Let ${\frak C} \subseteq {\cal P}({\cal P}(\omega))$ be the collection of all covers of $\omega$ (that is, ${\cal C} \in {\frak C}$ iff $\bigcup {\cal C} = \omega$.)


We define the following binary relation on ${\frak C} = $: For ${\cal A}, {\cal B} \in {\frak R}$ we say ${\cal A} \leq\_\text{r} {\cal B}$ if ${\cal A}$ is a refinement of ${\cal B}$, that is for all $A\in {\cal A}$ there is $B\in {\cal B}$ with $A\subseteq B$. The elements $\{\omega\}, \{\omega, \{0\}\}\in{\frak C}$ show that the relation $\leq\_\text{r}$ is not anti-symmetric, so we set ${\cal A}\simeq\_\text{r} {\cal B}$ if ${\cal A} \leq\_\text{r}{\cal B}$ and ${\cal B} \leq\_\text{r}{\cal A}$. So we get a poset $$({\frak C}/\simeq\_\text{r},\leq\_\text{r}),$$ where $\leq\_\text{r}$ applies to equivalence classes in the usual way. 


**Question.** Is the poset $({\frak C}/\simeq\_\text{r},\leq\_\text{r})$ a (complete) lattice?



Is this ordering on the set of all covers of $\omega$ a (complete) lattice?
Yes. The l.u.b. of $\mathcal A$ and $\mathcal B$ is $\mathcal A \cup \mathcal B$. The g.l.b. of $\mathcal A$ and $\mathcal B$ is $\{A\cap B: A\in\mathcal A , B\in\mathcal B\}$.


We can even generalize by replacing $\subseteq$ by an arbitrary meet-semilattice preordering, in which case 


* The l.u.b. of $\mathcal A$ and $\mathcal B$ is still $\mathcal A \cup \mathcal B$.
* The g.l.b. of $\mathcal A$ and $\mathcal B$ is now the equivalence closure of $\{A\wedge B: A\in\mathcal A , B\in\mathcal B\}$ where $A\wedge B$ is a representative of the meet of $A$ and $B$.


One example of this is to replace $\subseteq$ by Turing reducibility. Then $A\wedge B$ can be a set encoding both the information in $A$ and $B$. The resulting structure is then called the **Muchnik degrees** (if we ignore the covering $\omega$ condition I guess).


