$ \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}