Classifying Topological spaces by Kuratowski monoid I was going through the paper by  B. J. Gardner and M. Jackson. ``The Kuratowski closure-complement theorem ," New Zealand J. Math, 38 (2008),9-44. It deals with the Kuratowski Closure Complement theorem. While reading the paper I came  across the following three definitions (given in the paper):
$k(A)$ (the $k$-number of $A$) denotes the number of distinct sets obtainable  from $A$ by taking closures and complement, 
$ k((X,\tau))$ (the $k$-number of $(X, \tau)$ denotes $max\{k(A) \mid A \subseteq X\}$, 
$ K((X, \tau))$ (the $K$-number of $(X, \tau))$ denotes the number of distinct Kuratowski operators on $(X, \tau)$, that is, the order of the Kuratowski monoid of $(X, \tau)$.
Next the author discusses the theorem, originaly proven in  the paper "A. V. Chagrov, Kuratowski numbers, in Application of functional analysis in
approximation theory, 186–190, Kalinin. Gos. Univ., Kalinin (1982)'', a paper which I couldnt find, but also has been proven by Gardner and Jackson themselves. I state it as is given in the paper:
Theorem 2.1: The Kuratowski monoid of a non-empty topological space corresponds to either the diagram in Figure 1.1 or one of the 5 diagrams in Figure 2.1. Thus the possible $K$-numbers of a topological space are $14, 10, 8, 6, 2$.
I have the following doubts in regarding to the above discussion:
(i) In the definition of $K$-number of $(X,\tau)$ do we mean pairwise distinct operations in the monoid or does a there exists a single subset which distinguishes all the operations in hand (as is with the case of real line which gives 14 distinct sets for its subset)?
(ii) Does the Theorem 2.1 stated above classifies all topological space into six types means  "If we are given any  topological space $(X, \tau)$, by calculating its $K$ number can we put it in one of the above classes"? If it is so then if we consider real line equipped with  indiscrete topology then $K(\mathbb{R}$, indiscrete topology)=4 as closure of any subset is $X$ and interior of any subset is $\emptyset$, ( I may be wrong !!) then which class does it fall?
(iii) Does these bounds 10,8,6,2 attained for any subset of a particular topological space, that is, does there exits subsets for which we can distinguish 10,8,6, or 2 operations using a single subset?
I don't know may be these queries are very trivial and I may be messing up things, but I got confused in the nature of these classifications using the monoid structure, and the concept of $K$-number ( different from $k$-number). It would be very helpful if anyone could just give me an idea about this so that I could proceed with this.
 A: It can take a little time to digest the distinction between $K$-number and $k$-number when you first read about them.  You seem to already understand it pretty well since you are asking the right question: does a single subset distinguish all of the operators in the Kuratowski monoid?
Here are my answers to your three subquestions:
(i) The $K$-number of $(X,{\cal T})$ is the number of (pairwise) distinct operators in the Kuratowski monoid of $(X,{\cal T})$.  Two operators $o_1$ and $o_2$ in this monoid are distinct if and only if there exists $A\subset X$ such that $o_1(A)\neq o_2(A)$.  As the authors note, $K(X,{\cal T})\geq k(X,{\cal T})$.  A single subset $A$ distinguishes all of the operators in the Kuratowski monoid if and only if $K(X,{\cal T})=k(X,{\cal T})$.  The authors call any space satisfying this property a full space. As you mention, the reals under the usual topology is a full space, but there also exist topological spaces that are not full spaces.
(ii) Yes the Kuratowski monoid of every topological space satisfies exactly one of the six possible orderings.  We have $K({\mathbb R},\rm indiscrete\ topology)\geq6$ since the six operators $\{{\rm id},a,b,ab,i,ai\}$ are distinct in this space, where $a$ is complement, $b$ closure and $i$ interior: assuming $\emptyset\subsetneq A\subsetneq\mathbb R$ we have
$iA\subsetneq {\rm id}(A)\subsetneq bA$,
$abA\subsetneq aA\subsetneq aiA$,
$ab{\mathbb R}=ai{\mathbb R}\subsetneq{\rm id}(\mathbb R)=b{\mathbb R}=i{\mathbb R}$,
$b\emptyset=i\emptyset={\rm id}(\emptyset)\subsetneq a\emptyset=ai\emptyset$.
The above contain all ${6\choose2}=15$ inequalities needed to distinguish the six operators.  This proves that $K({\mathbb R},\rm indiscrete\ topology)\geq6$; the reverse inequality holds since $ib=b$ and $bi=i$. (Actually these last two identities together with $b\neq i$ suffice to prove the result since they characterize the Kuratowski monoid, but the inequalities make it easier to see.)
(iii) You appear to be asking roughly the same question that you asked in (i), namely, whether it is always the case that $K(X,{\cal T})=k(X,{\cal T})$.  As I mentioned above, the answer is no.  Searching Gardner and Jackson for an example of a space that is not full, I found this on page 28:
"An example where the sum on only two copies of $X$ will not suffice to provide a full space is the door space on $\{x, y, z\}$ with basis $\{\emptyset, \{x\}, \{y\}, \{x, y, z\}\}$."
Lastly here is an English translation of Chagrov's paper (originally published in Russian):
http://www.mathematrucker.com/mathtransit/cornucopia/1982_chagrov_english.pdf
