ is an open subset of $A$. Therefore $$i(A\cup B)=[i(A\cup B)\setminus iB]\cup[i(A\cup B)\cap iB]\subset iA\cup iB.$$ The reverse inclusion always holds, hence $i(A\cup B)=iA\cup iB$. The second equation holds similarly. This proves Claim 2.

Corollary 1. $bi(A\cup B)=biA\cup biB$ and $bi(A\cup aB)=biA\cup biaB$.

Proof. This follows immediately from Claim 2 since closure distributes across union.

The following claim allows us to rule out several $h$‑numbers.

Claim 3. $A\cup B=b(A\cup B)\;\Longleftrightarrow\;b(A\cup B)=bi(A\cup B)$.

Proof. $(\Rightarrow)$ By Claim 2, $$B\subset bB=ibB\subset ib(A\cup B)=i(A\cup B)=iA\cup iB\subset A\cup biB.$$ Hence $$b(A\cup B)=A\cup B\subset A\cup biB=biA\cup biB=bi(A\cup B).$$ Since $bi(A\cup B)\subset b(A\cup B)$ always, the result follows.

$(\Leftarrow)$ Let $i_{A\cup B}$ denote the interior operator in the subspace $A\cup B$ of $X$. By Theorem 2 in [1], we have $$\tag5A\cup B=i(A\cup B)\cup W$$ where $i(A\cup B)\cap W=i_{A\cup B}W=\varnothing$. Let $$U=aiB\cap aA.$$ Note that $U$ is open. By Claim 2, we have $$\tag6U\cap i(A\cup B)=(aiB\cap aA)\cap(iA\cup iB)=\varnothing.$$ Since $U\cap(A\cup B)$ is open in the subspace $A\cup B$ of $X$ and $i_{A\cup B}W=\varnothing$, equations $(5)$ and $(6)$ imply $U\cap W=\varnothing$. Thus $U\cap(A\cup B)=\varnothing$. Since $Y\cap aA\subset U\cap(A\cup B)$, we get $Y\subset A$. Hence, by Claim 1, $Y\cup Z=bY\subset bA=A$. Thus $$b(A\cup B)=bA\cup bB=A\cup a(iaB)=A\cup[iB\cup(Y\cup Z)]\subset A\cup B.$$ Since $A\cup B\subset b(A\cup B)$ always, this completes the proof of Claim 3.

Corollary 2. $h(A\cup B)\not\in\{21,22,23,24,27,28,29\}$.

Proof. This holds by Claim 3.

Corollary 3. If $h(A\cup B)=26$, then $Y\cup Z\subset iA$ and $V\cap iaB\neq\varnothing$.

Proof. It was shown in part $(\Leftarrow)$ above that $b(A\cup B)=bi(A\cup B)$ implies $Y\cup Z\subset A$. Since $Y\cup Z$ is clopen, this implies $Y\cup Z\subset iA$. Thus, $$\tag7V\cap(Y\cup Z)=\varnothing.$$ If $V\subset iB$, then $$i(A\cup B)=iA\cup iB=iA\cup(V\cup iB)=A\cup iB=biA\cup biB=bi(A\cup B).$$ Since this violates $h(A\cup B)=26$, we have $V\not\subset iB$. It follows by $(4)$ and $(7)$ that $V\cap iaB\neq\varnothing$. This completes the proof of Corollary 3.

