# Topological Spaces with 14-sets

Let $X$ a topological space and let's suppose there exists $A\subseteq X$ such that applying closure and complement succesively, 14 different sets can be made (This set is called a 14-set).

Does this kind of spaces receive some name (Different from "Space with 14-sets")?

Are there some kinds of requirements for the space to have a 14-set?

For example, I don't know, "a space must be Haussdorff to have a 14-set", which would be false if a nondiscrete finite space with 14-sets is found (For example).

• Wikipedia says that they are just called "14-sets". – Ian Dec 6 '15 at 23:13

Suppose $A$ is a 14-set in a topological space $X=(X, \tau)$, and $Y=(Y, \upsilon)$ is any other topological space. Consider the space $Z=X\oplus Y$ gotten by placing $X$ and $Y$ "side by side" - the underlying set is the disjoint union of $X$ and $Y$, and the topology is generated by $\tau\cup\upsilon$. ($Z$ is the coproduct of $X$ and $Y$.) Then in $Z$, $A$ is still a 14-set. (Proof sketch: let $f_i, g_i\in\{closure, complement\}$, $1\le i\le n$. Then by induction on $n$, we can show that $f_1f_2 . . .f_n(A)=g_1g_2...g_n(A)$ as interpreted in $X$ iff $f_1f_2...f_n(A)=g_1g_2...g_n(A)$ as interpreted in $Y$.)
So if $\pi$ is a property of 14-spaces (=space with a 14-set), then $\pi$ is "upwards hereditary" - $\pi$ holds of $X$ implies $\pi$ holds of $X\oplus Y$, regardless of what $Y$ is. This means that the following, for example, no separation axiom holds of every 14-space. This kills off such a broad class of possible properties that I'm going to tentatively say that the answer to your question is "no."
NOTE: This does not mean that if $X$ is a subspace of $Z$ which has a 14-set, then $Z$ has a 14-set. (We can have $X, Y$ partition a space $Z$ without having $Z\cong X\oplus Y$ - for example, consider the two-point indiscrete space.) In fact, I'm pretty sure this stronger statement is false, although I don't have a counterexample off the top of my head.