Existence of a closure of a subset

Question

Let $M$ be a set and $\circ$ an operation defined on subsets of $M$ in the following way:

1. $A \circ B \subseteq M$ for any $A \subseteq M$ and $B \subseteq M$;
2. If $A' \subseteq A$ and $B' \subseteq B$, then $A' \circ B' \subseteq A \circ B$.

Let us say a subset $C$ of $M$ is closed under $\circ$ if $C \circ C \subseteq C$.

Clearly, $M$ is closed under $\circ$.

Let us say a subset $C$ is a closure of a subset $S$ of $M$ if:

1. $C$ is closed under $\circ$ and $S \subseteq C$;
2. If $S \subseteq C' \subseteq C$ for a subset $C'$ closed under $\circ$, then $C' = C$.

Clearly, $M$ is a closure of itself.

It is easy to show that a closure of a subset $S$ of $M$, if exists, is unique:

• Assuming $C_1$ and $C_2$ are two closures of $S$;
• Denoting $C$ = $C_1 \cap C_2$;
• $C \subseteq C_1$, therefore $C \circ C \subseteq C_1 \circ C_1 \subseteq C_1$;
• $C \subseteq C_2$, therefore $C \circ C \subseteq C_2 \circ C_2 \subseteq C_2$;
• $C \circ C \subseteq C_1$ and $C \circ C \subseteq C_2$, therefore $C \circ C \subseteq C_1 \cap C_2$;
• $C \circ C \subseteq C$, therefore $C$ is closed under $\circ$;
• $C$ is closed under $\circ$, $S \subseteq C \subseteq C_1$, therefore $C = C_1$;
• $C$ is closed under $\circ$, $S \subseteq C \subseteq C_2$, therefore $C = C_2$.

But how can we prove that the closure exists for any subset of $M$?

If it is not true, what could be an additional requirement on $\circ$ to guarantee existence of the closure?

Answer 1

Let $S\subseteq M$ be arbitrary. Let $\mathcal{A}$ be the collection of all closed supersets of $S$. We know that $M\in\mathcal{A}$ so $\mathcal{A}\neq \emptyset$. Let us consider $C=\bigcap \mathcal{A}$.

First we will check that $C$ is a closed superset of $S$. Arguing as you did, for any $K\in \mathcal{A}$

• $C\subseteq K$, therefore $C\circ C\subseteq K\circ K \subseteq K$.

As $K$ was arbitrary this means $C\circ C\subseteq C$ and hence $C$ is a closed superset of $S$. By construction $C$ is also the smallest such set and hence the closure of $S$.