Kuratowski Operations I went through the concept of Kuratowski monoid in the paper by " B. J. Gardner and M. Jackson, The Kuratowski closure-complement theorem, New Zealand J. Math.  38 (2008), 9--44".  They consider the set of all distinct operators on the  topological space $( X,\tau )$ produced by composition of  closure and complement operators.   This set forms a monoid as $c ^2=I$ where $c, I$ denote the complement and identity operator respectively. This monoid is the Kuratowski monoid consisting of Kuratowski operators.
In the paper by  D. Sherman, Variations on Kuratowski's 14-set theorem,   Amer. Math. Monthly  117 (2010), no.2, 113--123. We have the following theorem:
Let $k, i ∈ End(P(X))$ be the closure and interior operators of a topological
space. Then the cardinality of the monoid generated by $k$ and $i$ is at most 7 
where $End(P(X))$ is the  collection of maps $P(X) \longrightarrow P(X)$, $X$ is a topological space. 
A doubt came in my mind which  is when the operations closure ($k$) and interior ($i$) are in consideration how come it forms a monoid ? Since $k^2=k$ and $i^2=i$ so no composition between closure and interior becomes identity. So is the presence of identity operation an assumption which is made  in all such equivalent versions  of Kuratowski's Theorem?
 A: The monoid consists of $I$, $k$, $i$, $ki$, $iki$, $ik$, $kik$, and no more, as $kiki = ki$ and $ikik=ik$ in all spaces, and indeed $i^2 = i, k^2 = k$.
A monoid does not have to have inverses (it would be a group if it had). We just have an associative operation, with an identity $1$ (which gives it its name,  I think, mono refers to the $1$). So in order to be a monoid, we need a 1, which is the identity. In general we do need the identity when we have $c$ (complement) as well, because then $cc = 1$ and it actually occurs as a product.
But consider a classical monoid of all finite strings over an alphabet $\Sigma$, denoted $\Sigma^\ast$, with concatenation $\ast$ as operation. There we also add $\varepsilon$, the empty string, to make this a monoid, while $\varepsilon$ will also never occur as a product, except in the trivial $\varepsilon\ast\varepsilon = \varepsilon$, or in the Kuratowski case $II = I$. 
A: The questioner is only asking how the identity element gets into the monoid (in Sherman's paper) that is generated by $k$ and $i$.
This is a good question.  Since it is framed within the context of Sherman's paper, it helps to consider the lead-in to Lemma 2.2 (which states that the cardinality of the monoid generated by $k$ and $i$ is at most 7):
"The advantage in the situation at hand is that we may invoke a familiar friend from group theory (or universal algebra, to those in the know): presentations. This just means that we will describe sets of operations in terms of generators and relations."
So we see here that Sherman is talking about monoids defined in terms of generators and relations.  Hence these are the monoids the questioner is asking about.
The questioner is clearly not aware that these particular monoids exist within a universe monoid consisting of all finite strings over some alphabet Σ under concatenation; for otherwise the question would not have been asked.
Since a string of length zero is a finite string (over any alphabet) which serves as the identity string under concatenation, the existence of the identity in a monoid defined in terms of generators (the alphabet Σ above) and relations is always implied by the definition (of the universe monoid). The given relations therefore need not produce an identity element.