Claim 4. If $h(A\cup B)=26$, then $$i(A\cap aB)\subsetneq bi(A\cap aB)\subsetneq(A\cap aB)\subsetneq b(A\cap aB).$$

Proof. Suppose $x\in V\cap iaB$ and $P$ is an open neighborhood of $x$. Since $V\subset A=biA$, the open neighborhood $P\cap iaB$ of $x$ contains a point $y$ in $iA$. Since $P$ was arbitrary, this implies $V\cap iaB\subset b(iA\cap iaB)=bi(A\cap aB)$. By Corollary 3 we have $V\cap iaB\neq\varnothing$. Since $(V\cap iaB)\cap i(A\cap aB)=\varnothing$, it follows that $$i(A\cap aB)\subsetneq bi(A\cap aB).$$ Note that $$bi(A\cap aB)=b(iA\cap iaB)\subset biA\cap biaB=A\cap iaB.$$ Since $Z\subset A$ by Corollary 3, we have $$\tag8A\cap aB=A\cap(iaB\cup Z)=(A\cap iaB)\cup Z.$$ Since $(A\cap iaB)\cap Z=\varnothing\neq Z$, it follows that $$bi(A\cap aB)\subsetneq (A\cap aB).$$ Note that $A\cap iaB$ is closed. By $(8)$ and Corollary 3, it follows that $$\eqalign{b(A\cap aB)&=b[(A\cap iaB)\cup Z]\cr&=b(A\cap iaB)\cup bZ\cr&=(A\cap iaB)\cup(Z\cup Y)\cr&=(A\cap iaB)\cup(A\cap Z)\cup Y\cr&=(A\cap aB)\cup Y.}$$ Since $(A\cap aB)\cap Y=\varnothing\neq Y$, we conclude $$(A\cap aB)\subsetneq b(A\cap aB).$$ This completes the proof of Claim 4.

Corollary 4. If $h(A\cup B)=26$, then $k(A\cap aB)\geq8$.

Proof. This holds by Claim 4 and Lemma 1.

Claim 5. If $bi(A\cup B)=i(A\cup B)\neq A\cup B$, then $$i(A\cup aB)\subsetneq bi(A\cup aB)\subsetneq (A\cup aB)\subsetneq b(A\cup aB).$$

Proof. By $(1)$, $(3)$, $(4)$ and Claim 2, we have

$$\eqalign{i(A\cup aB)&=iA\cup iaB,\cr\vsp bi(A\cup aB)&=biA\cup biaB\cr&=A\cup iaB\cr&=(iA\cup iaB)\cup V,\cr\vsp(A\cup aB)&=(iA\cup iaB)\cup V\cup Z,\cr\vsp b(A\cup aB)&=bA\cup baB\cr&=A\cup aiB\cr&=A\cup(iaB\cup Z\cup Y)\cr&=(iA\cup iaB)\cup V\cup Z\cup Y.\cr}$$

Since $iB$ and $i(A\cup B)$ are each clopen, by Claim 2 we have $$\tag9\eqalign{V\cup iA\cup iB=A\cup iB&=biA\cup biB\cr&=bi(A\cup B)=i(A\cup B)=iA\cup iB.}$$ Since $V\cap iA=\varnothing$, this implies $V\subset iB$. Hence, $V\cap(iA\cup iaB)=\varnothing$. Since $V\neq\varnothing$, this gives us $$i(A\cup aB)\subsetneq bi(A\cup aB).$$ By $(9)$ we have $A\cup iB=i(A\cup B)$. Hence, $$\eqalign{Y\subset A\;&\Longrightarrow\;A\cup(iB\cup Y)\subset A\cup iB\cr&\Longrightarrow\;A\cup B\subset i(A\cup B)\cr&\Longrightarrow\;A\cup B=i(A\cup B).}$$ Since $A\cup B\neq i(A\cup B)$, this implies $Y\not\subset A$. On the other hand, $$Z\subset A\;\Longrightarrow\;Y\cup Z=bZ\subset bA=A.$$ Since $Y\not\subset A$, this implies $Z\not\subset A$. Since $Z\cap iaB=\varnothing$, it follows that $$bi(A\cup aB)\subsetneq (A\cup aB).$$ Since $Y\not\subset A$ and $Y\cap aB=\varnothing$, it further follows that $$(A\cup aB)\subsetneq b(A\cup aB).$$ This completes the proof of Claim 5.

Corollary 5. If $h(A\cup B)\in\{20,25\}$, then $k(A\cup aB)\geq8$.

