Let $X\neq \emptyset$ be a set. We say that $U\subseteq {\cal P}(X)\setminus \{\emptyset\}$ is a *covering* of $X$ if $\bigcup U = X$. We call a covering *proper* if for $a\neq b\in U$ we have $a\not\subseteq b$.


Let $\text{Cov}(X)$ denote the collection of all 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 $a\subseteq b$.


This relation defines a pre-order on $\text{Cov}(X)$; it is easily seen that it is not anti-symmetric. However, the refinement relation is anti-symmetric on $\text{PropCov}(X)$, the set of proper coverings.


On $\text{Cov}(X)$ we set $A \simeq B$ if $A\leq B$ and $B\leq A$. It is easy to verify that the relation $\leq\_q $ on $\text{Cov}(X)/\simeq$ defined by $$[A]\_\simeq \leq\_q [B]\_\simeq \text{ if and only if } A \leq B \text{ in Cov}(X)$$ is a partially ordered set.


Are $(\text{Cov}(X)/\simeq, \leq\_q)$ and $(\text{PropCov}(X),\leq)$ isomorphic as posets?



Proper coverings as a quotient of the coverings
**Are $(\text{Cov}(X)/\simeq, \leq\_q)$ and
$(\text{PropCov}(X),\leq)$ isomorphic as posets?**

If $X$ is finite, then $(\text{Cov}(X)/\simeq, \leq\_q)$
and $(\text{PropCov}(X),\leq)$ 
are isomorphic, but if $X$ is infinite they are not.
To see why they are isomorphic when $X$ is finite it is enough to
note that, in the case of finite $X$, each $\simeq$-class
of covers has the property that all members share
the same maximal elements. Moreover, this common set 
of maximal elements constitutes a proper cover,
and it is the unique proper cover in this $\simeq$-class.
This bijection between $\simeq$-classes of covers
and proper covers preserves and reflects order, hence establishes
an isomorphism between $(\text{Cov}(X)/\simeq, \leq\_q)$ and
$(\text{PropCov}(X),\leq)$. 
To see why $(\text{Cov}(X)/\simeq, \leq\_q)$
and $(\text{PropCov}(X),\leq)$ 
are not isomorphic whene $X$ is infinite,
it will suffice to point out that 
$(\text{Cov}(X)/\simeq, \leq\_q)$ is a lattice
while $(\text{PropCov}(X),\leq)$ is not.

I first argue the easy part: 
$(\text{Cov}(X)/\simeq, \leq\_q)$
is a lattice.


Consider the following closure operator
on the set of covers:
given $B\in \text{Cov}(X)$, let $B^\* = \bigcup\_{A\leq B} A$.
It is easy to see that $B^\*$ is a cover of $X$, and that
an alternative definition for it is
$$
B^\* = (\bigcup\_{b\in B} {\mathcal P}(b))\setminus \{\emptyset\}.
$$
Moreover, $A\leq B$ holds iff $A\subseteq B^\*$ iff $A^\*\subseteq B^\*$,
so $A\simeq B$ iff $A^\*=B^\*$.


Now, a subset $Z\subseteq {\mathcal P}(X)$ is of the form $Z=B^\*$
iff (i) $\emptyset\notin Z$ and (ii) $Z\cup\{\emptyset\}$
is an order ideal of ${\mathcal P}(X)$ which contains
all the singleton subsets of $X$. This is enough to show that 
$(\text{Cov}(X)/\simeq, \leq\_q)$ is isomorphic to the
interval of the lattice of order ideals of ${\mathcal P}(X)$
above the order ideal consisting of singleton sets.
This is a distributive, algebraic lattice.



Now the less-easy part. I argue that 
$(\text{PropCov}(X),\leq)$ is not a lattice.
In fact, I will write out only the case
$X=\omega$, but afterwards I will explain how to extend the argument to
any superset of $\omega$.

Let's name some subsets of $\omega$:  

$E$: the set $\{0,2,4,\ldots\}$ 
of all even natural numbers.  

$E\_n$: the set $\{0,2,4,\ldots,2n\}$ 
of even numbers up to $2n$.  

$E\_n^\*$: the set $\{0,2,4,\ldots,2n,2n+1\}$ 
of even numbers up to $2n$
along with one odd number $2n+1$.
Now let's name some proper covers of $\omega$:  

$C\_E$: the proper cover consisting of $E$ and all odd singletons.  

$C\_{E^\*}$: The proper cover consisting of all $E\_n^\*$'s.  

$C\_n$: The proper cover including $E\_n$ and all singletons
of elements from $\omega \setminus E\_n$.
It is not hard to see that  

(1) $C\_E$ and $C\_{E^\*}$ are incomparable under $\leq$.
  

(2) $C\_n\leq C\_E$ and $C\_n\leq C\_{E^\*}$ for all $n$.  



My goal is to show that $C\_E$ and $C\_{E^\*}$ do not have a
meet (= greatest lower bound) in $(\text{PropCov}(X),\leq)$.
Suppose that $D$ is a proper cover of $\omega$
that is a candidate for the meet 
of $C\_E$ and $C\_{E^\*}$ in $(\text{PropCov}(X),\leq)$.
Then $D\leq C\_E$, $D\leq C\_{E^\*}$, and $D$ is above all other
common lower bounds of
$C\_E$ and $C\_{E^\*}$, for example $C\_n\leq D$ for all $n$.


Since $E\_n=\{0,2,4,\ldots,2n\}\in C\_n$,
there must be a set $d\_n\in D$ such that $E\_n\subseteq d\_n$.
Since $D\leq C\_E$, it follows that $d\_n$ consists of even
integers.
Since $D\leq C\_{E^\*}$, it follows that $d\_n$ is a finite set.
Thus $d\_n$ is properly contained in some
$E\_m=\{0,2,\ldots,2m\}$ for sufficiently
large $m$. But now repeat this argument to find
some $d\_m\in D$ containing $E\_m$. We now have
$d\_n\subsetneq E\_m \subseteq d\_m$, contradicting the
properness of $D$.
\\\\\


Finally, to extend the
proof that $(\text{PropCov}(\omega),\leq)$ is not a lattice
to a proof that $(\text{PropCov}(X),\leq)$ is not a lattice,
for any superset $X\supseteq \omega$, repeat the argument
above, but extend all the covers of $\omega$ which were used
in the argument to covers
of $X$ by adding to those covers
all singletons $\{x\}$, $x\in X\setminus\omega$.




















