This is a follow-up question to this question, prompted by a comment in Todd Trimble's answer.

Let $X\neq \emptyset$ be a set. We say that $U\subseteq {\cal P}(X)\setminus \{\emptyset\}$ is a proper covering if

• $\bigcup U = X$, and
• for $a\neq b\in U$ we have $a\not\subseteq b$.

Let $\text{Cov}(X)$ denote the collection of all (proper) coverings of $X$. For $A, B\in \text{Cov}(X)$ we set $A\leq B$ if $A$ refines $B$, that is for all $a\in A$ there is $b\in B$ such that $a\subseteq b$. (Note that if we consider all coverings instead of just the proper ones, we may lose anti-symmetry.)

Is $\text{Cov}(X)$ with the ordering defined above a lattice? Is it complete?

This is a follow-up question to this question, prompted by a comment in Todd Trimble's answer.

Let $X\neq \emptyset$ be a set. We say that $U\subseteq {\cal P}(X)\setminus \{\emptyset\}$ is a proper covering if

• $\bigcup U = X$, and
• for $a\neq b\in U$ we have $a\not\subseteq b$.

Let $\text{Cov}(X)$ denote the collection of all (proper) coverings of $X$. For $A, B\in \text{Cov}(X)$ we set $A\leq B$ if $A$ refines $B$, that is for all $a\in A$ there is $b\in B$ such that $a\subseteq b$. (Note that if we consider all coverings instead of just the proper ones, we may lose anti-symmetry.)

Is $\text{Cov}(X)$ with the ordering defined above a lattice? Is it complete?