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?