Proof. This holds by Claim 5 and Lemma 1.

Aside. It can be shown that

$\ \ \ \ \ \ \ (h(A)=26$ and $h(B)=25)\;\Longrightarrow\;h(A\cup B)\neq20$.

Claim 6. If $h(A\cup B)=30$, then $$i(A\cap B)\subsetneq bi(A\cap B)\subsetneq(A\cap B)\subsetneq b(A\cap B).$$

Proof. By $(1)$, $(2)$ and Claim 2, $$(iA\cup V)\cup(iB\cup Y)=A\cup B=i(A\cup B)=iA\cup iB.$$ Since $V\cap iA=Y\cap iB=\varnothing$, this implies $V\subset iB$ and $Y\subset iA$. Hence, $$\eqalign{bi(A\cap B)&=b(iA\cap iB)\cr&\subset biA\cap biB\cr&=A\cap iB\cr&=(iA\cap iB)\cup(V\cap iB)\cr&=(iA\cap iB)\cup V.}$$ Suppose $x\in V$ and $P$ is an open neighborhood of $x$. Since $x\in biA$, the open neighborhood $P\cap iB$ of $x$ contains a point $y\in iA$. Since $P$ was arbitrary, it follows that $V\subset b(iA\cap iB)=bi(A\cap B)$. Clearly $(iA\cap iB)\subset bi(A\cap B)$, hence $bi(A\cap B)=(iA\cap iB)\cup V$. Since $(iA\cap iB)\cap V=\varnothing\neq V$, this implies $$i(A\cap B)\subsetneq bi(A\cap B).$$ Since $Y\subset iA$, we have $V\cap Y=\varnothing$. It follows that $$\eqalign{A\cap B&=(iA\cup V)\cap(iB\cup Y)\cr&=(iA\cap iB)\cup V\cup Y\cr&=bi(A\cap B)\cup Y.}$$ Since $(iA\cap iB)\cap Y=\varnothing\neq Y$, this implies $$bi(A\cap B)\subsetneq(A\cap B).$$ By Corollary 3 and the expression above for $A\cap B$, we have $$\eqalign{b(A\cap B)&=b[bi(A\cap B)\cup Y]\cr&=bi(A\cap B)\cup bY\cr&=bi(A\cap B)\cup Y\cup Z\cr&=(A\cap B)\cup Z.}$$ Since $(A\cap B)\cap Z=\varnothing\neq Z$, this implies $$(A\cap B)\subsetneq b(A\cap B).$$ This completes the proof of Claim 6.

Corollary 6. If $h(A\cup B)=30$, then $k(A\cap B)\geq8$.

Proof. This holds by Claim 6 and Lemma 1.

Claim 7. If $h(A\cup B)\geq20$, then $k(\XT)\geq8$.

Proof. This holds by Corollaries 2, 4, 5, and 6.

If $h(A\cup B)<20$, then $k(A\cup B)\geq8$. Conclude $k(\XT)\geq8.$ $\blacksquare$

Remark. The Apr 30 2019 update to the question mentions that if $A\subset X$ exists with $h(A)=23$, then $X$ contains a subset with $h$‑number 14. It turns out that the converse is also true. Taking complements, we get the mildly interesting result that a topological space $X$ contains a subset with $h$‑number in $\{14,16,23,24\}$ iff it contains subsets with all four of those $h$‑numbers. It turns out that no other nontrivial theorems of the form

$\ \ \ \ \ \ \ (X$ contains $A$ such that $h(A)=x)\;\Longrightarrow\;(X$ contains $B$ such that $h(B)=y)$

exist besides the one above.

References.

[1] T. A. Chapman, A further note on closure and interior operators, Amer. Math. Monthly, 69 1962, 524‑529.

[2] B. J. Gardner and M. Jackson, The Kuratowski closure-complement theorem, New Zealand J. Math., 38 2008, 9‑44.

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