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# How many different kinds of sets can you get with complement and closure?
Consider a topological space $(X,\tau)$, where $\tau$ is the collection of all open sets in $X$.
Let us investigate the following problem. If we start with an arbitrary subset $A\subset X$, how many different sets can we create from $A$ by applying the operations complement $A\mapsto A^c$ and closure $A\mapsto A^k$ in some finite order?
That is, starting with some $A\subset X$, how big could the following collection of sets get:
$$
\operatorname{Ku}(A) = \{A,A^c,A^k,(A^c)^k, (A^k)^c,((A^k)^c)^k,\ldots\}
$$
This is a curious theorem of **Kuratowski**, so let us denote such collection as $\operatorname{Ku}(A)$. Hence we want to know if we can say anything about the size of $\operatorname{Ku}(A)$.
===
For any given subsets $A,B\subset X$, we have have the following natural set operations:
(S1) Union: $A\cup B$
(S2) Intersection: $A\cap B$
(S3) Complementation: $A^c = X-A$
And since $X$ is a topological space, we also have the following topological operations
(T1) Closure: $A^k$
(T2) Interior: $A^o$
There are many equivalent ways to define closure and interior. For example we can define closure of set $A$ in $X$ as the collection of all **adherent points** of $A$ in $X$, where $p$ is an adherent point of $A$ if every open neighborhood $U\in\tau$ about $p$ intersects $A$. Formally $$
A^k = \{p\in X: \forall_{ U\in\tau} (p \in U \implies U\cap A\neq\varnothing)\}
$$
And one way to define interior of $A$ is the collection of all **interior points** of $A$ in $X$, where $p$ is an interior point of $A$ if there exists some open neighborhood $U\in\tau$ such that $p\in U \subset X$. Formally $$
A^o = \{p\in X:\exists_{U\in \tau} (p \in U \text{ and } U\subset A) \}
$$
![[---images/---assets/---icons/question-icon.svg]] To get a handle of things, consider the set $A$ in usual reals, where $A = \{0\} \cup (\mathbb Q \cap [1,2])$. How big is $\operatorname{Ku}(A)$?
![[---images/---assets/---icons/question-icon.svg]] Show $A^o \subset A$ and $A\subset A^k$.
![[---images/---assets/---icons/question-icon.svg]] Show both interior and closure are **monotone**, that is, if $A\subset B$, then $A^o \subset B^o$ and $A^k \subset B^k$.
![[---images/---assets/---icons/question-icon.svg]] Show $(A^o)^c = (A^c)^k$. As well as $(A^c)^o = (A^k)^c$. Algebraically, if we adopt a "composition on the right" notation, namely writing $f(g(x))$ as $xgf$, then we have $Aoc = Ack$, namely $oc=ck$. and $co=kc$ here.
![[---images/---assets/---icons/question-icon.svg]] Show $oo = o$ and $kk=k$, namely interior and closure operations are **idempotent**.
By the way, if we denote $1$ as the identity operation, then we have $cc=1$, that twice complementation is doing nothing. So complement $c$ is an **involution**.
/// Hint. ///
Recall that in a topology $\tau$, arbitrary union of open sets remain open, and intersection of two open sets is open.
Also, one has equivalent definition that $A^o$ is the union of all open sets $U$ such that $U\subset A$.
Finally, if one can show $oo=o$, one gets $kk=k$ for free. Why?
///
![[---images/---assets/---icons/question-icon.svg]] Show $koko = ko$. And then show $kckckckc = kckc$.
![[---images/---assets/---icons/question-icon.svg]] Give a reasonable upperbound bound to $\operatorname{Ku}(A)$, that is independent of the underlying topological space.
![[---images/---assets/---icons/question-icon.svg]] Can you realize the upperbound of $\operatorname{Ku}(A)$ for some subset $A$ of the reals $\mathbb R$ with usual Euclidean topology?