We define an equivalence relation on $\mathcal{P}(\omega)$: for $x,y\in\mathcal{P}(\omega)$ we say $$x\simeq\_{fin} y \text{ iff there is } n \in \omega \text{ such that }
x\setminus \{0,\ldots,n\} = y \setminus \{0,\ldots,n\}.$$


The set $\mathcal{P}(\omega)/\simeq\_{fin}$ is usually written as $\mathcal{P}(\omega)/fin$. For $[x], [y]\in \mathcal{P}(\omega)/fin$ we say $[x]\leq[y]$ iff there is $n \in \omega$ such that $(x\setminus \{ 0,\ldots,n\}) \subseteq (y \setminus \{0,\ldots,n\})$. It is a routine verification that this relation is well-defined.


Does $\mathcal{P}(\omega)/fin$ contain an anti-chain of cardinality $2^{\aleph\_0}$?



Antichain on $\mathcal{P}(\omega)/fin$ of cardinality $2^{\aleph_0}$?
Yes. This is an easy exercise:


For every $r\in\Bbb R$ fix some sequence of rational numbers $r\_n$ such that $\lim r\_n=r$. Now enumerate $\Bbb Q$ as $\{q\_n\mid n\in\Bbb N\}$ and consider $A\_r=\{k\mid\exists n:q\_k=r\_n\}$.


Then given $r\neq r'$, the sequences $r\_n$ and $r'\_n$ must be disjoint from some point onwards, because $\Bbb R$ is Hausdorff. Therefore $A\_r$ and $A\_{r'}$ have finite intersection. And so $\{[A\_r]\mid r\in\Bbb R\}$ is an antichain of size $2^{\aleph\_0}$.


And behold, by choosing $r\_n$ to be $q\_k$, such that $k$ is the least for which $0<|q\_k-r|<\frac1n$, the entire thing doesn't require a smidgen of choice too!


