Can closure and complement generate 14 distinct operations on a topological space if no subset generates more than 6 distinct sets?

$$\newcommand{\XT}{(X,\mathcal{T})}$$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.

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

• 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. – Gerry Myerson Apr 30 '19 at 23:18
• A silly typo: surely $\mathcal T \setminus \varnothing$ should be $\mathcal T \setminus \{\varnothing\}$? – LSpice Jun 2 '19 at 23:55
• @LSpice thanks, good catch. It's been corrected. – mathematrucker Jun 3 '19 at 0:41

$$\newcommand{\XT}{(X,\mathcal{T})}\newcommand{\vsp}{\raise7pt\hbox{\phantom-}}$$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.

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.