While reading your question I was reminded of Kuratowski's closure-complement problem. Here we start with an arbitrary subset $S$ of a topological space, and are allowed to apply the two operations of closure and complement. It turns out that for any $S$, we get at most 14 distinct sets by applying these two operations. If we let $k$ denote the closure operator and $c$ denote the complement operator, then the following relations imply the result

1. $kk=k$
2. $cc=id$, and
3. $kckckck=kck$.

So, to answer the question, if we set $s=ck$ (take the closure of a set and then take the complement of the result), we get the relation $ssss=ss$.

I'll end by mentioning that there do exist $S$ where all 14 sets are possible. For example, under the usual topology of the reals,

$(0,1) \cup (1,2) \cup \{3\} \cup ([4,5] \cap \mathbb{Q})$

is one such set.

While reading your question I was reminded of Kuratowski's closure-complement problem. Here we start with an arbitrary subset $X$ of a topological space, and are allowed to apply the two operations of closure and complement. It turns out that for any $X$, we get at most 14 distinct sets by applying these two operations. If we let $k$ denote the closure operator and $c$ denote the complement operator, then the following relations imply the result

1. $kk=k$
2. $cc=id$, and
3. $kckckck=kck$.

So, to answer the question, if we set $s=ck$ (take the closure of a set and then take the complement of the result), we get the relation $ssss=ss$.

I'll end by mentioning that there do exist $X$ where all 14 sets are possible. For example, under the usual topology of the reals,

$(0,1) \cup (1,2) \cup \{3\} \cup ([4,5] \cap \mathbb{Q})$

is one such set.

This isn't really an answer to your question, but while reading it I was reminded of Kuratowski's closure-complement problem. Here we start with an arbitrary subset $S$ of a topological space, and are allowed to apply the two operations of closure and complement. It turns out that for any $S$, we get at most 14 distinct sets by applying these two operations. If we let $k$ denote the closure operator and $c$ denote the complement operator, then the following relations imply the result

1. $kk=k$
2. $cc=id$, and
3. $kckckck=kck$.

There do exist $S$ where all 14 sets are possible. For example, under the usual topology of the reals,

$(0,1) \cup (1,2) \cup \{3\} \cup ([4,5] \cap \mathbb{Q})$

is one such set.

While reading your question I was reminded of Kuratowski's closure-complement problem. Here we start with an arbitrary subset $S$ of a topological space, and are allowed to apply the two operations of closure and complement. It turns out that for any $S$, we get at most 14 distinct sets by applying these two operations. If we let $k$ denote the closure operator and $c$ denote the complement operator, then the following relations imply the result

1. $kk=k$
2. $cc=id$, and
3. $kckckck=kck$.

So, to answer the question, if we set $s=ck$ (take the closure of a set and then take the complement of the result), we get the relation $ssss=ss$.

I'll end by mentioning that there do exist $S$ where all 14 sets are possible. For example, under the usual topology of the reals,

$(0,1) \cup (1,2) \cup \{3\} \cup ([4,5] \cap \mathbb{Q})$

is one such set.